Are there "more" complex numbers than reals? It seems so, every real has
its y, or imaginary, component set to zero. Therefore for each real there
is an infinity of infinite embedding's for it wrt any real with a non-zero
y axis? Fair enough, or really dumb? A little stupid? What do you think?
Are there "more" complex numbers than reals?
It seems so,
every real has its y, or imaginary, component
set to zero.
Therefore
for each real
there is an infinity of infinite embedding's
for it wrt any real with a non-zero y axis?
Fair enough, or really dumb?
A little stupid?
What do you think?
A mathematician, to whom this is a whole new topic, would start by
asking you what you mean by "more". Without that, they could not
possibly answer you.
So, what do you mean by "more" when applied to
sets like C and R?
Le 09/07/2024 14:37, Ben Bacarisse a crit :
A mathematician, to whom this is a whole new topic, would start by
asking you what you mean by "more". Without that, they could not
possibly answer you.
Good mathematicians could.
So, what do you mean by "more" when applied to
sets like C and R?
Proper subsets have less elements than their supersets.
In a sense there are 'more' since the reals are all on the x axis line
whereas the 2D R x R space is filled with complex numbers. R is
contained in C. In another sense they are the same size set, C being
basically R by R in the same sense as Q being Z by Z).
Are there any other sizes of sets between countable Q and uncountable
R?
How about between uncountable R and uncountable C?
Are there "more" complex numbers than reals? It seems so, every real has
its y, or imaginary, component set to zero. Therefore for each real
there is an infinity of infinite embedding's for it wrt any real with a non-zero y axis? Fair enough, or really dumb? A little stupid? What do
you think?
How can you compare them if they are not even in the same dimension?
On 7/9/2024 12:38 PM, sobriquet wrote:
Op 09/07/2024 om 07:24 schreef Chris M. Thomasson:
Are there "more" complex numbers than reals? It seems so, every real
has its y, or imaginary, component set to zero. Therefore for each
real there is an infinity of infinite embedding's for it wrt any real
with a non-zero y axis? Fair enough, or really dumb? A little stupid?
What do you think?
How can you compare them if they are not even in the same dimension?
It seems that instead of comparing the real number line to the complex
plane (or the Cartesian plane for that matter), you might as well
compare a unit of length to a square unit of area.
Numerically they might be the same, but they are not identical, because
a line has no area, so a unit of length 1 has an area of 0 units squared.
Are the complex numbers just as infinitely dense as the reals are? Or
are there somehow "more" of them wrt the density of the reals vs, say,
the rationals and/or the naturals? Is the "density" of an uncountable infinity the same for every uncountable infinity? The density of the
complex numbers and the reals is the same?
On 7/9/2024 1:29 PM, Chris M. Thomasson wrote:
On 7/9/2024 1:27 PM, sobriquet wrote:
Op 09/07/2024 om 22:05 schreef Chris M. Thomasson:
On 7/9/2024 12:38 PM, sobriquet wrote:
Op 09/07/2024 om 07:24 schreef Chris M. Thomasson:
Are there "more" complex numbers than reals? It seems so, every
real has its y, or imaginary, component set to zero. Therefore for >>>>>> each real there is an infinity of infinite embedding's for it wrt
any real with a non-zero y axis? Fair enough, or really dumb? A
little stupid? What do you think?
How can you compare them if they are not even in the same dimension? >>>>> It seems that instead of comparing the real number line to the
complex plane (or the Cartesian plane for that matter), you might
as well compare a unit of length to a square unit of area.
Numerically they might be the same, but they are not identical,
because
a line has no area, so a unit of length 1 has an area of 0 units
squared.
Are the complex numbers just as infinitely dense as the reals are?
Or are there somehow "more" of them wrt the density of the reals vs,
say, the rationals and/or the naturals? Is the "density" of an
uncountable infinity the same for every uncountable infinity? The
density of the complex numbers and the reals is the same?
How do you define sets exactly?
Is there a specific set that corresponds to sqrt(2)?
Does this set have an infinite number of elements analogous to the
sqrt(2) having an infinite decimal expansion?
It seems that the existence of something like sqrt(2) is already
rather dubious.
In reality, things are finite and space and time might also be finite
(composed of atoms of space and time that can't be subdivided with
the parts retaining their original spatial and temporal properties).
So if the concept of irrational numbers like sqrt(2) turns out to be
nonsense, that means the concept of the number line is nonsense
likewise and we don't even need to consider higher dimensional nonsense. >>>
Wrt the sqrt of two. Well, every square already has it in its
diagonals, right?
The sqrt of 2 can scale to any square. Right?
Op 09/07/2024 om 22:05 schreef Chris M. Thomasson:
On 7/9/2024 12:38 PM, sobriquet wrote:
Op 09/07/2024 om 07:24 schreef Chris M. Thomasson:
Are there "more" complex numbers than reals? It seems so, every real
has its y, or imaginary, component set to zero. Therefore for each
real there is an infinity of infinite embedding's for it wrt any real
with a non-zero y axis? Fair enough, or really dumb? A little stupid?
What do you think?
How can you compare them if they are not even in the same dimension?
It seems that instead of comparing the real number line to the complex
plane (or the Cartesian plane for that matter), you might as well
compare a unit of length to a square unit of area.
Numerically they might be the same, but they are not identical, because
a line has no area, so a unit of length 1 has an area of 0 units squared.
Are the complex numbers just as infinitely dense as the reals are? Or
are there somehow "more" of them wrt the density of the reals vs, say,
the rationals and/or the naturals? Is the "density" of an uncountable
infinity the same for every uncountable infinity? The density of the
complex numbers and the reals is the same?
How do you define sets exactly?
Is there a specific set that corresponds to sqrt(2)?
Does this set have an infinite number of elements analogous to the
sqrt(2) having an infinite decimal expansion?
It seems that the existence of something like sqrt(2) is already rather dubious.
In reality, things are finite and space and time might also be finite (composed of atoms of space and time that can't be subdivided with
the parts retaining their original spatial and temporal properties).
So if the concept of irrational numbers like sqrt(2) turns out to be nonsense, ....
.... that means the concept of the number line is nonsense likewise and
we don't even need to consider higher dimensional nonsense.
sobriquet <dohduhdah@yahoo.com> wrote:
Op 09/07/2024 om 22:05 schreef Chris M. Thomasson:
On 7/9/2024 12:38 PM, sobriquet wrote:
Op 09/07/2024 om 07:24 schreef Chris M. Thomasson:
Are there "more" complex numbers than reals? It seems so, every real >>>>> has its y, or imaginary, component set to zero. Therefore for each
real there is an infinity of infinite embedding's for it wrt any real >>>>> with a non-zero y axis? Fair enough, or really dumb? A little stupid? >>>>> What do you think?
How can you compare them if they are not even in the same dimension?
It seems that instead of comparing the real number line to the complex >>>> plane (or the Cartesian plane for that matter), you might as well
compare a unit of length to a square unit of area.
Numerically they might be the same, but they are not identical, because >>>> a line has no area, so a unit of length 1 has an area of 0 units squared.
Are the complex numbers just as infinitely dense as the reals are? Or
are there somehow "more" of them wrt the density of the reals vs, say,
the rationals and/or the naturals? Is the "density" of an uncountable
infinity the same for every uncountable infinity? The density of the
complex numbers and the reals is the same?
How do you define sets exactly?
By the axioms of set theory.
Is there a specific set that corresponds to sqrt(2)?
There exists a Dedekind cut of the set of rational numbers which is
sqrt(2).
Does this set have an infinite number of elements analogous to the
sqrt(2) having an infinite decimal expansion?
No. There is no such analogy. The infinite decimal expansion is
entirely irrelevant.
It seems that the existence of something like sqrt(2) is already rather
dubious.
Not at all. You remind me of John Gabriel, who continually asserted the non-existence of mathematical entities, without being able to say what he meant by this non-existence. The real number sqrt(2) exists by the
axioms of set theory and real numbers.
In reality, things are finite and space and time might also be finite
(composed of atoms of space and time that can't be subdivided with
the parts retaining their original spatial and temporal properties).
So if the concept of irrational numbers like sqrt(2) turns out to be
nonsense, ....
You're several hundred years too late for that debate. The concept of irrational numbers is fully formed, without known inconsistency.
.... that means the concept of the number line is nonsense likewise and
we don't even need to consider higher dimensional nonsense.
The number line isn't nonsense. You're in danger of aligning yourself
with this newsgroup's cranks, past and present.
So basically what you're saying is that reality is irrelevant with
respect to the concepts in the mathematical realm?
Can we really expect a concept of continuity to exist in the
mathematical realm even if it turns out that we find solid evidence for
the claim that everything in the empirical realm (energy, matter, time, space, information) turns out to be discrete?
Are you saying N J Wildberger is a crank?
I'm not a mathematician myself, but I'm interested in science and math
and the arguments given by N J Wildberger seem reasonable.
How do you define sets exactly?
Is there a specific set that corresponds to sqrt(2)?
Does this set have an infinite number of elements analogous to the
sqrt(2) having an infinite decimal expansion?
It seems that the existence of something like sqrt(2) is already rather dubious.
In reality, things are finite and space and time might also be finite (composed of atoms of space and time that can't be subdivided with
the parts retaining their original spatial and temporal properties).
So if the concept of irrational numbers like sqrt(2) [etc.]
Op 09/07/2024 om 23:33 schreef Alan Mackenzie:
sobriquet <dohduhdah@yahoo.com> wrote:
Op 09/07/2024 om 22:05 schreef Chris M. Thomasson:
On 7/9/2024 12:38 PM, sobriquet wrote:
Op 09/07/2024 om 07:24 schreef Chris M. Thomasson:
Are there "more" complex numbers than reals? It seems so, every real >>>>>> has its y, or imaginary, component set to zero. Therefore for each >>>>>> real there is an infinity of infinite embedding's for it wrt any real >>>>>> with a non-zero y axis? Fair enough, or really dumb? A little stupid? >>>>>> What do you think?
How can you compare them if they are not even in the same
dimension? It seems that instead of comparing the real number line
to the complex plane (or the Cartesian plane for that matter), you
might as well compare a unit of length to a square unit of area.
Numerically they might be the same, but they are not identical,
because a line has no area, so a unit of length 1 has an area of 0
units squared.
Are the complex numbers just as infinitely dense as the reals are? Or
are there somehow "more" of them wrt the density of the reals vs, say, >>>> the rationals and/or the naturals? Is the "density" of an uncountable
infinity the same for every uncountable infinity? The density of the
complex numbers and the reals is the same?
How do you define sets exactly?
By the axioms of set theory.
Is there a specific set that corresponds to sqrt(2)?
There exists a Dedekind cut of the set of rational numbers which is
sqrt(2).
Does this set have an infinite number of elements analogous to the
sqrt(2) having an infinite decimal expansion?
No. There is no such analogy. The infinite decimal expansion is
entirely irrelevant.
It seems that the existence of something like sqrt(2) is already rather
dubious.
Not at all. You remind me of John Gabriel, who continually asserted the
non-existence of mathematical entities, without being able to say what he
meant by this non-existence. The real number sqrt(2) exists by the
axioms of set theory and real numbers.
In reality, things are finite and space and time might also be finite
(composed of atoms of space and time that can't be subdivided with
the parts retaining their original spatial and temporal properties).
So if the concept of irrational numbers like sqrt(2) turns out to be
nonsense, ....
You're several hundred years too late for that debate. The concept of
irrational numbers is fully formed, without known inconsistency.
.... that means the concept of the number line is nonsense likewise and
we don't even need to consider higher dimensional nonsense.
The number line isn't nonsense. You're in danger of aligning yourself
with this newsgroup's cranks, past and present.
So basically what you're saying is that reality is irrelevant with
respect to the concepts in the mathematical realm?
Can we really expect a concept of continuity to exist in the
mathematical realm even if it turns out that we find solid evidence for
the claim that everything in the empirical realm (energy, matter, time, space, information) turns out to be discrete?
Are you saying N J Wildberger is a crank? https://www.youtube.com/watch?v=jlnBo3APRlU
I'm not a mathematician myself, but I'm interested in science and math
and the arguments given by N J Wildberger seem reasonable.
Seems to boil down to:
Is uncountable infinity the same "size", as any other uncountable
infinity? Say reals vs. complex numbers...
Seems to boil down to:
Is uncountable infinity the same "size", as any other uncountable
infinity? Say reals vs. complex numbers...
sobriquet <dohduhdah@yahoo.com> wrote:
Op 09/07/2024 om 23:33 schreef Alan Mackenzie:
sobriquet <dohduhdah@yahoo.com> wrote:
Op 09/07/2024 om 22:05 schreef Chris M. Thomasson:
On 7/9/2024 12:38 PM, sobriquet wrote:
Op 09/07/2024 om 07:24 schreef Chris M. Thomasson:
Are there "more" complex numbers than reals? It seems so, every real >>>>>>> has its y, or imaginary, component set to zero. Therefore for each >>>>>>> real there is an infinity of infinite embedding's for it wrt any real >>>>>>> with a non-zero y axis? Fair enough, or really dumb? A little stupid? >>>>>>> What do you think?
How can you compare them if they are not even in the same
dimension? It seems that instead of comparing the real number line >>>>>> to the complex plane (or the Cartesian plane for that matter), you >>>>>> might as well compare a unit of length to a square unit of area.
Numerically they might be the same, but they are not identical,
because a line has no area, so a unit of length 1 has an area of 0 >>>>>> units squared.
Are the complex numbers just as infinitely dense as the reals are? Or >>>>> are there somehow "more" of them wrt the density of the reals vs, say, >>>>> the rationals and/or the naturals? Is the "density" of an uncountable >>>>> infinity the same for every uncountable infinity? The density of the >>>>> complex numbers and the reals is the same?
How do you define sets exactly?
By the axioms of set theory.
Is there a specific set that corresponds to sqrt(2)?
There exists a Dedekind cut of the set of rational numbers which is
sqrt(2).
Does this set have an infinite number of elements analogous to the
sqrt(2) having an infinite decimal expansion?
No. There is no such analogy. The infinite decimal expansion is
entirely irrelevant.
It seems that the existence of something like sqrt(2) is already rather >>>> dubious.
Not at all. You remind me of John Gabriel, who continually asserted the >>> non-existence of mathematical entities, without being able to say what he >>> meant by this non-existence. The real number sqrt(2) exists by the
axioms of set theory and real numbers.
In reality, things are finite and space and time might also be finite
(composed of atoms of space and time that can't be subdivided with
the parts retaining their original spatial and temporal properties).
So if the concept of irrational numbers like sqrt(2) turns out to be
nonsense, ....
You're several hundred years too late for that debate. The concept of
irrational numbers is fully formed, without known inconsistency.
.... that means the concept of the number line is nonsense likewise and >>>> we don't even need to consider higher dimensional nonsense.
The number line isn't nonsense. You're in danger of aligning yourself
with this newsgroup's cranks, past and present.
So basically what you're saying is that reality is irrelevant with
respect to the concepts in the mathematical realm?
The concepts in the mathematical realm are an integral part of reality.
Can we really expect a concept of continuity to exist in the
mathematical realm even if it turns out that we find solid evidence for
the claim that everything in the empirical realm (energy, matter, time,
space, information) turns out to be discrete?
Yes. Just as we have 2 + 2 = 4 (in the abstract) we have continuity
(again, in the abstract). There's a well developed theory of continuity lacking any inconsistency, so far as people are aware.
Are you saying N J Wildberger is a crank?
https://www.youtube.com/watch?v=jlnBo3APRlU
No, I'm not familiar with the said lady/gentleman. I don't have access
to youtube.
I'm not a mathematician myself, but I'm interested in science and math
and the arguments given by N J Wildberger seem reasonable.
OK, that's fair enough. I have a degree in maths, although I'm not
otherwise a mathematician. Sometimes it needs to be said that the mathematical basics are firmly established as being true, and are not at
all a matter of opinion. Just like Newton's laws of motion or the
roundness of the Earth or the Theory of Special Relativity are no longer matters of opinion.
sobriquet was thinking very hard :
Op 09/07/2024 om 23:33 schreef Alan Mackenzie:
sobriquet <dohduhdah@yahoo.com> wrote:
Op 09/07/2024 om 22:05 schreef Chris M. Thomasson:
On 7/9/2024 12:38 PM, sobriquet wrote:
Op 09/07/2024 om 07:24 schreef Chris M. Thomasson:
Are there "more" complex numbers than reals? It seems so, every real >>>>>>> has its y, or imaginary, component set to zero. Therefore for each >>>>>>> real there is an infinity of infinite embedding's for it wrt any >>>>>>> real
with a non-zero y axis? Fair enough, or really dumb? A little
stupid?
What do you think?
How can you compare them if they are not even in the same dimension? >>>>>> It seems that instead of comparing the real number line to the
complex
plane (or the Cartesian plane for that matter), you might as well
compare a unit of length to a square unit of area.
Numerically they might be the same, but they are not identical,
because
a line has no area, so a unit of length 1 has an area of 0 units
squared.
Are the complex numbers just as infinitely dense as the reals are? Or >>>>> are there somehow "more" of them wrt the density of the reals vs, say, >>>>> the rationals and/or the naturals? Is the "density" of an uncountable >>>>> infinity the same for every uncountable infinity? The density of the >>>>> complex numbers and the reals is the same?
How do you define sets exactly?
By the axioms of set theory.
Is there a specific set that corresponds to sqrt(2)?
There exists a Dedekind cut of the set of rational numbers which is
sqrt(2).
Does this set have an infinite number of elements analogous to the
sqrt(2) having an infinite decimal expansion?
No. There is no such analogy. The infinite decimal expansion is
entirely irrelevant.
It seems that the existence of something like sqrt(2) is already rather >>>> dubious.
Not at all. You remind me of John Gabriel, who continually asserted the >>> non-existence of mathematical entities, without being able to say
what he
meant by this non-existence. The real number sqrt(2) exists by the
axioms of set theory and real numbers.
In reality, things are finite and space and time might also be finite
(composed of atoms of space and time that can't be subdivided with
the parts retaining their original spatial and temporal properties).
So if the concept of irrational numbers like sqrt(2) turns out to be
nonsense, ....
You're several hundred years too late for that debate. The concept of
irrational numbers is fully formed, without known inconsistency.
.... that means the concept of the number line is nonsense likewise and >>>> we don't even need to consider higher dimensional nonsense.
The number line isn't nonsense. You're in danger of aligning yourself
with this newsgroup's cranks, past and present.
So basically what you're saying is that reality is irrelevant with
respect to the concepts in the mathematical realm?
Can we really expect a concept of continuity to exist in the
mathematical realm even if it turns out that we find solid evidence
for the claim that everything in the empirical realm (energy, matter,
time, space, information) turns out to be discrete?
Are you saying N J Wildberger is a crank?
https://www.youtube.com/watch?v=jlnBo3APRlU
I'm not a mathematician myself, but I'm interested in science and math
and the arguments given by N J Wildberger seem reasonable.
One should keep in mind that we are not building reality from
mathematical objects.
On 7/9/2024 10:30 AM, Ben Bacarisse wrote:
WM <wolfgang.mueckenheim@tha.de> writes:
Le 09/07/2024 14:37, Ben Bacarisse a crit :
A mathematician, to whom this is a whole new topic, would start by
asking you what you mean by "more". Without that, they could not
possibly answer you.
Good mathematicians could.
So, what do you mean by "more" when applied to
sets like C and R?
Proper subsets have less elements than their supersets.
Let's see if Chris is using that definition. I think he's cleverer than
you so he will probably want to be able to say that {1,2,3} has "more"
elements than {4,5}.
I was just thinking that there seems to be "more" reals than natural
numbers. Every natural number is a real, but not all reals are natural numbers.
So, wrt the complex. Well... Every complex number has a x, or real
component. However, not every real has a y, or imaginary component...
Fair enough? Or still crap? ;^o
Le 09/07/2024 14:37, Ben Bacarisse a crit :
A mathematician, to whom this is a whole new topic, would start by
asking you what you mean by "more". Without that, they could not
possibly answer you.
Good mathematicians could.
So, what do you mean by "more" when applied to
sets like C and R?
Proper subsets have less elements than their supersets.
Regards, WM
Either spacetime is continuous or it's not.
sobriquet <dohduhdah@yahoo.com> wrote or quoted:
Either spacetime is continuous or it's not.
I wouldn't put it that way. For starters, continuity is something
that's defined for mappings (functions) of topological spaces.
Spacetime isn't a function, it's not even a mathematical object!
So what's the deal with "continuity of spacetime" anyway?
WM explained on 7/9/2024 :
Le 09/07/2024 à 14:37, Ben Bacarisse a écrit :
A mathematician, to whom this is a whole new topic, would start by
asking you what you mean by "more". Without that, they could not
possibly answer you.
Good mathematicians could.
So, what do you mean by "more" when applied to
sets like C and R?
Proper subsets have less elements than their supersets.
Proper subsets do not have all of the elements that are in their
supersets. This is not the same as "less" or fewer elements
WM <wolfgang.mueckenheim@tha.de> writes:
Le 09/07/2024 à 14:37, Ben Bacarisse a écrit :
A mathematician, to whom this is a whole new topic, would start by
asking you what you mean by "more". Without that, they could not
possibly answer you.
Good mathematicians could.
So, what do you mean by "more" when applied to
sets like C and R?
Proper subsets have less elements than their supersets.
Let's see if Chris is using that definition. I think he's cleverer than
you
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
On 7/9/2024 10:30 AM, Ben Bacarisse wrote:
WM <wolfgang.mueckenheim@tha.de> writes:
Le 09/07/2024 à 14:37, Ben Bacarisse a écrit :
A mathematician, to whom this is a whole new topic, would start by
asking you what you mean by "more". Without that, they could not
possibly answer you.
Good mathematicians could.
So, what do you mean by "more" when applied to
sets like C and R?
Proper subsets have less elements than their supersets.
Let's see if Chris is using that definition. I think he's cleverer than >>> you so he will probably want to be able to say that {1,2,3} has "more"
elements than {4,5}.
I was just thinking that there seems to be "more" reals than natural
numbers. Every natural number is a real, but not all reals are natural
numbers.
You are repeating yourself. What do you mean by "more"? Can you think
if a general rule -- a test maybe -- that could be applied to any two
set to find one which has more elements?
So, wrt the complex. Well... Every complex number has a x, or real
component. However, not every real has a y, or imaginary component...
Fair enough? Or still crap? ;^o
So you are using WM's definition based on subsets? That's a shame.
One consequence is that you can't say if the set of even numbers has
more or fewer elements than {1,3,5} because {1,3,5} is not a subset of
the even numbers, and the set of even numbers is not a subset of
{1,3,5}. They just can't be compared using your (and WM's) notion of
"more".
"In physics, spacetime is a mathematical model that fuses the three
Quantities like matter and energy initially seemed to be continuous,
since they typically consist of extremely large collections of discrete >elements (like molecules, atoms or particles).
So gradually we came to realize that the concept of a continuous
quantity that can be subdivided indefinitely is unrealistic.
We don't know if this holds as well for time and space, but it might be
that everything in reality that can be quantified ultimately consists of >discrete elements. If that turns out to be the case, that renders
discussions about the concept of continuity irrelevant, because at that
point it seems to be about a concept that only exists in our imagination.
So it's a bit like discussing how many angels can dance on the head of a
pin (assuming that angels don't really exist).
Sometimes it needs to be said that the
mathematical basics are firmly established as being true, and are not at
all a matter of opinion.
Le 10/07/2024 à 00:20, Alan Mackenzie a écrit :Yeeeeees.
Sometimes it needs to be said that the
mathematical basics are firmly established as being true,
and are not at all a matter of opinion.
One of them is this ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
The Number of UnitFractions between 0 and x can only increase by 1 at
every x.
[...] Hence there must be <bla>
"Why String Theory" (2016), Joseph Conlon.
Am 10.07.2024 um 18:28 schrieb WM:
Le 10/07/2024 à 00:20, Alan Mackenzie a écrit :Yeeeeees.
Sometimes it needs to be said that the
mathematical basics are firmly established as being true,
and are not at all a matter of opinion.
One of them is this ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
The Number of UnitFractions between 0 and x can only increase by 1 at
every x.
No,
Am 10.07.2024 um 18:28 schrieb WM:
Le 10/07/2024 à 00:20, Alan Mackenzie a écrit :
Sometimes it needs to be said that the
mathematical basics are firmly established as being true, and are not
at all a matter of opinion.
One of them is this ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Yeeeeees.
Op 09/07/2024 om 22:33 schreef Chris M. Thomasson:
Either spacetime is continuous or it's not.
So far it seems things usually are not continuous
(stuff like energy and space can't be subdivided indefinitely).
So it seems reasonable to assume that
the concept of continuity is nonsense.
If spacetime is not continuous,
what basis do we have to the concept of continuity?
But it physics, we need high energy particle accelerators to study
small distances.
Le 10/07/2024 à 18:35, Moebius a écrit :No. :-)
Am 10.07.2024 um 18:28 schrieb WM:
Le 10/07/2024 à 00:20, Alan Mackenzie a écrit :
Sometimes it needs to be said that the
mathematical basics are firmly established as being true, and are
not at all a matter of opinion.
One of them is this ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Yeeeeees.
The Number of UnitFractions between 0 and x can only increase by 1 at
every x.
No,
Yes.
It can increase from 0 only to 1 because <bla>
Well, it missed an infinite number of reals between 1 and 2. So, the
reals are denser than the naturals. Fair enough?
It just seems to have "more", so to speak.
Perhaps using the word "more" is just wrong.
However, the density of an infinity makes sense to me. Not sure why, it
just does...
The set of evens and odds has an [countably] infinite number of elements. Just like
the set of naturals.
Well, can we zoom out forever? Zooming in forever,
Am 10.07.2024 um 23:26 schrieb Chris M. Thomasson:
Well, can we zoom out forever? Zooming in forever,
In the comtext of the real and/or complex numbers we can!
That's just the beauty of math! Isn't it?
On 7/10/2024 2:42 PM, Moebius wrote:
Am 10.07.2024 um 23:26 schrieb Chris M. Thomasson:Yeah, I know we can zoom in forever in math for sure; fractals are neat
Well, can we zoom out forever? Zooming in forever,
In the comtext of the real and/or complex numbers we can!
That's just the beauty of math! Isn't it?
and fun all in one. No doubt. However, wrt the physical world... Can we
zoom in forever?
Or for that matter, zoom out forever?
On 7/10/2024 2:42 PM, Moebius wrote:
Am 10.07.2024 um 23:26 schrieb Chris M. Thomasson:
Well, can we zoom out forever? Zooming in forever,
In the comtext of the real and/or complex numbers we can!
That's just the beauty of math! Isn't it?
Yeah, I know we can zoom in forever in math for sure; fractals are neat
and fun all in one. No doubt. However, wrt the physical world... Can we
zoom in forever? Or for that matter, zoom out forever?
Op 10/07/2024 om 23:42 schreef Moebius:
Am 10.07.2024 um 23:26 schrieb Chris M. Thomasson:
Well, can we zoom out forever? Zooming in forever,
In the comtext of the real and/or complex numbers we can!
That's just the beauty of math! Isn't it?
In our imagination we can cook and eat ourselves. In practice, it
wouldn't really work out.
On 7/10/2024 2:53 PM, Moebius wrote:
Am 10.07.2024 um 23:44 schrieb Chris M. Thomasson:
[So] let's do math, not physics. Ok?
Or for that matter, zoom out forever?
Who nows?! [...]
Touche! Thanks for your patience with my questions. Your are a nice
person to converse with. Thanks.
On 7/10/2024 3:16 PM, Moebius wrote:
Am 11.07.2024 um 00:10 schrieb Chris M. Thomasson:
On 7/10/2024 2:53 PM, Moebius wrote:
Am 10.07.2024 um 23:44 schrieb Chris M. Thomasson:
[So] let's do math, not physics. Ok?
Or for that matter, zoom out forever?
Who nows?! [...]
Touche! Thanks for your patience with my questions. Your are a nice
person to converse with. Thanks.
Actually, I'm a brutal (but honest) one (sorry abot that).
"He who has ears to hear, let him hear. Anyone with ears to hear
should listen and understand!"
:-P
Fair enough! Hey, at least you have not annihilated me to a point where
I have two assholes instead of one... You do have the ability for
patience. Thanks again.
On 7/10/2024 3:00 PM, sobriquet wrote:
Op 10/07/2024 om 23:44 schreef Chris M. Thomasson:
On 7/10/2024 2:42 PM, Moebius wrote:
Am 10.07.2024 um 23:26 schrieb Chris M. Thomasson:
Well, can we zoom out forever? Zooming in forever,
In the comtext of the real and/or complex numbers we can!
That's just the beauty of math! Isn't it?
Yeah, I know we can zoom in forever in math for sure; fractals are
neat and fun all in one. No doubt. However, wrt the physical world...
Can we zoom in forever? Or for that matter, zoom out forever?
We might think we're zooming in forever.. but we're just stuck in a loop.
https://www.desmos.com/calculator/4mivqc0iht
Now, that makes my brain want to bleed a little bit. We hit a "limit"
and loop back on it? Something akin to that?
No rule is better than a foolish rule, if it yields nonsense like Cantor's "bijections".
On 7/9/2024 4:45 PM, Ben Bacarisse wrote:
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
On 7/9/2024 10:30 AM, Ben Bacarisse wrote:
WM <wolfgang.mueckenheim@tha.de> writes:
Le 09/07/2024 14:37, Ben Bacarisse a crit :
A mathematician, to whom this is a whole new topic, would start by >>>>>> asking you what you mean by "more". Without that, they could not
possibly answer you.
Good mathematicians could.
So, what do you mean by "more" when applied to
sets like C and R?
Proper subsets have less elements than their supersets.
Let's see if Chris is using that definition. I think he's cleverer than >>>> you so he will probably want to be able to say that {1,2,3} has "more" >>>> elements than {4,5}.
I was just thinking that there seems to be "more" reals than natural
numbers. Every natural number is a real, but not all reals are natural
numbers.
You are repeating yourself. What do you mean by "more"? Can you think
if a general rule -- a test maybe -- that could be applied to any two
set to find one which has more elements?
natural numbers: 1, 2, 3, ...
Well, it missed an infinite number of reals between 1 and 2. So, the reals are denser than the naturals. Fair enough? It just seems to have "more", so to speak. Perhaps using the word "more" is just wrong. However, the density of an infinity makes sense to me. Not sure why, it just does...
So, wrt the complex. Well... Every complex number has a x, or real
component. However, not every real has a y, or imaginary component...
Fair enough? Or still crap? ;^o
So you are using WM's definition based on subsets? That's a shame. WM
is not a reasonable person to agree with!
One consequence is that you can't say if the set of even numbers has
more or fewer elements than {1,3,5} because {1,3,5} is not a subset of
the even numbers, and the set of even numbers is not a subset of
{1,3,5}. They just can't be compared using your (and WM's) notion of
"more".
The set of evens and odds has an infinite number of elements. Just like the set of naturals.
WM <wolfgang.mueckenheim@tha.de> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des Unendlichen" at [Technische] Hochschule Augsburg.)
No rule is better than a foolish rule, if it yields nonsense like Cantor's >> "bijections".
Is that why you still can't define set membership, difference and
equality in WMaths such that you could prove one of the most surprising results of WMaths: that sets E and P exist such that E in P and P \ {E}
= P?
{a, b, c} vs { 3, 4, 5 }
Both have the same number of elements, [...]
On 7/10/2024 5:24 PM, Moebius wrote:
Am 11.07.2024 um 02:16 schrieb Chris M. Thomasson:
{a, b, c} vs { 3, 4, 5 }
Both have the same number of elements, [...]
HOW do you know that? Please define (for any sets A, B):
A and B /have the same number of elements/ iff ___________________ .
(i.e. fill out the blanks). :-)
Hint: That's what Ben Bacarisse is asking for.
Sure, it's "obvious" for us. But how would you define "have the same
number of elements" (in mathematical terms) such that it can be
DEDUCED (!) für certain sets A and B?
________________________________________
Ok, I'm slighty vicious now... :-)
If a = b = c, {a, b, c} still has "the same number of elements" as {3,
4, 5 }? :-P
I see {a, b, c} and {3, 4, 5} and think three elements.
Am 11.07.2024 um 02:28 schrieb Chris M. Thomasson:
On 7/10/2024 5:24 PM, Moebius wrote:
Am 11.07.2024 um 02:16 schrieb Chris M. Thomasson:I see {a, b, c} and {3, 4, 5} and think three elements.
{a, b, c} vs { 3, 4, 5 }
Both have the same number of elements, [...]
HOW do you know that? Please define (for any sets A, B):
A and B /have the same number of elements/ iff ___________________ . >>>
(i.e. fill out the blanks). :-)
Hint: That's what Ben Bacarisse is asking for.
Sure, it's "obvious" for us. But how would you define "have the same
number of elements" (in mathematical terms) such that it can be DEDUCED
(!) fr certain sets A and B?
________________________________________
Ok, I'm slighty vicious now... :-)
If a = b = c, {a, b, c} still has "the same number of elements" as {3,
4, 5 }? :-P
Even if a = b = c = 1?
C'mon man! :-P
Am 11.07.2024 um 02:28 schrieb Chris M. Thomasson:
On 7/10/2024 5:24 PM, Moebius wrote:
Ok, I'm slighty vicious now... :-)
If a = b = c, {a, b, c} still has "the same number of elements" as
{3, 4, 5}? :-P
I see {a, b, c} and {3, 4, 5} and think three elements.
Even if a = b = c = 1?
C'mon man! :-P
Moebius <invalid@example.invalid> writes:
Am 11.07.2024 um 02:28 schrieb Chris M. Thomasson:
On 7/10/2024 5:24 PM, Moebius wrote:
Am 11.07.2024 um 02:16 schrieb Chris M. Thomasson:I see {a, b, c} and {3, 4, 5} and think three elements.
{a, b, c} vs { 3, 4, 5 }
Both have the same number of elements, [...]
HOW do you know that? Please define (for any sets A, B):
A and B /have the same number of elements/ iff ___________________ .
(i.e. fill out the blanks). :-)
Hint: That's what Ben Bacarisse is asking for.
Sure, it's "obvious" for us. But how would you define "have the same
number of elements" (in mathematical terms) such that it can be DEDUCED >>>> (!) für certain sets A and B?
________________________________________
Ok, I'm slighty vicious now... :-)
If a = b = c, {a, b, c} still has "the same number of elements" as {3, >>>> 4, 5 }? :-P
Even if a = b = c = 1?
C'mon man! :-P
Please, that's a red herring, and you know it! No where did I say that
a, b and c stood for anything (i.e. that they might be variables in the
maths sense). I this sort of context they are just distinct symbols.
On 7/10/2024 4:53 PM, Ben Bacarisse wrote:
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
On 7/9/2024 4:45 PM, Ben Bacarisse wrote:I am trying to get you to come up with a definition. If it is all about
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
On 7/9/2024 10:30 AM, Ben Bacarisse wrote:
WM <wolfgang.mueckenheim@tha.de> writes:
Le 09/07/2024 14:37, Ben Bacarisse a crit :
A mathematician, to whom this is a whole new topic, would start by >>>>>>>> asking you what you mean by "more". Without that, they could not >>>>>>>> possibly answer you.
Good mathematicians could.
So, what do you mean by "more" when applied to
sets like C and R?
Proper subsets have less elements than their supersets.
Let's see if Chris is using that definition. I think he's cleverer than >>>>>> you so he will probably want to be able to say that {1,2,3} has "more" >>>>>> elements than {4,5}.
I was just thinking that there seems to be "more" reals than natural >>>>> numbers. Every natural number is a real, but not all reals are natural >>>>> numbers.
You are repeating yourself. What do you mean by "more"? Can you think >>>> if a general rule -- a test maybe -- that could be applied to any two
set to find one which has more elements?
natural numbers: 1, 2, 3, ...
Well, it missed an infinite number of reals between 1 and 2. So, the reals >>> are denser than the naturals. Fair enough? It just seems to have "more", so >>> to speak. Perhaps using the word "more" is just wrong. However, the density >>> of an infinity makes sense to me. Not sure why, it just does...
"missing" things then you can't compare the sizes of sets like {a,b,c}
and {3,4,5} as both "miss" all of the members of the others.
{a, b, c} vs { 3, 4, 5 }
Both have the same number of elements,
both have a monotonically increasing
value wrt its elements wrt, ect...
Ok, let's reformulate my statement:
| Even if a = b = c?
Satisfied now?!
For example, you probably
think, intuitively, that there are "more" reals then integers.
Am 11.07.2024 um 02:35 schrieb Ben Bacarisse:
Moebius <invalid@example.invalid> writes:
Am 11.07.2024 um 02:28 schrieb Chris M. Thomasson:Please, that's a red herring, and you know it! No where did I say that
On 7/10/2024 5:24 PM, Moebius wrote:
Am 11.07.2024 um 02:16 schrieb Chris M. Thomasson:I see {a, b, c} and {3, 4, 5} and think three elements.
{a, b, c} vs { 3, 4, 5 }
Both have the same number of elements, [...]
HOW do you know that? Please define (for any sets A, B):
A and B /have the same number of elements/ iff ___________________ .
(i.e. fill out the blanks). :-)
Hint: That's what Ben Bacarisse is asking for.
Sure, it's "obvious" for us. But how would you define "have the same >>>>> number of elements" (in mathematical terms) such that it can be DEDUCED >>>>> (!) fr certain sets A and B?
________________________________________
Ok, I'm slighty vicious now... :-)
If a = b = c, {a, b, c} still has "the same number of elements" as {3, >>>>> 4, 5 }? :-P
Even if a = b = c = 1?
C'mon man! :-P
a, b and c stood for anything (i.e. that they might be variables in the
maths sense). I this sort of context they are just distinct symbols.
Of course, "a", "b" and "c" are three distinct symols. (Hell!)
But you dont't think that they should DENOTE some mathematical objects (in
a mathematical context)?*)
Huh?!
Are you doing math with symbols without any denotaton? Strange! (Really.) [I'm sort of clueless now.]
Ok, let's reformulate my statement:
| Even if a = b = c?
Satisfied now?!
What about dropping the word more in favor of density, or granularity of
an infinite set? The reals are denser, or more granular than the
rationals and naturals?
Am 11.07.2024 um 02:51 schrieb Chris M. Thomasson:
What about dropping the word more in favor of density, or granularity
of an infinite set? The reals are denser, or more granular than the
rationals and naturals?
Yeah, makes sense.
On 7/10/2024 4:29 PM, sobriquet wrote:
Op 11/07/2024 om 00:12 schreef Chris M. Thomasson:
On 7/10/2024 3:00 PM, sobriquet wrote:
Op 10/07/2024 om 23:44 schreef Chris M. Thomasson:
On 7/10/2024 2:42 PM, Moebius wrote:
Am 10.07.2024 um 23:26 schrieb Chris M. Thomasson:
Well, can we zoom out forever? Zooming in forever,
In the comtext of the real and/or complex numbers we can!
That's just the beauty of math! Isn't it?
Yeah, I know we can zoom in forever in math for sure; fractals are
neat and fun all in one. No doubt. However, wrt the physical
world... Can we zoom in forever? Or for that matter, zoom out forever? >>>>
We might think we're zooming in forever.. but we're just stuck in a
loop.
https://www.desmos.com/calculator/4mivqc0iht
Now, that makes my brain want to bleed a little bit. We hit a "limit"
and loop back on it? Something akin to that?
Computers have finite computational resources. So if you keep zooming
in (or out), you eventually run into these computational limitations
(or perhaps it just takes too long, but lets assume that science has
solved the issue of aging and you can keep living as long as you please).
A loop can give the impression of zooming in indefinitely, but you're
not really zooming in indefinitely in that case.
I think I kind of see what you are getting at. Humm... We have got
pretty deep zooms with fractals. Humm... You seem to be saying
interpolate from a deep zoom A to a deep zoom B if B is close enough to
a limit where B can "run out of resources" on a computer? If the
closeness factor exceeds a certain threshold, we simply loop from A to
B? The viewer might not be able to notice this aspect wrt the fact that
we are looping and not zooming anymore, so to speak? Fractal experts
should be able to detect this. I think I would be able to. Fwiw, here is
a deepish zoom:
https://youtu.be/Xjy_HSUujaw
I do not notice any artificial looping things in there...
Also, be sure to check this one out:
https://youtu.be/S530Vwa33G0
So elegant! No artificial looping.
Am 11.07.2024 um 03:06 schrieb Moebius:
Am 11.07.2024 um 02:51 schrieb Chris M. Thomasson:
What about dropping the word more in favor of density, or granularity
of an infinite set? The reals are denser, or more granular than the
rationals and naturals?
Deep zoom is kind of relative compared to infinity.
10^9 iterations seems deep, but even 10^(googleplex^googleplex)
iterations is insignificantly small in comparison to infinity.
There are tricks to work with big numbers on computers, but that doesn't
mean that you magically can get beyond any limitations. If the numbers
get big/small enough, you run out of memory or you run out of time.
Infinite zoom would only work on a hypothetical computer that has
unlimited memory and unlimited time and those only exist in our
imagination.
Am 11.07.2024 um 02:59 schrieb Ben Bacarisse: [nonsense]
Fuck you!
EOD.
Am 11.07.2024 um 03:08 schrieb Moebius:
Am 11.07.2024 um 03:06 schrieb Moebius:
Am 11.07.2024 um 02:51 schrieb Chris M. Thomasson:
What about dropping the word more in favor of density, or
granularity of an infinite set? The reals are denser, or more
granular than the rationals and naturals?
Just to make that clear, in math we say that the _cardinality_ of IR is larger than the _cardinality_ of Q or IN.
See: https://en.wikipedia.org/wiki/Cardinality
Read it, man!
Infinite zoom would only work on a hypothetical computer that has
unlimited memory and unlimited time and [...]
On 7/10/2024 5:46 PM, Ben Bacarisse wrote:
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
On 7/10/2024 4:53 PM, Ben Bacarisse wrote:That will fall down for infinite sets unless, by decree, you state that
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
On 7/9/2024 4:45 PM, Ben Bacarisse wrote:I am trying to get you to come up with a definition. If it is all about >>>> "missing" things then you can't compare the sizes of sets like {a,b,c} >>>> and {3,4,5} as both "miss" all of the members of the others.
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
On 7/9/2024 10:30 AM, Ben Bacarisse wrote:
WM <wolfgang.mueckenheim@tha.de> writes:
Le 09/07/2024 14:37, Ben Bacarisse a crit :
A mathematician, to whom this is a whole new topic, would start by >>>>>>>>>> asking you what you mean by "more". Without that, they could not >>>>>>>>>> possibly answer you.
Good mathematicians could.
So, what do you mean by "more" when applied to
sets like C and R?
Proper subsets have less elements than their supersets.
Let's see if Chris is using that definition. I think he's cleverer than
you so he will probably want to be able to say that {1,2,3} has "more" >>>>>>>> elements than {4,5}.
I was just thinking that there seems to be "more" reals than natural >>>>>>> numbers. Every natural number is a real, but not all reals are natural >>>>>>> numbers.
You are repeating yourself. What do you mean by "more"? Can you think >>>>>> if a general rule -- a test maybe -- that could be applied to any two >>>>>> set to find one which has more elements?
natural numbers: 1, 2, 3, ...
Well, it missed an infinite number of reals between 1 and 2. So, the reals
are denser than the naturals. Fair enough? It just seems to have "more", so
to speak. Perhaps using the word "more" is just wrong. However, the density
of an infinity makes sense to me. Not sure why, it just does...
{a, b, c} vs { 3, 4, 5 }
Both have the same number of elements,
your meaning of "more" makes all infinite sets have the same number of
elements.
What about dropping the word more in favor of density, or granularity of an infinite set? The reals are denser, or more granular than the rationals and reals? Crap?
Am 11.07.2024 um 03:06 schrieb sobriquet:
Deep zoom is kind of relative compared to infinity.
10^9 iterations seems deep, but even 10^(googleplex^googleplex)
iterations is insignificantly small in comparison to infinity.
There are tricks to work with big numbers on computers, but that doesn't
mean that you magically can get beyond any limitations. If the numbers
get big/small enough, you run out of memory or you run out of time.
Infinite zoom would only work on a hypothetical computer that has
unlimited memory and unlimited time and those only exist in our
imagination.
Exacty!
So what?
Hint: Mathematical objects "only exist in our imagination", imho. (You
see, I'm not a platonist/mathematical realist.)
Le 09/07/2024 à 14:37, Ben Bacarisse a écrit :
A mathematician, to whom this is a whole new topic, would start by
asking you what you mean by "more". Without that, they could not
possibly answer you.
Good mathematicians could.
So, what do you mean by "more" when applied to
sets like C and R?
Proper subsets have less elements than their supersets.
Regards, WM
Am 11.07.2024 um 02:59 schrieb Ben Bacarisse: [nonsense]
Fuck you!
EOD.
On 7/9/2024 3:29 PM, FromTheRafters wrote:
Chris M. Thomasson formulated the question :[...]
On 7/9/2024 10:30 AM, Ben Bacarisse wrote:Seems is a funny word. Does there not 'seem' to be 'more' naturals than
WM <wolfgang.mueckenheim@tha.de> writes:
Le 09/07/2024 14:37, Ben Bacarisse a crit :
A mathematician, to whom this is a whole new topic, would start by >>>>>> asking you what you mean by "more". Without that, they could not
possibly answer you.
Good mathematicians could.
So, what do you mean by "more" when applied to
sets like C and R?
Proper subsets have less elements than their supersets.
Let's see if Chris is using that definition. I think he's cleverer than >>>> you so he will probably want to be able to say that {1,2,3} has "more" >>>> elements than {4,5}.
I was just thinking that there seems to be "more" reals than natural
numbers. Every natural number is a real, but not all reals are natural
numbers.
primes? Intuition fails, these sets are of the same cardinality.
I do think that the number of primes is infinite in the sense that they are all in the naturals.
Every prime is a natural, but not every natural is a prime? Strange
thoughts? Is there a countable number of infinite primes, just like
there is a countable number of infinite naturals? Fair enough?
Well, some people would claim that the human mind has capabilities
beyond what a computer can do.
sobriquet pretended :
Op 11/07/2024 om 00:12 schreef Chris M. Thomasson:
On 7/10/2024 3:00 PM, sobriquet wrote:
Op 10/07/2024 om 23:44 schreef Chris M. Thomasson:
On 7/10/2024 2:42 PM, Moebius wrote:
Am 10.07.2024 um 23:26 schrieb Chris M. Thomasson:
Well, can we zoom out forever? Zooming in forever,
In the comtext of the real and/or complex numbers we can!
That's just the beauty of math! Isn't it?
Yeah, I know we can zoom in forever in math for sure; fractals are
neat and fun all in one. No doubt. However, wrt the physical
world... Can we zoom in forever? Or for that matter, zoom out forever? >>>>
We might think we're zooming in forever.. but we're just stuck in a
loop.
https://www.desmos.com/calculator/4mivqc0iht
Now, that makes my brain want to bleed a little bit. We hit a "limit"
and loop back on it? Something akin to that?
Computers have finite computational resources. So if you keep zooming
in (or out), you eventually run into these computational limitations
(or perhaps it just takes too long, but lets assume that science has
solved the issue of aging and you can keep living as long as you please).
A loop can give the impression of zooming in indefinitely, but you're
not really zooming in indefinitely in that case.
Why not? Doesn't the scope of displayable pixels shrink commensurate
with the amount of zoom?
sobriquet <dohduhdah@yahoo.com> wrote or quoted:
Well, some people would claim that the human mind has capabilities
beyond what a computer can do.
Language and the mind can refer to something like "infinity" and
speak about it. When writing numbers using the notation 10^-10, then
10^-(10^10), and so on, you never will arrive at something that is
infinitesimally small. But what mathematicians do instead, they say:
"let epsilon > 0.". That is /any/ epsilon (it just have to be > 0).
So, for any given /fixed/ number > 0 (like 10^-(10^10)), you now
can choose an epsilon that is much smaller! And computers can
do that too - it just depends on how they are programmed.
Then, in topology, we have "open neighborhoods". They have no
specific size at all, yet they capture properties of a topological
space like the continuity of functions, convergence etc.
Ben Bacarisse was thinking very hard :
Moebius <invalid@example.invalid> writes:
Am 11.07.2024 um 02:28 schrieb Chris M. Thomasson:
On 7/10/2024 5:24 PM, Moebius wrote:
Am 11.07.2024 um 02:16 schrieb Chris M. Thomasson:
{a, b, c} vs { 3, 4, 5 }
Both have the same number of elements, [...]
HOW do you know that? Please define (for any sets A, B):
A and B /have the same number of elements/ iff
___________________ .
(i.e. fill out the blanks). :-)
Hint: That's what Ben Bacarisse is asking for.
Sure, it's "obvious" for us. But how would you define "have the same >>>>> number of elements" (in mathematical terms) such that it can be
DEDUCED
(!) für certain sets A and B?
________________________________________
Ok, I'm slighty vicious now... :-)
If a = b = c, {a, b, c} still has "the same number of elements" as {3, >>>>> 4, 5 }? :-P
I see {a, b, c} and {3, 4, 5} and think three elements.
Even if a = b = c?
C'mon man! :-P
Please, that's a red herring, and you know it! No where did I say that
a, b and c stood for anything (i.e. that they might be variables in the
maths sense). I this sort of context they are just distinct symbols.
Indeed!
I sometimes try to steer WM away from 'math' symbols in sets
like asking for a bijection of something like {elephant, rhinoceros,
dune buggy} and {circle, square, megaphone}.
Is there a countable number of infinite[ly many] primes, just like there is a countable number of infinite[ly many] naturals?
You can do that, but then you can't specify what it is you're talking
about exactly. So if two people are talking about an epsilon and a delta
that are positive numbers that are so small we can't represent them >numerically, we can't really figure out which of the two is smaller.
On 7/10/2024 6:08 PM, Moebius wrote:
Am 11.07.2024 um 03:06 schrieb Moebius:
Am 11.07.2024 um 02:51 schrieb Chris M. Thomasson:
What about dropping the word more in favor of density, or
granularity of an infinite set? The reals are denser, or more
granular than the rationals and naturals?
Yeah, makes sense.
On the other hand there are (indeed) "more" reals (in a certain sense)
than naturals and/or rationals. :-)
So this "more" allows for more "granularity". :-P
Can there be a granularity "score" for any infinite set?
sobriquet <dohduhdah@yahoo.com> wrote or quoted:
You can do that, but then you can't specify what it is you're talking
about exactly. So if two people are talking about an epsilon and a delta
that are positive numbers that are so small we can't represent them
numerically, we can't really figure out which of the two is smaller.
We can sometimes figure out which of the two is smaller.
The word "delta" in isolation does not have a meaning.
The name "delta" has to be properly introduced somehow into a text.
And sometimes this way of introduction gives us a relationship to
epsilon that will allow us to figure out which of the two is smaller.
Other times, delta may be helpful to show something where one is
not required to know whether it's larger or smaller than epsilon.
For example, let f(x):=x. Then I can show that for every
epsilon > 0, I can find a delta>0, so that x<delta implies
f(x)<epsilon. To prove this, I choose delta:=epsilon/2;
then x<delta implies x<epsilon/2, so f(x)<f(epsilon/2)=
epsilon/2<epsilon. My choice "delta:=epsilon/2" makes it
clear that in this case delta is smaller than epsilon.
So, we can talk about numbers without the need to represent
them numerically, and computers can do that too. It's called
"symbolic mathematics" and can be done with software such as
Maxima or using modern AI chatbots.
On 7/10/2024 5:31 PM, Moebius wrote:
Am 11.07.2024 um 02:28 schrieb Chris M. Thomasson:
On 7/10/2024 5:24 PM, Moebius wrote:
Am 11.07.2024 um 02:16 schrieb Chris M. Thomasson:
{a, b, c} vs { 3, 4, 5 }
Both have the same number of elements, [...]
HOW do you know that? Please define (for any sets A, B):
A and B /have the same number of elements/ iff _______________ . >>>>
(i.e. fill out the blanks). :-)
Hint: That's what Ben Bacarisse is asking for.
Sure, it's "obvious" for us.
Yes?But how would you define "have the same number of elements" (in mathematical
terms) such that it can be DEDUCED (!) für certain sets A and B?
________________________________________
Ok, I'm slighty wicked now... :-)
If a = b = c, {a, b, c} still has "the same number of elements" as
{3, 4, 5 }? :-P
I see {a, b, c} and {3, 4, 5} and think three elements.
Even if a = b = c?
C'mon man! :-P
Well, In my programming mind, { a, b, c } and { 3, 4, 5 } have the same number of elements. Is this bad?
Am 11.07.2024 um 02:33 schrieb Chris M. Thomasson:
On 7/10/2024 5:31 PM, Moebius wrote:
Am 11.07.2024 um 02:28 schrieb Chris M. Thomasson:
On 7/10/2024 5:24 PM, Moebius wrote:
Am 11.07.2024 um 02:16 schrieb Chris M. Thomasson:
{a, b, c} vs { 3, 4, 5 }
Both have the same number of elements, [...]
HOW do you know that? Please define (for any sets A, B):
A and B /have the same number of elements/ iff _______________ . >>>>>
(i.e. fill out the blanks). :-)
Hint: That's what Ben Bacarisse is asking for.
Sure, it's "obvious" for us.
Actually, it's not, since we don't know if a, b, c are "pairwise
distinct" (see below). Sorry about that.
Yes?But how would you define "have the same number of elements" (in
mathematical terms) such that it can be DEDUCED (!) für certain
sets A and B?
________________________________________
Ok, I'm slighty wicked now... :-)
If a = b = c, {a, b, c} still has "the same number of elements" as
{3, 4, 5 }? :-P
I see {a, b, c} and {3, 4, 5} and think three elements.
Even if a = b = c?
C'mon man! :-P
Well, In my programming mind, { a, b, c } and { 3, 4, 5 } have the
same number of elements. Is this bad?
Yes it's bad. :-)
If a = b = c, then {a, b, c} only contains 1 element.
Don't mix up the symbols used for denoting some objects with the objects denoted by the symbols.
In this case there are 3 (different) symbols, namely "a", "b", "c", but
it is not excluded that they all denote the same object, say, the number
1. Then {a, b, c} = {1}.
I hope "in [your] programming mind" [you] know the difference beween using
a, b, b
and
"a", "b", "c".
:-)
Hint (Python):
a = 1
print(a)
print("a")
:-P
On 7/10/2024 5:28 PM, Chris M. Thomasson wrote:
On 7/10/2024 5:24 PM, Moebius wrote:
If a = b = c, {a, b, c} still has "the same number of elements" as
{3, 4, 5 }? :-P
I see {a, b, c} and {3, 4, 5} and think three elements.
Then I start to examine how the elements are different and their
potential similarities, if any.
For some reason, { a, b, c } and { 3, 4, 5 } makes me think of monotonically increasing.
Am 11.07.2024 um 02:29 schrieb Chris M. Thomasson:
On 7/10/2024 5:28 PM, Chris M. Thomasson wrote:
On 7/10/2024 5:24 PM, Moebius wrote:
If a = b = c, {a, b, c} still has "the same number of elements" as
{3, 4, 5 }? :-P
Hint: In this case card({a, b, c}) = 1.
Or with other words: Ex(x e {a, b, c} & Ay(y e {a, b, c} -> x = y)).
Using a special quantifier: E!x(x e {a, b, c}) ("There is exactly one x
such that x is in {a, b, c}.")
I see {a, b, c} and {3, 4, 5} and think three elements.
You see (!) there
terms in "{a, b, c}" (namely "a", "b" and "c") and 3
terms in "{3, 4, 5}" (namely "3", "4" and "5").
Then I start to examine how the elements are different and their
potential similarities, if any.
Right. In this case (i.e. an arithmetic context) we may safely assume
that 3 =/= 3, 3 =/= 5 and 4 =/= 5. :-P
On the other hand, since we don't know anything concerning a, b and c,
all we can state/say is:
1 <= card({a, b, c}) <= 3
(while card({3, 4, 5}) = 3.)
For some reason, { a, b, c } and { 3, 4, 5 } makes me think of
monotonically increasing.
Concerning { 3, 4, 5 } the reason is, that indeed 3 < 4 < 5, though { 3,
4, 5 } = { 5, 4, 3 } = ... etc.
But concerning { a, b, c } there simply is NO (rational) reason for
assuming that.
a may be pi
b may be 0
c may be -i
Then {a, b, c} = {pi, 0, -i}. See?!
Or:
a may be 0
b may be 0
c may be 0
Then {a, b, c} = {0}. etc.
So how can we "compare" sets concerning their "size" (""number of elements"")? :-)
How about using a "measuring stick" (sort of)?
After serious thinking Moebius wrote :
Am 11.07.2024 um 12:34 schrieb FromTheRafters:
Ben Bacarisse was thinking very hard :
Moebius <invalid@example.invalid> writes:
Am 11.07.2024 um 02:28 schrieb Chris M. Thomasson:
On 7/10/2024 5:24 PM, Moebius wrote:
Am 11.07.2024 um 02:16 schrieb Chris M. Thomasson:
{a, b, c} vs { 3, 4, 5 }
Both have the same number of elements, [...]
HOW do you know that? Please define (for any sets A, B):
A and B /have the same number of elements/ iff
___________________ .
(i.e. fill out the blanks). :-)
Hint: That's what Ben Bacarisse is asking for.
Sure, it's "obvious" for us. But how would you define "have the same >>>>>>> number of elements" (in mathematical terms) such that it can be
DEDUCED
(!) für certain sets A and B?
________________________________________
Ok, I'm slighty vicious now... :-)
If a = b = c, {a, b, c} still has "the same number of elements"
as {3,
4, 5 }? :-P
I see {a, b, c} and {3, 4, 5} and think three elements.
Even if a = b = c?
C'mon man! :-P
Please, that's a red herring, and you know it! No where did I say that >>>> a, b and c stood for anything (i.e. that they might be variables in the >>>> maths sense). I this sort of context they are just distinct symbols.
Indeed!
Nonsense. (See below.)
You have stated this nonsense before, I simply disagree with a set in
roster form having duplicates.
On 7/10/2024 10:08 AM, Jim Burns wrote:
Is spacetime _actually_ continuous?
We don't think so, because
our current best theories develop problems
at small enough scale.
We expect some so.far.unknown theory
to make itself felt somewhere around
distances = 1 in natural units,
that is, units in which c = G = ℏ = 1,
the Planck length 1.6×10⁻³⁵ meter
Well, can we zoom out forever?
Zooming in forever, well
not sure if that is possible with
our current state of the art of
our understanding of the universe.
sobriquet <dohduhdah@yahoo.com> wrote or quoted:
You can do that, but then you can't specify what it is you're talking
about exactly. So if two people are talking about an epsilon and a delta
that are positive numbers that are so small we can't represent them
numerically, we can't really figure out which of the two is smaller.
We can sometimes figure out which of the two is smaller.
The word "delta" in isolation does not have a meaning.
The name "delta" has to be properly introduced somehow into a text.
And sometimes this way of introduction gives us a relationship to
epsilon that will allow us to figure out which of the two is smaller.
Other times, delta may be helpful to show something where one is
not required to know whether it's larger or smaller than epsilon.
For example, let f(x):=x. Then I can show that for every
epsilon > 0, I can find a delta>0, so that x<delta implies
f(x)<epsilon. To prove this, I choose delta:=epsilon/2;
then x<delta implies x<epsilon/2, so f(x)<f(epsilon/2)=
epsilon/2<epsilon. My choice "delta:=epsilon/2" makes it
clear that in this case delta is smaller than epsilon.
So, we can talk about numbers without the need to represent
them numerically, and computers can do that too. It's called
"symbolic mathematics" and can be done with software such as
Maxima or using modern AI chatbots.
On 7/11/2024 11:53 AM, Jim Burns wrote:
The rationals "are fractal".
They allow for unbounded zooming.
However,
not all splits of rationals
are situated (have a point at)
in the rationals.
Well, I use reals for my fractals.
If I am using complex numbers,
both of their parts (x, y) or
(real, imaginary) if you will,
use reals...
;^)
All splits of the reals
are situated in the reals.
In that sense, the reals are complete.
Well, I use reals for my fractals. If I am using complex numbers, both
of their parts (x, y) or (real, imaginary) if you will, use reals...
Are the gaps [between] prime numbers "random" wrt their various length's?
(2, 3) has no gap wrt the naturals, however, (3, 5) does [...].
https://en.wikipedia.org/wiki/Countable_set
We can index the primes:
[0] = 2
[1] = 3
[2] = 5
[3] = 7
...
On 7/11/2024 5:16 AM, sobriquet wrote:
Op 11/07/2024 om 12:49 schreef FromTheRafters:
sobriquet pretended :
Op 11/07/2024 om 00:12 schreef Chris M. Thomasson:
On 7/10/2024 3:00 PM, sobriquet wrote:
Op 10/07/2024 om 23:44 schreef Chris M. Thomasson:
On 7/10/2024 2:42 PM, Moebius wrote:
Am 10.07.2024 um 23:26 schrieb Chris M. Thomasson:
Well, can we zoom out forever? Zooming in forever,
In the comtext of the real and/or complex numbers we can!
That's just the beauty of math! Isn't it?
Yeah, I know we can zoom in forever in math for sure; fractals
are neat and fun all in one. No doubt. However, wrt the physical >>>>>>> world... Can we zoom in forever? Or for that matter, zoom out
forever?
We might think we're zooming in forever.. but we're just stuck in
a loop.
https://www.desmos.com/calculator/4mivqc0iht
Now, that makes my brain want to bleed a little bit. We hit a
"limit" and loop back on it? Something akin to that?
Computers have finite computational resources. So if you keep
zooming in (or out), you eventually run into these computational
limitations (or perhaps it just takes too long, but lets assume that
science has solved the issue of aging and you can keep living as
long as you please).
A loop can give the impression of zooming in indefinitely, but
you're not really zooming in indefinitely in that case.
Why not? Doesn't the scope of displayable pixels shrink commensurate
with the amount of zoom?
Because if you actually zoom in (so stopping the variable that is
looping) by using the + button on the right, eventually you will see
that the graphics get messed up as a result of running into the
computational limitations.
Right. Zooming in using floats vs doubles is a pretty big difference
right off the bat. For really deep zooms that doubles cannot handle, we
tend to use arbitrary precision libraries, slow, but doable. Then of
course we can run into memory issues wrt said libraries...
On 7/11/2024 1:31 PM, Jim Burns wrote:
Floating point numbers aren't real numbers.
They don't even include all of the rationals.
Floating point numbers are what I use on the good ol' computer to create
a bunch of my work. Sometimes I can get away with using floats, other
times I need to use doubles to get the higher precision needed.
Sometimes, I need to bust out an arbitrary precision lib... ;^)
On 7/11/2024 1:35 PM, Moebius wrote:
Am 11.07.2024 um 21:47 schrieb Chris M. Thomasson:
Well, I use reals for my fractals. If I am using complex numbers,
both of their parts (x, y) or (real, imaginary) if you will, use
reals...
Still, the complex numbers allow for more profound results in certain
(mathematical) fields.
See: https://en.wikipedia.org/wiki/Complex_analysis
Moreover, there are some quite interesting results in the context of
physics, btw.:
"Quantum Physics Falls Apart without Imaginary Numbers"
https://www.scientificamerican.com/article/quantum-physics-falls-
apart-without-imaginary-numbers/
"Quantum physics needs complex numbers"
https://www.researchgate.net/
publication/348802613_Quantum_physics_needs_complex_numbers
"Quantum theory based on real numbers can be experimentally falsified"
https://www.nature.com/articles/s41586-021-04160-4
Absolutely. By the way, have you ever played around with the so-called triplex numbers that can be used to generate the Mandelbulb?
Fwiw, here is a result I got from using Mandelbulbs as a base object for
one of my experimental biomorphic algorithms:
https://youtu.be/XpbPzrSXOgk
Here is another experiment using the same biomorphic algorithm, just
using different base shapes:
https://youtu.be/TLd64a4gdZQ
Pretty interesting results... :^)
Op 11/07/2024 om 15:03 schreef Stefan Ram:. . .
So, we can talk about numbers without the need to representOk, in some cases we can, but in other cases we might not be able to do
them numerically, and computers can do that too. It's called
"symbolic mathematics" and can be done with software such as
Maxima or using modern AI chatbots.
that if the only way to really say anything about how they relate to one >another would be to actually inspect their representation.
Not everything can be done with symbolic math and often we're forced
to work with numerical approximations instead.
On 7/11/2024 1:31 PM, Jim Burns wrote:
On 7/11/2024 3:47 PM, Chris M. Thomasson wrote:
On 7/11/2024 11:53 AM, Jim Burns wrote:
The rationals "are fractal".
They allow for unbounded zooming.
However,
not all splits of rationals
are situated (have a point at)
in the rationals.
Well, I use reals for my fractals.
If I am using complex numbers,
both of their parts (x, y) or
(real, imaginary) if you will,
use reals...
;^)
In computing,
floating point numbers are often called real numbers.
I don't know if you mean "floating point" here.
Floating point numbers aren't real numbers.
They don't even include all of the rationals.
Floating point numbers are
what I use on the good ol' computer
to create a bunch of my work.
Sometimes I can get away with using floats,
other times I need to use doubles
to get the higher precision needed.
Sometimes,
I need to bust out an arbitrary precision lib...
;^)
All splits of the reals
are situated in the reals.
In that sense, the reals are complete.
On 7/9/24 12:40 PM, WM wrote:
Le 09/07/2024 à 14:37, Ben Bacarisse a écrit :
A mathematician, to whom this is a whole new topic, would start by
asking you what you mean by "more". Without that, they could not
possibly answer you.
Good mathematicians could.
So, what do you mean by "more" when applied to
sets like C and R?
Proper subsets have less elements than their supersets.
But that logic leads to inconsistencies with infinite sets.
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
{a, b, c} vs { 3, 4, 5 }
Both have the same number of elements,
That will fall down for infinite sets unless, by decree, you state that
your meaning of "more" makes all infinite sets have the same number of elements.
Moebius <invalid@example.invalid> writes:
Am 11.07.2024 um 02:28 schrieb Chris M. Thomasson:
On 7/10/2024 5:24 PM, Moebius wrote:
Am 11.07.2024 um 02:16 schrieb Chris M. Thomasson:I see {a, b, c} and {3, 4, 5} and think three elements.
{a, b, c} vs { 3, 4, 5 }
Both have the same number of elements, [...]
HOW do you know that? Please define (for any sets A, B):
A and B /have the same number of elements/ iff ___________________ .
(i.e. fill out the blanks). :-)
Hint: That's what Ben Bacarisse is asking for.
Sure, it's "obvious" for us. But how would you define "have the same
number of elements" (in mathematical terms) such that it can be DEDUCED >>>> (!) für certain sets A and B?
________________________________________
Ok, I'm slighty vicious now... :-)
If a = b = c, {a, b, c} still has "the same number of elements" as {3, >>>> 4, 5 }? :-P
Even if a = b = c = 1?
C'mon man! :-P
Please, that's a red herring, and you know it! No where did I say that
a, b and c stood for anything (i.e. that they might be variables in the
maths sense). I this sort of context they are just distinct symbols.
Le 11/07/2024 02:46, Ben Bacarisse a crit :
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
{a, b, c} vs { 3, 4, 5 }That will fall down for infinite sets unless, by decree, you state that
Both have the same number of elements,
your meaning of "more" makes all infinite sets have the same number of
elements.
There are some rules for comparing sets which are not subset and superset, namely symmetry:
[...] we can use fractals to store/load data...
WM <wolfgang.mueckenheim@tha.de> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des Unendlichen"
at Hochschule Augsburg.)
Le 11/07/2024 à 02:46, Ben Bacarisse a écrit :
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
{a, b, c} vs { 3, 4, 5 }That will fall down for infinite sets unless, by decree, you state that
Both have the same number of elements,
your meaning of "more" makes all infinite sets have the same number of
elements.
There are some rules for comparing sets which are not subset and superset, >> namely symmetry:
Still nothing about defining set membership, equality and difference in WMaths though.
You've been dodging that one for years. Without sound
definitions of those things WMaths is pretty useless.
Le 11/07/2024 à 02:46, Ben Bacarisse a écrit :What is their number?
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
There are some rules for comparing sets which are not subset and{a, b, c} vs { 3, 4, 5 }That will fall down for infinite sets unless, by decree, you state that
Both have the same number of elements,
your meaning of "more" makes all infinite sets have the same number of
elements.
superset, namely symmetry:
The real numbers in intervals of same length like (n, n+1] are
equinumerous.
Further there is a rule of construction: The rational numbers are |ℚ| = 2|ℕ|^2 + 1.No, they are countable: bijective to the naturals.
The real numbers are infinitely more than the rational numbers becauseConsider the set of even numbers. Clearly they are bijective to the
every rational multiplied by an irrational is irrational.
Of course the complex numbers are infinitely many more than the reals.
That's the subset rule.
These rules have not lead to any contradiction, to my knowledge. Please
try.
Consider the set of even numbers. Clearly they are bijective to the
naturals, yet a subset of them. How many are there?
Le 13/07/2024 02:12, Ben Bacarisse a crit :
WM <wolfgang.mueckenheim@tha.de> writes:
(AKA Dr. Wolfgang Mckenheim or Mueckenheim who teaches "Geschichte des
Unendlichen"
and "Kleine Geschichte der Mathematik"
at Hochschule Augsburg.)
Meanwhile Technische Hochschule Augsburg.
Le 11/07/2024 02:46, Ben Bacarisse a crit :Still nothing about defining set membership, equality and difference in
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
{a, b, c} vs { 3, 4, 5 }That will fall down for infinite sets unless, by decree, you state that >>>> your meaning of "more" makes all infinite sets have the same number of >>>> elements.
Both have the same number of elements,
There are some rules for comparing sets which are not subset and superset, >>> namely symmetry:
WMaths though.
Are my rules appearing too reasonable for a believer in equinumerosity of prime numbers and algebraic numbers?
WM <wolfgang.mueckenheim@tha.de> writes:
Le 13/07/2024 à 02:12, Ben Bacarisse a écrit :
WM <wolfgang.mueckenheim@tha.de> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des >>> Unendlichen"
and "Kleine Geschichte der Mathematik"
Optional, I hope.
at Hochschule Augsburg.)
Meanwhile Technische Hochschule Augsburg.
A sound name change that reflects the technical college's focus.
Le 11/07/2024 à 02:46, Ben Bacarisse a écrit :Still nothing about defining set membership, equality and difference in
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
{a, b, c} vs { 3, 4, 5 }That will fall down for infinite sets unless, by decree, you state that >>>>> your meaning of "more" makes all infinite sets have the same number of >>>>> elements.
Both have the same number of elements,
There are some rules for comparing sets which are not subset and superset, >>>> namely symmetry:
WMaths though.
Are my rules appearing too reasonable for a believer in equinumerosity of
prime numbers and algebraic numbers?
You can define equinumerosity any way you like.
But you can't claim the
"surprising" result of WMaths that E in P and P \ {E} = P
Presumably that's why you teach history courses now -- you can avoid
having to write down even the most basic definitions of WMaths sets.
Le 14/07/2024 à 00:55, joes a écrit :That's fucking useless. I want SOME number.
Am Fri, 12 Jul 2024 17:13:09 +0000 schrieb WM:
It is dark, cannot be expressed in finite numbers. But certainly it isThere are some rules for comparing sets which are not subset andWhat is their number?
superset, namely symmetry:
The real numbers in intervals of same length like (n, n+1] are
equinumerous.
larger than |ℕ|.
No, the reals for example are uncountable.In fact every interval contains uncountably many numbers.Countable and uncountable are nonsense notions.
All sets have the same number of elements? Laughable.Further there is a rule of construction: The rational numbers areNo, they are countable: bijective to the naturals.
|ℚ| = 2|ℕ|^2 + 1.
Only if you wrongly regard |N| as a natural number itself.And what would this expression mean if you can't manipulate it?We have some measure for relative size.
I don't understand. What is the same?This proves the bijection wrong. The reason is that all elements of the bijection have ℵo successors. The bijection concerns only theOf course the complex numbers are infinitely many more than the reals.Consider the set of even numbers. Clearly they are bijective to the
That's the subset rule.
These rules have not lead to any contradiction, to my knowledge.
Please try.
naturals,
potentially infinite collections, they are the same for all well-ordered sets.
Can you define the division? What does precision have to do with it?yet a subset of them. How many are there?Even integers |ℕ|, even naturals |ℕ|/2, both with precision of one or two.
Am Fri, 12 Jul 2024 17:13:09 +0000 schrieb WM:
There are some rules for comparing sets which are not subset andWhat is their number?
superset, namely symmetry:
The real numbers in intervals of same length like (n, n+1] are
equinumerous.
In fact every interval contains uncountably many numbers.
Further there is a rule of construction: The rational numbers are |ℚ| =No, they are countable: bijective to the naturals.
2|ℕ|^2 + 1.
And what would this expression mean if you can't manipulate it?
The real numbers are infinitely more than the rational numbers because
every rational multiplied by an irrational is irrational.
Of course the complex numbers are infinitely many more than the reals.Consider the set of even numbers. Clearly they are bijective to the
That's the subset rule.
These rules have not lead to any contradiction, to my knowledge. Please
try.
naturals,
yet a subset of them. How many are there?
Am Sun, 14 Jul 2024 13:28:49 +0000 schrieb WM:
Le 14/07/2024 à 00:55, joes a écrit :That's fucking useless. I want SOME number.
Am Fri, 12 Jul 2024 17:13:09 +0000 schrieb WM:It is dark, cannot be expressed in finite numbers. But certainly it is
There are some rules for comparing sets which are not subset andWhat is their number?
superset, namely symmetry:
The real numbers in intervals of same length like (n, n+1] are
equinumerous.
larger than |ℕ|.
In fact every interval contains uncountably many numbers.Countable and uncountable are nonsense notions.
No, the reals for example are uncountable.All sets have the same number of elements? Laughable.Further there is a rule of construction: The rational numbers areNo, they are countable: bijective to the naturals.
|ℚ| = 2|ℕ|^2 + 1.
The bijection concerns only theI don't understand. What is the same?
potentially infinite collections, they are the same for all well-ordered
sets.
Can you define the division? What does precision have to do with it?yet a subset of them. How many are there?Even integers |ℕ|, even naturals |ℕ|/2, both with precision of one or
two.
Le 14/07/2024 à 15:45, joes a écrit :
Am Sun, 14 Jul 2024 13:28:49 +0000 schrieb WM:
even naturals |ℕ|/2, [...] with precision of one or [...]
Can you define the division? What does precision have to do with it?
If the elements of ℕ are distributed equally into two sets, then each
one has |ℕ|/2 elements, +/- 1.
Le 14/07/2024 à 03:30, Ben Bacarisse a écrit :
the "surprising" result of WMaths that E in P and P \ {E} = P
Le 14/07/2024 à 15:45, joes a écrit :
Am Sun, 14 Jul 2024 13:28:49 +0000 schrieb WM:
Le 14/07/2024 à 00:55, joes a écrit :
[The rational numbers] are countable: bijective to the naturals.
All sets have the same number of elements? [WM]
No, the reals for example are uncountable.
Every infinite set is uncountable.
The "surprising" result of Ben Bacarisse's math: card({a, b, c}) = 3, even for a = b = c.
Le 14/07/2024 03:30, Ben Bacarisse a crit :
WM <wolfgang.mueckenheim@tha.de> writes:
Le 13/07/2024 02:12, Ben Bacarisse a crit :Optional, I hope.
WM <wolfgang.mueckenheim@tha.de> writes:
(AKA Dr. Wolfgang Mckenheim or Mueckenheim who teaches "Geschichte des >>>> Unendlichen"
and "Kleine Geschichte der Mathematik"
A sound name change that reflects the technical college's focus.at Hochschule Augsburg.)
Meanwhile Technische Hochschule Augsburg.
You can define equinumerosity any way you like.Le 11/07/2024 02:46, Ben Bacarisse a crit :Still nothing about defining set membership, equality and difference in >>>> WMaths though.
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
{a, b, c} vs { 3, 4, 5 }That will fall down for infinite sets unless, by decree, you state that >>>>>> your meaning of "more" makes all infinite sets have the same number of >>>>>> elements.
Both have the same number of elements,
There are some rules for comparing sets which are not subset and superset,
namely symmetry:
Are my rules appearing too reasonable for a believer in equinumerosity of >>> prime numbers and algebraic numbers?
And I can prove that Cantor's way leads astray.
But you can't claim the
"surprising" result of WMaths that E in P and P \ {E} = P
That refers to potential infinity and dark elements. Visible elements form only a potentially infinite collection.
Presumably that's why you teach history courses now -- you can avoid
having to write down even the most basic definitions of WMaths sets.
At the end of the course I talk about the present state of the art.
Am 14.07.2024 um 16:22 schrieb WM:
If the elements of ℕ are distributed equally into two sets, then each
one has |ℕ|/2 elements, +/- 1.
Faszinierend, Mückenheim.
Und wenn wir IN in abzählbar unendlich viele abzählbar unendliche Teilmengen zerlegen, wieviele Elemente hat dann so eine Teilmenge im
Schnitt?
Also
T_1 = {1, 2, 4, 7, ...}
T_2 = {3, 5, 8, 12, ...}
T_3 = {6, 9, 13,18 ...}
:
|T_1| = ?
|T_2| = ?
|T_3| = ?
usw.
Und sollte dann nicht auch
|T_1| + |T_2| + |T_3| + ... = |IN| gelten?
Wir können es aber auch einfacher machen:
Sei P die Menge der Primzahlen. |P| = ? und |IN \ P| = ?
Auch hier sollte dann wohl
|P| + |IN \ P| = |IN|
sein. :-)
Am 14.07.2024 um 17:25 schrieb Moebius:
Am 14.07.2024 um 16:22 schrieb WM:
If the elements of ℕ are distributed equally into two sets, then each
one has |ℕ|/2 elements, +/- 1.
Faszinierend, Mückenheim.
Und wenn wir IN in abzählbar unendlich viele abzählbar unendliche
Teilmengen zerlegen, wieviele Elemente hat dann so eine Teilmenge im
Schnitt?
Also
T_1 = {1, 2, 4, 7, ...}
T_2 = {3, 5, 8, 12, ...}
T_3 = {6, 9, 13,18 ...}
:
|T_1| = ?
|T_2| = ?
|T_3| = ?
usw.
Und sollte dann nicht auch
|T_1| + |T_2| + |T_3| + ... = |IN| gelten?
|T_i| ist dann wohl |IN|/oo (für alle i e IN), oder so etwas, also vermutlich recht klein!
0 kann es schlecht sein, denn 0 + 0 + 0 + ... ist vermutlich (selbst in Mückenhausen) immer noch 0, nein?
Wenn es aber > 0 ist, so ist
|T_1| + |T_2| + |T_3| + ...
vermutlich ziemlich groß. Wie groß ist eigentlich |IN|, und was genau
für ein mathematisches Objekt soll das sein, Mückenheim?
Haben Sie das schon iw. definiert? Ist es eine natürliche, reelle oder anderswie geartete Zahl? Viell. sogar eine Kardinalzahl? Wo genau haben
Sie diese Art von Zahlen denn definiert?
Oder ist es nicht viel eher so, dass Sie einfach nur saudummen
Scheißdreck daherreden?
Wir können es aber auch einfacher machen:
Sei P die Menge der Primzahlen. |P| = ? und |IN \ P| = ?
Auch hier sollte dann wohl
|P| + |IN \ P| = |IN|
sein. :-)
Hier erheben sich die gleichen Fragen. (a) Welche mathem. Objekte werden
denn durch |P| und |IN \ P| bezeichnet? (b) Wie sind denn diese Objekte DEFINIERT? (c) Wie beweisen Sie dann (auf der Basis dieser
Definitionen), dass |P| + |IN \ P| = |IN| ist?
Wie soll ich es sagen, Mückenheim: Ihr dummes Geschwalle reicht als
Beweis dafür nicht aus.
Am 14.07.2024 um 16:22 schrieb WM:
Le 14/07/2024 à 15:45, joes a écrit :
Am Sun, 14 Jul 2024 13:28:49 +0000 schrieb WM:
even naturals |ℕ|/2, [...] with precision of one or [...]
Can you define the division? What does precision have to do with it?
If the elements of ℕ are distributed equally into two sets, then each
one has |ℕ|/2 elements, +/- 1.
Faszinierend, Mückenheim.
Und wenn wir IN in abzählbar unendlich viele anzählbar unendliche Teilmengen zerlegen, wieviele Elemente hat dann so eine Teilmenge im
Schnitt?
Also
T_1 = {1, 2, 4, 7, ...}
T_2 = {3, 5, 8, 12, ...}
T_3 = {6, 9, 13,18 ...}
:
|T_1| = ?
|T_2| = ?
|T_3| = ?
usw.
Und sollte dann nicht auch
|T_1| + |T_2| + |T_3| + ... = |IN| gelten?
____________________________
Wir können es aber auch einfacher machen:
Sei P die Menge der Primzahlen. |P| = ? und |IN \ P| = ?
Auch hier sollte dann wohl
|P| + |IN \ P| = |IN|
sein. :-)
WM <wolfgang.mueckenheim@tha.de> writes:
You can define equinumerosity any way you like.
And I can prove that Cantor's way leads astray.
But no journal will touch it. I can't remember which crank excuse you
use to explain that.
Presumably that's why you teach history courses now -- you can avoid
having to write down even the most basic definitions of WMaths sets.
At the end of the course I talk about the present state of the art.
Do you cite the journal that has published your proof that Cantor is
wrong?
Do you give the "proper" definitions for set membership,
difference and equality once you admit that those in your textbook are
only approximations? Do you present a proof of the "surprising" result
that sets E and P exist with E in P and P \ {E} = P?
Fortunately for you, your college has no degree program in mathematics
so none of your students know better. Unfortunately for your students,
you don't know better.
Sie sagten doch, dass mit IN = {1, 2, 3, ...} gilt:
|IN \ {1}| = |IN| - 1
Demnach müsste wohl gelten:
|IN \ {1} \ {3}| = |IN|- 1 - 1 = |IN| - {1 + 1}
usw.
Also schließlich:
|IN \ {1} \ {3} \ {5} \ ... | = |IN| - 1 - 1 - 1 - ... = |IN| - {1 + 1 +
1 + ...}.
Andererseits behaupten Sie auch, dass
|IN \ {1} \ {3} \ {5} \ ... | = |G| = |IN/2| ist.
Demnach müsste also
1 + 1 + 1 + ... = |IN|/2 sein.
Das ist interessant!
Müsste dann aber nicht auch
|IN \ {1} \ {4} \ {6} \ ... | = |P| = |IN/2| sein?
Wegen:
|IN \ {1} \ {4} \ {6} \ ... | = |IN| - 1 - 1 - 1 - ... = |IN| - {1 + 1 +
1 + ...} = |IN| - |IN|/2 = |IN|/2.
Le 15/07/2024 à 00:39, Ben Bacarisse a écrit :Standing in the face of the establishment is a sure sign of a crackpot.
WM <wolfgang.mueckenheim@tha.de> writes:
Simple: The journals are owned by matheologians and stupids. I haveBut no journal will touch it. I can't remember which crank excuse youYou can define equinumerosity any way you like.And I can prove that Cantor's way leads astray.
use to explain that.
never tried to address them.
Further all that stuff including this proof has been published as a
book.
Cutting down to different platforms, I see only one book and one article."Does Set Theory Cause Perceptual Problems?", viXra 2017-02-26Do you cite the journal that has published your proof that Cantor isPresumably that's why you teach history courses now -- you can avoidAt the end of the course I talk about the present state of the art.
having to write down even the most basic definitions of WMaths sets.
wrong?
"Transfinity - A Source Book", SSRN-Elsevier (April 2024)
"Proof of the existence of dark numbers (bilingual version)",
OSFPREPRINTS (Nov 2022)
"Dark numbers", Academia.edu (2020)
"Dark numbers", Quora (May 2023)
"Sequences and Limits", Advances in Pure Mathematics 5, 2015, pp. 59 -
61.
"Transfinity - A Source Book", ELIVA Press, Chisinau 2024.
And that is why no one uses it.Do you give the "proper" definitions for set membership,That cannot be done for potentially infinite collections because they
have no fixed membership.
Oh really? What do your students say?difference and equality once you admit that those in your textbook areThere has not yet been any disprove of my simplest proof (that I told
only approximations? Do you present a proof of the "surprising" result
that sets E and P exist with E in P and P \ {E} = P?
you recently and that you were wise enaugh to let it uncommented). The
only daredevil who tried it, Jim Burns, has to assume that by exchangig
one of the elements can disappear. No reason to pay attention. And the nonsense you once tried to sell to my former students has been rejected
by them flatly.
Am Mon, 15 Jul 2024 13:26:25 +0000 schrieb WM:
Cutting down to different platforms, I see only one book and one article."Does Set Theory Cause Perceptual Problems?", viXra 2017-02-26Do you cite the journal that has published your proof that Cantor isPresumably that's why you teach history courses now -- you can avoid >>>>> having to write down even the most basic definitions of WMaths sets.At the end of the course I talk about the present state of the art.
wrong?
"Transfinity - A Source Book", SSRN-Elsevier (April 2024)
"Proof of the existence of dark numbers (bilingual version)",
OSFPREPRINTS (Nov 2022)
"Dark numbers", Academia.edu (2020)
"Dark numbers", Quora (May 2023)
"Sequences and Limits", Advances in Pure Mathematics 5, 2015, pp. 59 -
61.
"Transfinity - A Source Book", ELIVA Press, Chisinau 2024.
The others count as selfpublished and haha, quora.
And that is why no one uses it.Do you give the "proper" definitions for set membership,That cannot be done for potentially infinite collections because they
have no fixed membership.
And theOh really? What do your students say?
nonsense you once tried to sell to my former students has been rejected
by them flatly.
Le 14/07/2024 à 17:25, Moebius a écrit :
Und wenn wir IN in abzählbar unendlich viele abzählbar unendliche
Teilmengen _zerlegen_, wieviele Elemente hat dann so eine Teilmenge im
Schnitt?
Also [z. B.]
T_1 = {1, 2, 4, 7, ...}
T_2 = {3, 5, 8, 12, ...}
T_3 = {6, 9, 13,18 ...}
:
|T_1| = ?
|T_2| = ?
|T_3| = ?
usw. [ad infinitum.]
Rech[n]en kann ich
Und sollte dann nicht auch
|T_1| + |T_2| + |T_3| + ... = |IN| gelten?
Nein, denn es fehlen die Zahlen 10, 14, 15, 19, 20, ...
Le 15/07/2024 à 00:39, Ben Bacarisse a écrit :
[...]
The only daredevil who tried it, Jim Burns,
has to assume that
by exchangig one of the elements can disappear.
No reason to pay attention.
On 7/15/2024 9:26 AM, WM wrote:
Le 15/07/2024 à 00:39, Ben Bacarisse a écrit :
[...]
The only daredevil who tried it, Jim Burns,
has to assume that
by exchangig one of the elements can disappear.
Assume that ℕⁿᵒᵗᐧᵂᴹ is the set of finiteⁿᵒᵗᐧᵂᴹ ordinals.
If α is finiteⁿᵒᵗᐧᵂᴹ, then α⁺¹ = α∪{α} is finiteⁿᵒᵗᐧᵂᴹ.
If α ∈ ℕⁿᵒᵗᐧᵂᴹ, then α⁺¹ ∈ ℕⁿᵒᵗᐧᵂᴹ.
f(α) = α⁺¹ is 1.to.1
¬∃β≠α: β⁺¹ = α⁺¹
f: ℕⁿᵒᵗᐧᵂᴹ → ℕⁿᵒᵗᐧᵂᴹ\{0}: 1.to.1 ℕⁿᵒᵗᐧᵂᴹ ∋ 0 ∉ f(ℕⁿᵒᵗᐧᵂᴹ)
'Bye, Bob!
No reason to pay attention.
E pur si muove.
Le 15/07/2024 à 15:41, joes a écrit :
Am Mon, 15 Jul 2024 13:26:25 +0000 schrieb WM:
And theOh really? What do your students say?
nonsense you once tried to sell to my former students has been rejected
by them flatly.
That was some years ago. I don't remember the details, only the result. Probably the idea was discussed that an inclusion-monotonic sequence of infinite terms could have an empty intersection. Every sensible student recognizes that this is impossible. As long as all terms contain an
infinite subset of the first set.
On 7/15/2024 9:26 AM, WM wrote:
No reason to pay attention.
Am 15.07.2024 um 20:31 schrieb Python:
Le 15/07/2024 à 16:46, WM a écrit :
Probably the idea was discussed that an inclusion-monotonic sequence
of infinite terms could have an empty intersection.
Which is an extremely trivial state of afairs, Mückenheim.
Hint: There is no natural number in the intersection of all "endsegments".
Extremely trivial reason: For each and every n e IN: n !e {n+1, n+2,
n+3, ...}. In other words, An e IN: n !e INTERSECTION_(n e IN) {n+1,
n+2, n+3, ...}.
Le 15/07/2024 à 16:46, WM a écrit :
Probably the idea was discussed that an inclusion-monotonic
sequence of infinite terms could have an empty intersection.
Every sensible student
Am 15.07.2024 um 19:04 schrieb Jim Burns:
On 7/15/2024 9:26 AM, WM wrote:
No reason to pay attention.
That's why
he's reading (parts of) all your posts,
of course.
On 7/15/2024 9:26 AM, WM wrote:
No reason to pay attention.
Am 15.07.2024 um 20:31 schrieb Python:
Le 15/07/2024 à 16:46, WM a écrit :
Probably the idea was discussed that an inclusion-monotonic
sequence of infinite terms could have an empty intersection.
Which is an extremely trivial state of afairs,
Hint: There is no natural number in the intersection of all "endsegments".
Am 15.07.2024 um 19:04 schrieb Jim Burns:
On 7/15/2024 9:26 AM, WM wrote:
No reason to pay attention.
That's why he's reading (parts of) all your posts, of course.
Hint: In (classical) mathematics {n+1, n+2, n+3, ...} is an infinite set
for all n in IN.
Did you ever meet a "sensible student" who claimed that there is an n in
IN such that {n+1, n+2, n+3, ...} is finite?
Le 15/07/2024 à 21:08, Moebius a écrit :
Hint: In (classical) mathematics {n+1, n+2, n+3, ...} is an infinite
set for all n in IN.
If always <bla bla bla>
Did you ever meet a "sensible student" who claimed that there is an n
in IN such that {n+1, n+2, n+3, ...} is finite?
There are many students and matematicians who believe that all natural
numbrs can be removed or simply applied, for instance for enzmerating purposes.
Le 15/07/2024 à 20:53, Moebius a écrit :
Am 15.07.2024 um 20:31 schrieb Python:
Le 15/07/2024 à 16:46, WM a écrit :
Probably the idea was discussed that an inclusion-monotonic sequence
of infinite terms could have an empty intersection.
Which is an extremely trivial state of afairs,
Hint: There is no natural number in the intersection of all
"endsegments".
True.
Am 16.07.2024 um 15:08 schrieb WM:
Question:
Did you ever meet a "sensible student" who claimed that there is an n
in IN such that {n+1, n+2, n+3, ...} is finite?
There are many students and mathematicians who believe that all natural
numbrs can be removed or simply applied, for instance for enzmerating
purposes.
MAN, my question was: Did you ever meet a "sensible student" who claimed
that there is an n in IN such that {n+1, n+2, n+3, ...} is finite?
Le 16/07/2024 à 16:06, Moebius a écrit :
Am 16.07.2024 um 15:08 schrieb WM:
Question:
Did you ever meet a "sensible student" who claimed that there is an
n in IN such that {n+1, n+2, n+3, ...} is finite?
There are many students and mathematicians who believe that all
natural numbrs can be removed or simply applied, for instance for
enzmerating purposes.
MAN, my question was: Did you ever meet a "sensible student" who
claimed that there is an n in IN such that {n+1, n+2, n+3, ...} is
finite?
{n+1, n+2, n+3, ...} is what remains.
Am 15.07.2024 um 14:38 schrieb WM:
Le 14/07/2024 à 17:25, Moebius a écrit :
Und wenn wir IN in abzählbar unendlich viele abzählbar unendliche
Teilmengen _zerlegen_, wie viele Elemente hat dann so eine Teilmenge
im Schnitt?
Also [z. B.]
T_1 = {1, 2, 4, 7, ...}
T_2 = {3, 5, 8, 12, ...}
T_3 = {6, 9, 13,18 ...}
:
Die PUNKTE (:) hier bedeuten, dass wir über eine UNENDLICHE Menge
(Folge) von Teilmengen sprechen, Du Depp.
Also T_1, T_2, T_3, und so weiter, ad infinitum.
Also über eine Folge (T_n)_(n e IN), wo T_n c IN für jedes n e IN und
für alle n e IN, m e IN: n =/= m -> T_n n T_m = { }.
|T_1| = ?
|T_2| = ?
|T_3| = ?
usw. [ad infinitum.]
Rech[n]en kann ich
Ja, schön, Mückenheim, aber lass uns beim Thema bleiben.
Und sollte dann nicht auch
|T_1| + |T_2| + |T_3| + ... = |IN| gelten?
Nein, denn es fehlen die Zahlen 10, 14, 15, 19, 20, ...
Nein, die fehlen nicht, Du Depp.
Gefragt ist hier nach SUM_(n=1..oo) |T_n|.
Und für jedes n e IN gibt es ein k e IN, so dass n e T_k ist - nach der
zwar nicht explizit angegebenen aber offensichtlichen Konstruktion der
T_i (i e IN).
Hinweis:
T_1 = {1, 2, 4, 7, 11, 16, ...}
T_2 = {3, 5, 8, 12, 17, ..}
T_3 = {6, 9, 13, 18 ...}
T_4 = {10, 14, 19, ...}
T_5 = {15, 20, ...}
:
Das Schema ist offensichtlich, und ziemlich trivial.
Das kleinste Element der Menge T_i ist i*(i+1)/2.
Nennen wir das kleinste Element der Menge T_i "T_i_1" und das nächstgrößere "T_i_2" usw.
Dann gilt für alle i e IN: T_i_2 = T_i_1 + i.
Den Rest überlasse ich Dir zur Übung.
___________________________________________________________
ALSO nochmal:
Wegen
1. U_(n e IN) T_n = IN, und
2. T_i n T_j = {} für alle i =/= j
sollte also wohl
|T_1| + |T_2| + |T_3| + ... = SUM_(n=1..oo) |T_n| = |IN| gelten.
|T_i| ist dann wohl |IN|/oo (für alle i e IN), oder so etwas, also vermutlich recht klein!
0 kann es schlecht sein, denn 0 + 0 + 0 + ... ist vermutlich (selbst in Mückenhausen) immer noch 0, nein?
Wenn es aber > 0 ist, so ist
|T_1| + |T_2| + |T_3| + ... = SUM_(n=1..oo) |T_n|
vermutlich ziemlich groß. Wie groß ist eigentlich |IN| [und wie groß
sind die |T_i| für i e IN], und was genau für ein mathematisches Objekt [mathematische Objekte] soll das [sollen die] sein, Mückenheim?
Haben Sie das schon iw. definiert? Ist es eine [sind es] natürliche,
reelle oder anderswie geartete Zahl[en]? Viell. sogar eine Kardinalzahl [Kardinalzahlen]? Wo genau haben Sie diese Art von Zahlen denn definiert?
Oder ist es nicht viel eher so, dass Sie einfach nur saudummen
Scheißdreck daherreden?
Le 15/07/2024 à 20:53, Moebius a écrit :
Am 15.07.2024 um 20:31 schrieb Python:
Le 15/07/2024 à 16:46, WM a écrit :
Probably the idea was discussed that
an inclusion-monotonic sequence of infinite terms
could have an empty intersection.
Which is an extremely trivial state of afairs,
Hint:
There is no natural number in
the intersection of all "endsegments".
True.
But you claim an empty intersection of
all infinite endsegments,
i.e. endsegments which keep
an infinite number of naturals in common with E(1).
That is false.
On 7/16/2024 9:03 AM, WM wrote:
Le 15/07/2024 à 20:53, Moebius a écrit :
Am 15.07.2024 um 20:31 schrieb Python:
Le 15/07/2024 à 16:46, WM a écrit :
Probably the idea was discussed that
an inclusion-monotonic sequence of infinite terms
could have an empty intersection.
Which is an extremely trivial state of afairs,
Hint:
There is no natural number in
the intersection of all "endsegments".
True.
And each end segment is infiniteⁿᵒᵗᐧᵂᴹ.
There is no natural number in
the intersection of all infiniteⁿᵒᵗᐧᵂᴹ end segments.
But you claim an empty intersection of
all infinite endsegments,
Each natural number is not.in at least one infiniteⁿᵒᵗᐧᵂᴹ end segments.
Ach, Mückenheim, wie kann ein (halbwegs) gebildeter Mensch so dumm sein
wie Sie?
What numbers constitute the [...] endsegment which is common to all
infinity endsegments.
Am 16.07.2024 um 21:33 schrieb WM:
What numbers constitute the [...] endsegment which is common to all
infinite endsegments.
Hint: No numbers, since there is no endsegment "which which is common to
all infinity endsegments".
Am 15.07.2024 um 18:02 schrieb Moebius:
Am 15.07.2024 um 14:38 schrieb WM:
Le 14/07/2024 à 17:25, Moebius a écrit :
Kommt da nochmal etwas, Mückenheim?
Und wenn wir IN in abzählbar unendlich viele abzählbar unendliche
Teilmengen _zerlegen_, wie viele Elemente hat dann so eine Teilmenge
im Schnitt?
Also [z. B.]
T_1 = {1, 2, 4, 7, ...}
T_2 = {3, 5, 8, 12, ...}
T_3 = {6, 9, 13,18 ...}
:
Die PUNKTE (:) hier bedeuten, dass wir über eine UNENDLICHE Menge
(Folge) von Teilmengen sprechen, Du Depp.
Also T_1, T_2, T_3, und so weiter, ad infinitum.
Also über eine Folge (T_n)_(n e IN), wo T_n c IN für jedes n e IN und
für alle n e IN, m e IN: n =/= m -> T_n n T_m = { }.
|T_1| = ?
|T_2| = ?
|T_3| = ?
usw. [ad infinitum.]
Rech[n]en kann ich
Ja, schön, Mückenheim, aber lass uns beim Thema bleiben.
Und sollte dann nicht auch
|T_1| + |T_2| + |T_3| + ... = |IN| gelten?
Nein, denn es fehlen die Zahlen 10, 14, 15, 19, 20, ...
Nein, die fehlen nicht, Du Depp.
Gefragt ist hier nach SUM_(n=1..oo) |T_n|.
Und für jedes n e IN gibt es ein k e IN, so dass n e T_k ist - nach
der zwar nicht explizit angegebenen aber offensichtlichen Konstruktion
der T_i (i e IN).
Hinweis:
T_1 = {1, 2, 4, 7, 11, 16, ...}
T_2 = {3, 5, 8, 12, 17, ..}
T_3 = {6, 9, 13, 18 ...}
T_4 = {10, 14, 19, ...}
T_5 = {15, 20, ...}
:
Das Schema ist offensichtlich, und ziemlich trivial.
Das kleinste Element der Menge T_i ist i*(i+1)/2.
Nennen wir das kleinste Element der Menge T_i "T_i_1" und das
nächstgrößere "T_i_2" usw.
Dann gilt für alle i e IN: T_i_2 = T_i_1 + i.
Den Rest überlasse ich Dir zur Übung.
___________________________________________________________
ALSO nochmal:
Wegen
1. U_(n e IN) T_n = IN, und
2. T_i n T_j = {} für alle i =/= j
sollte also wohl
|T_1| + |T_2| + |T_3| + ... = SUM_(n=1..oo) |T_n| = |IN| gelten.
|T_i| ist dann wohl |IN|/oo (für alle i e IN), oder so etwas, also
vermutlich recht klein!
0 kann es schlecht sein, denn 0 + 0 + 0 + ... ist vermutlich (selbst
in Mückenhausen) immer noch 0, nein?
Wenn es aber > 0 ist, so ist
|T_1| + |T_2| + |T_3| + ... = SUM_(n=1..oo) |T_n|
vermutlich ziemlich groß. Wie groß ist eigentlich |IN| [und wie groß
sind die |T_i| für i e IN], und was genau für ein mathematisches
Objekt [mathematische Objekte] soll das [sollen die] sein, Mückenheim?
Haben Sie das schon iw. definiert? Ist es eine [sind es] natürliche,
reelle oder anderswie geartete Zahl[en]? Viell. sogar eine
Kardinalzahl [Kardinalzahlen]? Wo genau haben Sie diese Art von Zahlen
denn definiert?
Oder ist es nicht viel eher so, dass Sie einfach nur saudummen
Scheißdreck daherreden?
T_1 = {1, 2, 4, 7, 11, 16, ...}
T_2 = {3, 5, 8, 12, 17, ..}
T_3 = {6, 9, 13, 18 ...}
T_4 = {10, 14, 19, ...}
T_5 = {15, 20, ...}
:
Wegen
1. U_(n e IN) T_n = IN, und
2. T_i n T_j = {} für alle i =/= j
sollte also wohl
|T_1| + |T_2| + |T_3| + ... = SUM_(n=1..oo) |T_n| = |IN| gelten.
Am 16.07.2024 um 17:37 schrieb Moebius:
Am 15.07.2024 um 18:02 schrieb Moebius:
Am 15.07.2024 um 14:38 schrieb WM:
Le 14/07/2024 à 17:25, Moebius a écrit :
Kommt da nochmal etwas, Mückenheim?
Kommt da nochmal etwas, Mückenheim?
Und wenn wir IN in abzählbar unendlich viele abzählbar unendliche
Teilmengen _zerlegen_, wie viele Elemente hat dann so eine
Teilmenge im Schnitt?
Also [z. B.]
T_1 = {1, 2, 4, 7, ...}
T_2 = {3, 5, 8, 12, ...}
T_3 = {6, 9, 13,18 ...}
:
Die PUNKTE (:) hier bedeuten, dass wir über eine UNENDLICHE Menge
(Folge) von Teilmengen sprechen, Du Depp.
Also T_1, T_2, T_3, und so weiter, ad infinitum.
Also über eine Folge (T_n)_(n e IN), wo T_n c IN für jedes n e IN und
für alle n e IN, m e IN: n =/= m -> T_n n T_m = { }.
|T_1| = ?
|T_2| = ?
|T_3| = ?
usw. [ad infinitum.]
Rech[n]en kann ich
Ja, schön, Mückenheim, aber lass uns beim Thema bleiben.
Und sollte dann nicht auch
|T_1| + |T_2| + |T_3| + ... = |IN| gelten?
Nein, denn es fehlen die Zahlen 10, 14, 15, 19, 20, ...
Nein, die fehlen nicht, Du Depp.
Gefragt ist hier nach SUM_(n=1..oo) |T_n|.
Und für jedes n e IN gibt es ein k e IN, so dass n e T_k ist - nach
der zwar nicht explizit angegebenen aber offensichtlichen
Konstruktion der T_i (i e IN).
Hinweis:
T_1 = {1, 2, 4, 7, 11, 16, ...}
T_2 = {3, 5, 8, 12, 17, ..}
T_3 = {6, 9, 13, 18 ...}
T_4 = {10, 14, 19, ...}
T_5 = {15, 20, ...}
:
Das Schema ist offensichtlich, und ziemlich trivial.
Das kleinste Element der Menge T_i ist i*(i+1)/2.
Nennen wir das kleinste Element der Menge T_i "T_i_1" und das
nächstgrößere "T_i_2" usw.
Dann gilt für alle i e IN: T_i_2 = T_i_1 + i.
Den Rest überlasse ich Dir zur Übung.
___________________________________________________________
ALSO nochmal:
Wegen
1. U_(n e IN) T_n = IN, und
2. T_i n T_j = {} für alle i =/= j
sollte also wohl
|T_1| + |T_2| + |T_3| + ... = SUM_(n=1..oo) |T_n| = |IN| gelten.
|T_i| ist dann wohl |IN|/oo (für alle i e IN), oder so etwas, also
vermutlich recht klein!
0 kann es schlecht sein, denn 0 + 0 + 0 + ... ist vermutlich (selbst
in Mückenhausen) immer noch 0, nein?
Wenn es aber > 0 ist, so ist
|T_1| + |T_2| + |T_3| + ... = SUM_(n=1..oo) |T_n|
vermutlich ziemlich groß. Wie groß ist eigentlich |IN| [und wie groß
sind die |T_i| für i e IN], und was genau für ein mathematisches
Objekt [mathematische Objekte] soll das [sollen die] sein, Mückenheim?
Haben Sie das schon iw. definiert? Ist es eine [sind es] natürliche,
reelle oder anderswie geartete Zahl[en]? Viell. sogar eine
Kardinalzahl [Kardinalzahlen]? Wo genau haben Sie diese Art von
Zahlen denn definiert?
Oder ist es nicht viel eher so, dass Sie einfach nur saudummen
Scheißdreck daherreden?
Le 16/07/2024 à 20:58, Jim Burns a écrit :
On 7/16/2024 9:03 AM, WM wrote:
Le 15/07/2024 à 20:53, Moebius a écrit :
Am 15.07.2024 um 20:31 schrieb Python:
Le 15/07/2024 à 16:46, WM a écrit :
Probably the idea was discussed that
an inclusion-monotonic sequence of
infinite terms
could have an empty intersection.
Which is an extremely trivial state of afairs,
Hint:
There is no natural number in
the intersection of all "endsegments".
True.
And each end segment is infiniteⁿᵒᵗᐧᵂᴹ.
Each number that can be used for bijections
has an infinite endsegment.
The ℵo terms of the infinite endsegments
cannot be deleted in steps.
There is no natural number in the intersection of
all infiniteⁿᵒᵗᐧᵂᴹ end segments.
What about
the infinitely many numbers which are remainimg from E(1)?
But you claim an empty intersection of
all infinite endsegments,
Each natural number is
not.in at least one infiniteⁿᵒᵗᐧᵂᴹ end segments.
What numbers constitute
the infinite endsegment which is
common to all infinity endsegments.
Le 17/07/2024 à 19:11, Jim Burns a écrit :
An infiniteⁿᵒᵗᐧᵂᴹ end segment
common to all infiniteⁿᵒᵗᐧᵂᴹ end segments
not.exists.
Because
On 7/16/2024 3:33 PM, WM wrote:
What about
the infinitely many numbers which are remainimg from E(1)?
None of them remain in all.infiniteⁿᵒᵗᐧᵂᴹ (all) end segments.
An infiniteⁿᵒᵗᐧᵂᴹ end segment
common to all infiniteⁿᵒᵗᐧᵂᴹ end segments
not.exists.
Le 15/07/2024 00:39, Ben Bacarisse a crit :
WM <wolfgang.mueckenheim@tha.de> writes:
But no journal will touch it. I can't remember which crank excuse youYou can define equinumerosity any way you like.
And I can prove that Cantor's way leads astray.
use to explain that.
Simple: The journals are owned by matheologians and stupids. I have never tried to address them.
Further all that stuff including this proof has been published as a book.
Do you cite the journal that has published your proof that Cantor isPresumably that's why you teach history courses now -- you can avoid
having to write down even the most basic definitions of WMaths sets.
At the end of the course I talk about the present state of the art.
wrong?
I could do so:
"Does Set Theory Cause Perceptual Problems?", viXra 2017-02-26 "Not enumerating all positive rational numbers", viXra 2017-02-26 "The union is not the limit.", viXra 2017-03-06 "Failure of the Diagonal Argument",
viXra 2017-03-13 "Set Theory or Slipper Animalcule: Who Wins?", viXra 2017-03-13 "Proof of the existence of dark numbers (bilingual version)", viXra (Nov 2022)
"Shortest Proof of Dark Numbers", viXra (May 2023)
"Seven Internal Contradictions of Set Theory", viXra (Dec 2023)
"Transfinity - A Source Book", SSRN-Elsevier (April 2024)
"Proof of the existence of dark numbers (bilingual version)", OSFPREPRINTS (Nov 2022)
"Dark natural numbers in set theory", ResearchGate, October 2019
"Dark natural numbers in set theory" II, ResearchGate, October 2019 "Transfinity - A Source Book", ResearchGate, October 2019 "What scatters
the space?", MResearchGate, May 2020 "Countability Contradicted", ResearchGate, February 2022
"Proof of the existence of dark numbers (bilingual version)", ResearchGate, Nov 2022
"The seven deadly sins of set theory", ResearchGate, Dec 2023
"Dark numbers", Academia.edu (2020) "Transfinity - A Source Book", Academia.edu (31 Dec 2020) "Countability contradicted", Academia.edu (Feb 2022) "Proof of the existence of dark numbers (bilingual version)", Academia.edu (Nov 2022)
"The seven deadly sins of set theory", Academia.edu (Dec 2023)
"Dark numbers", Quora (May 2023)
"Sequences and Limits", Advances in Pure Mathematics 5, 2015, pp. 59 - 61. "Transfinity - A Source Book", ELIVA Press, Chisinau 2024.
But I do not quote all that (some of the above with over 1000 reads - more than usual for maths journals) like I do not quote Newton's or Euler's or Gauss' or Cauchy's original essays.
Do you give the "proper" definitions for set membership,
That cannot be done for potentially infinite collections because they have
no fixed membership.
difference and equality once you admit that those in your textbook are
only approximations? Do you present a proof of the "surprising" result
that sets E and P exist with E in P and P \ {E} = P?
WM <wolfgang.mueckenheim@tha.de> writes:
Le 15/07/2024 à 00:39, Ben Bacarisse a écrit :
WM <wolfgang.mueckenheim@tha.de> writes:
But no journal will touch it. I can't remember which crank excuse youYou can define equinumerosity any way you like.
And I can prove that Cantor's way leads astray.
use to explain that.
Simple: The journals are owned by matheologians and stupids. I have never
tried to address them.
*ticks crank excuse bingo card*
Further all that stuff including this proof has been published as a book.
Do you cite the journal that has published your proof that Cantor isPresumably that's why you teach history courses now -- you can avoid >>>>> having to write down even the most basic definitions of WMaths sets.
At the end of the course I talk about the present state of the art.
wrong?
I could do so:
"Does Set Theory Cause Perceptual Problems?", viXra 2017-02-26 "Not
enumerating all positive rational numbers", viXra 2017-02-26 "The union is >> not the limit.", viXra 2017-03-06 "Failure of the Diagonal Argument",
viXra 2017-03-13 "Set Theory or Slipper Animalcule: Who Wins?", viXra
2017-03-13 "Proof of the existence of dark numbers (bilingual version)",
viXra (Nov 2022)
"Shortest Proof of Dark Numbers", viXra (May 2023)
"Seven Internal Contradictions of Set Theory", viXra (Dec 2023)
"Transfinity - A Source Book", SSRN-Elsevier (April 2024)
"Proof of the existence of dark numbers (bilingual version)", OSFPREPRINTS >> (Nov 2022)
"Dark natural numbers in set theory", ResearchGate, October 2019
"Dark natural numbers in set theory" II, ResearchGate, October 2019
"Transfinity - A Source Book", ResearchGate, October 2019 "What scatters
the space?", MResearchGate, May 2020 "Countability Contradicted",
ResearchGate, February 2022
"Proof of the existence of dark numbers (bilingual version)", ResearchGate, >> Nov 2022
"The seven deadly sins of set theory", ResearchGate, Dec 2023
"Dark numbers", Academia.edu (2020) "Transfinity - A Source Book",
Academia.edu (31 Dec 2020) "Countability contradicted", Academia.edu (Feb
2022) "Proof of the existence of dark numbers (bilingual version)",
Academia.edu (Nov 2022)
"The seven deadly sins of set theory", Academia.edu (Dec 2023)
"Dark numbers", Quora (May 2023)
"Sequences and Limits", Advances in Pure Mathematics 5, 2015, pp. 59 - 61. >> "Transfinity - A Source Book", ELIVA Press, Chisinau 2024.
So no peer reviewed publications. We knew that, of course.
But I do not quote all that (some of the above with over 1000 reads - more >> than usual for maths journals) like I do not quote Newton's or Euler's or
Gauss' or Cauchy's original essays.
Do you give the "proper" definitions for set membership,
That cannot be done for potentially infinite collections because they have >> no fixed membership.
Ah, OK. So there is no way to prove any theorems about WMaths sets
because there are no definitions of even the most basic terms.
difference and equality once you admit that those in your textbook are
only approximations? Do you present a proof of the "surprising" result
that sets E and P exist with E in P and P \ {E} = P?
So how did you know that for the E and P we talked about six years ago
that E in P and P \ {E} = P? Was it all just how you felt at the time?
Is it still true today, or have the sets changed in the last six years
and it's no longer true? I have to ask because in WMaths set membership difference and equality can't be defined, so all anyone can do to find
out what you laughingly call "proper mathematics" says is to ask you.
You (apparently) "know" (or knew) that E in P even though you can't
prove it.
Le 17/07/2024 à 19:11, Jim Burns a écrit :
On 7/16/2024 3:33 PM, WM wrote:
What about the infinitely many numbers
which are remainimg from E(1)?
None of them remain in
all.infiniteⁿᵒᵗᐧᵂᴹ (all) end segments.
What remains to keep the endsegments infinite?
An infiniteⁿᵒᵗᐧᵂᴹ end segment
common to all infiniteⁿᵒᵗᐧᵂᴹ end segments
not.exists.
Because of inclusion monotony
this is a wrong result.
On 7/17/2024 1:57 PM, WM wrote:
What remains to keep the endsegments infinite?
| For each number, there is a number after
proves false
| There is a number after all numbers.
You (WM) aren't referring to ℕⁿᵒᵗᐧᵂᴹ
WM <wolfgang.mueckenheim@tha.de> writes:
So no peer reviewed publications. We knew that, of course.
But I do not quote all that (some of the above with over 1000 reads - more >> than usual for maths journals) like I do not quote Newton's or Euler's or
Gauss' or Cauchy's original essays.
Do you give the "proper" definitions for set membership,
That cannot be done for potentially infinite collections because they have >> no fixed membership.
Ah, OK. So there is no way to prove any theorems about WMaths sets
because there are no definitions of even the most basic terms.
difference and equality once you admit that those in your textbook are
only approximations? Do you present a proof of the "surprising" result
that sets E and P exist with E in P and P \ {E} = P?
So how did you know that for the E and P we talked about six years ago
that E in P and P \ {E} = P? Was it all just how you felt at the time?
Is it still true today, or have the sets changed in the last six years
and it's no longer true? I have to ask because in WMaths set membership difference and equality can't be defined, so all anyone can do to find
out what you laughingly call "proper mathematics" says is to ask you.
You (apparently) "know" (or knew) that E in P even though you can't
prove it.
Le 18/07/2024 à 01:54, Ben Bacarisse a écrit :
WM <wolfgang.mueckenheim@tha.de> writes:
So no peer reviewed publications. We knew that, of course.
I refuse to call contemporaries my peers
Le 18/07/2024 à 13:18, Jim Burns a écrit :
On 7/17/2024 1:57 PM, WM wrote:
What remains to keep the endsegments infinite?
| For each number, there is a number after
proves false
| There is a number after all numbers.
The claim is that
after every n defining the endsegment E(n)
there are ℵo natnumbers remaining from E(1)
in all infinite endsegments.
You (WM) aren't referring to ℕⁿᵒᵗᐧᵂᴹ
I am referring to mathematics.
Most of my "peers" are too stupid to comprehend this,
not all, fortunately, and no students.
The sequence of endsegments is inclusion monotonic.
Every infinite endsegment has
an infinite intersection with all infinite endsegments.
ℵ₀.many remain in E(n)
0.many remain in all.infinite (all) end segments
Le 18/07/2024 à 20:49, Jim Burns a écrit :
ℵ₀.many remain in E(n)
0.many remain in all.infinite (all) end segments
Each infinite endsegment has infinitely many numbers.
How many are not in all predecessors?
On 7/18/2024 4:50 PM, WM wrote:
It is insufficient to be in all _predecessor_ end.segments.
On 7/18/2024 4:50 PM, WM wrote:
Each infinite endsegment has infinitely many numbers.
"Each infinite endsegment has infinitely many numbers." (!!!)
How many are not in all predecessors?>You are confused about
the intersection of all endsegments.
Le 18/07/2024 à 01:54, Ben Bacarisse a écrit :
WM <wolfgang.mueckenheim@tha.de> writes:
So no peer reviewed publications. We knew that, of course.
I refuse to call contemporaries my peers who are too stupid to understand questions like the following:
If it is impossible to explain the increase of NUF(x) in agreement with
basic mathematics (i.e. by not more than 1 at any x > 0), then the axioms
of natural numbers are incompatible with basic mathematics. We can change this basic mathematics
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
or the axiom
∀n ∈ ℕ: ∃ n+1 ∈ ℕ
or pursue inconsistent mathematics.
What would you propose?
It's "strange" that WM "forgets" about the infinitely man "successors".
WM <wolfgang.mueckenheim@tha.de> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte
des Unendlichen" and "Kleine Geschichte der Mathematik" at Technische Hochschule Augsburg.)
Le 18/07/2024 à 01:54, Ben Bacarisse a écrit :
WM <wolfgang.mueckenheim@tha.de> writes:
So no peer reviewed publications. We knew that, of course.
I refuse to call contemporaries my peers who are too stupid to understand
questions like the following:
If it is impossible to explain the increase of NUF(x) in agreement with
basic mathematics (i.e. by not more than 1 at any x > 0), then the axioms
of natural numbers are incompatible with basic mathematics. We can change
this basic mathematics
You pretend to offer an alternative
but you admit that even the most
basic operations can't be defined because WMsets have no fixed
membership.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
or the axiom
∀n ∈ ℕ: ∃ n+1 ∈ ℕ
or pursue inconsistent mathematics.
What would you propose?
I would propose that you prove an inconsistency.
You dislike axioms and rules of
deduction
Either way, you
never show a proof.
It's "strange" that WM "forgets" about the infinitely man "successors".
On 7/18/2024 4:50 PM, WM wrote:
Each infinite endsegment has infinitely many numbers.
How many are not in all predecessors?
Each infinite end.segment has infinitely many numbers.
Each is not in all.infinite (all) end segments.
It is insufficient to be in all _predecessor_ end.segments.
Le 18/07/2024 à 23:51, Jim Burns a écrit :
It is insufficient to be in all _predecessor_ end.segments.
Infinitely many numbers are in all predecessors and in all *infinite* successors.
Am 19.07.2024 um 16:07 schrieb WM:
Le 18/07/2024 à 23:51, Jim Burns a écrit :
It is insufficient to be in all _predecessor_ end.segments.
Infinitely many numbers are in all predecessors and in all *infinite*
successors.
Nope.
1. ALL end.segments are infinite (by definition).
2. NO natural number is in all (infinite) end.segments.
Hint: For each and every natural number n: n is not in the (infinite) end.segment {n+1, n+2, n+3, ...}.
Because NUF(x) doesn't have a finite value of ANY finite x > 0, and thus
is just UNDEFINED, [...]
Le 19/07/2024 à 01:33, Ben Bacarisse a écrit :
WM <wolfgang.mueckenheim@tha.de> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte
des Unendlichen" and "Kleine Geschichte der Mathematik" at Technische
Hochschule Augsburg.)
Le 18/07/2024 à 01:54, Ben Bacarisse a écrit :
WM <wolfgang.mueckenheim@tha.de> writes:
So no peer reviewed publications. We knew that, of course.
I refuse to call contemporaries my peers who are too stupid to
understand
questions like the following:
If it is impossible to explain the increase of NUF(x) in agreement with
basic mathematics (i.e. by not more than 1 at any x > 0), then the
axioms
of natural numbers are incompatible with basic mathematics. We can
change
this basic mathematics
You pretend to offer an alternative
Above all I have disproved set theory.
but you admit that even the most
basic operations can't be defined because WMsets have no fixed
membership.
On the contrary, actually infinite sets have an absolute fixed
membership. Only the definable elements of potentially infinite
collection have none. An example is Hilbert's hotel which can expand.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
or the axiom
∀n ∈ ℕ: ∃ n+1 ∈ ℕ
or pursue inconsistent mathematics.
What would you propose?
I would propose that you prove an inconsistency.
Try to excplain how NUF(x) can increase from 0 to infinity without
violating ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Then you will understand the inconsistency.
You dislike axioms and rules of
deduction
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
is all you need. Do you believe it without deriving it from axioms?
Either way, youTry to understand and repeat the above. Then you will understand.
never show a proof.
Regards, WM
Am 19.07.2024 um 17:51 schrieb Richard Damon:But aleph_0 isn't a finite value, which are the only sorts of numbers
Because NUF(x) doesn't have a finite value of ANY finite x > 0, and
thus is just UNDEFINED, [...]
Nonsense. It's defined (in the context of set theory) with
NUF: IR --> {0, aleph_0}
x |-> card({u e {1/n : n e IN} : u <= x}) .
Then NUF(x) = 0 for all x e IR, x <= 0
and NUF(x) = aleph_0 for all x e IR, x > 0.
On 7/19/24 12:14 PM, Moebius wrote:
Am 19.07.2024 um 17:51 schrieb Richard Damon:
Because NUF(x) doesn't have a finite value of ANY finite x > 0, and
thus is just UNDEFINED, [...]
[...] It's defined (in the context of set theory) with
NUF: IR --> {0, aleph_0}
x |-> card({u e {1/n : n e IN} : u <= x}) .
Then NUF(x) = 0 for all x e IR, x <= 0
and NUF(x) = aleph_0 for all x e IR, x > 0.
But aleph_0 isn't a finite value, which are the only sorts of numbers
that WMs system can use.
Once you allow transfinite values, then [...] NUF(x) can jump from 0 to aleph_0.
Hint: For each and every natural number n: n is not in the (infinite)
end.segment {n+1, n+2, n+3, ...}.
Das gilt FÜR ALLE natürlichen Zahlen n.
On 7/19/24 9:49 AM, WM wrote:
Try to excplain how NUF(x) can increase from 0 to infinity without
violating ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Then you will understand the inconsistency.
Because NUF(x) doesn't have a finite value of ANY finite x > 0,
and thus
is just UNDEFINED,
because its verbal definition makes in incorrect
assumption that there *IS* a smallest unit fraction,
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
is all you need. Do you believe it without deriving it from axioms?
Which means that there can not be a smallest unit fraction, as if you
think think that 1/n is the smallest unit fraction, it shows that
1/(n+1) will be smaller, but still greater than 0.
Try to understand and repeat the above. Then you will understand.
And you logic only works if you assume that there must be a largest
natural number, even if you don't know what it is.
Your "Darkness" is just your inability to handle the unboundedness of
the infinite set of Natural Numbers.
On 7/19/24 12:14 PM, Moebius wrote:
Am 19.07.2024 um 17:51 schrieb Richard Damon:But aleph_0 isn't a finite value, which are the only sorts of numbers
Because NUF(x) doesn't have a finite value of ANY finite x > 0, and
thus is just UNDEFINED, [...]
Nonsense. It's defined (in the context of set theory) with
NUF: IR --> {0, aleph_0}
x |-> card({u e {1/n : n e IN} : u <= x}) .
Then NUF(x) = 0 for all x e IR, x <= 0
and NUF(x) = aleph_0 for all x e IR, x > 0.
that WMs system can use.
Le 18/07/2024 à 23:51, Jim Burns a écrit :
On 7/18/2024 4:50 PM, WM wrote:
Each infinite endsegment has infinitely many numbers.
How many are not in all predecessors?
Each infinite end.segment has infinitely many numbers.
Each is not in all.infinite (all) end segments.
It is insufficient to be in all _predecessor_ end.segments.
Infinitely many numbers are
in all predecessors and
in all *infinite* successors.
Infinitely many of them are the same
because of inclusion monotony.
Infinite endsegments with empty intersection
would require an complete exchange of thenumbers.
But that is excluded by inclusion montony.
You are wrong.
Infinitely many numbers are
in all predecessors and
in all *infinite* successors.
Infinitely many of them are the same
because of inclusion monotony.
On 7/19/2024 10:07 AM, WM wrote:
Infinitely many numbers are
in all predecessors and
in all *infinite* successors.
Infinitely many of them are the same
because of inclusion monotony.
Infinite endsegments with empty intersection
would require an complete exchange of the numbers.
But that is excluded by inclusion montony.
You are wrong.
The natural numbers ℕⁿᵒᵗᐧᵂᴹ are
well.ordered with step.up and non.0.step.down.
If you say I am wrongᵂᴹ about that,
what you (WM) are calling "the natural numbers"
aren't the natural numbers.
Infinitely many of them are the same
because of inclusion monotony.
Quantifier shift.
Le 19/07/2024 à 19:10, Richard Damon a écrit :
On 7/19/24 12:14 PM, Moebius wrote:
Am 19.07.2024 um 17:51 schrieb Richard Damon:But aleph_0 isn't a finite value, which are the only sorts of numbers
Because NUF(x) doesn't have a finite value of ANY finite x > 0, and
thus is just UNDEFINED, [...]
Nonsense. It's defined (in the context of set theory) with
NUF: IR --> {0, aleph_0}
x |-> card({u e {1/n : n e IN} : u <= x}) .
Then NUF(x) = 0 for all x e IR, x <= 0
and NUF(x) = aleph_0 for all x e IR, x > 0.
that WMs system can use.
No, NUF(x) = ℵo for every definable x > 0.
Regards, WM
Am 19.07.2024 um 16:07 schrieb WM:
Le 18/07/2024 à 23:51, Jim Burns a écrit :
It is insufficient to be in all _predecessor_ end.segments.
Infinitely many numbers are in all predecessors and in all *infinite*
successors.
Nope.
1. ALL end.segments are infinite (by definition).
2. NO natural number is in all (infinite) end.segments.
Hint: For each and every natural number n: n is not in the (infinite) end.segment {n+1, n+2, n+3, ...}.
Le 19/07/2024 à 16:18, Moebius a écrit :They are not the same ones.
Am 19.07.2024 um 16:07 schrieb WM:
Le 18/07/2024 à 23:51, Jim Burns a écrit :
2. NO natural number is in all (infinite) end.segments.But infinitely many are in all those endsegments which contain
infinitely many natnumbers.
Quoted for contradiction.Hint: For each and every natural number n: n is not in the (infinite)Of course. But that does not concern infinite endsegments.
end.segment {n+1, n+2, n+3, ...}.
It is not guarantedd that n+1 exists for every n.If that were a result, NUF can't be defined and is useless.
Le 19/07/2024 à 17:51, Richard Damon a écrit :
On 7/19/24 9:49 AM, WM wrote:
That is your desired result, not an argment.Try to excplain how NUF(x) can increase from 0 to infinity withoutBecause NUF(x) doesn't have a finite value of ANY finite x > 0,
violating ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Then you will understand the inconsistency.
How do you know that number?and thus is just UNDEFINED,Not at all. NUF(x) is the number of unit fractions between 0 and x.
I am left speechless.It is not guarantedd that n+1 exists for every n.∀n ∈ ℕ: 1/n - 1/(n+1) > 0Which means that there can not be a smallest unit fraction, as if you
is all you need. Do you believe it without deriving it from axioms?
think think that 1/n is the smallest unit fraction, it shows that
1/(n+1) will be smaller, but still greater than 0.
Yes we can.Your "Darkness" is just your inability to handle the unboundedness ofYou can't handle it either. NUF(x) is well defined. Your denial is
the infinite set of Natural Numbers.
wrong.
On 7/20/24 8:37 AM, WM wrote:
Le 19/07/2024 à 19:10, Richard Damon a écrit :
Which means your logic includes transfinite numbers,But aleph_0 isn't a finite value, which are the only sorts of numbers
that WMs system can use.
No, NUF(x) = ℵo for every definable x > 0.
so there is the gap
between 0 and the finite numbers where NUF(x) increases from 0 to ℵo.
So, NUF can jump from 0 to 1 at 1/ℵo which if you consider ℵo to be a number that represents a count, then 1/ℵo is close enough to a unit fraction.
Am Sat, 20 Jul 2024 12:19:26 +0000 schrieb WM:
But infinitely many are in all those endsegments which containThey are not the same ones.
infinitely many natnumbers.
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM
Not at all. NUF(x) is the number of unit fractions between 0 and x.How do you know that number?
I am left speechless.It is not guarantedd that n+1 exists for every n.∀n ∈ ℕ: 1/n - 1/(n+1) > 0Which means that there can not be a smallest unit fraction, as if you
is all you need. Do you believe it without deriving it from axioms?
think think that 1/n is the smallest unit fraction, it shows that
1/(n+1) will be smaller, but still greater than 0.
Le 20/07/2024 à 15:07, Richard Damon a écrit :
On 7/20/24 8:37 AM, WM wrote:
Le 19/07/2024 à 19:10, Richard Damon a écrit :
Which means your logic includes transfinite numbers,But aleph_0 isn't a finite value, which are the only sorts of
numbers that WMs system can use.
No, NUF(x) = ℵo for every definable x > 0.
That is my premise.
so there is the gap between 0 and the finite numbers where NUF(x)
increases from 0 to ℵo.
Yes, between 0 and the definable numbers x > 0 there is a gap.
So, NUF can jump from 0 to 1 at 1/ℵo which if you consider ℵo to be a
number that represents a count, then 1/ℵo is close enough to a unit
fraction.
I would not call it 1/ℵo because in my system ℵo means infinitely many with respect to all infinites, but it is not forbidden. The dark number
x where NUF(x) = 1 is certainly in the range of 1/ℵo.
Regards, WM
On 7/20/24 9:37 AM, WM wrote:
Le 20/07/2024 à 15:07, Richard Damon a écrit :
On 7/20/24 8:37 AM, WM wrote:
Le 19/07/2024 à 19:10, Richard Damon a écrit :
Which means your logic includes transfinite numbers,But aleph_0 isn't a finite value, which are the only sorts of
numbers that WMs system can use.
No, NUF(x) = ℵo for every definable x > 0.
That is my premise.
And thus in includes all the natural numbers as defined.
I would not call it 1/ℵo because in my system ℵo means infinitely many >> with respect to all infinites, but it is not forbidden. The dark number
x where NUF(x) = 1 is certainly in the range of 1/ℵo.
But the point is that the number isn't in the range of the finite
numbers, like the conventional unit fractions.
Your "Dark numbers" are just the transfinite numbers,
ALL the members of the Natural, Rational, and Real numbers are
definable.
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM
<wolfgang.mueckenheim@tha.de>:
NUF(x) is the number of unit fractions between 0 and x.
How do you know that number?
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is all you need. [WM]
Which means that there can not be a smallest unit fraction, as if you
think that 1/n is the smallest unit fraction, it shows that
1/(n+1) will be smaller [...]
It is not guarantedd that n+1 exists for every n.
I am left speechless.
On 7/20/24 9:37 AM, WM wrote:...
Le 20/07/2024 à 15:07, Richard Damon a écrit :
On 7/20/24 8:37 AM, WM wrote:
Le 19/07/2024 à 19:10, Richard Damon a écrit :
Which means your logic includes transfinite numbers, so there is the gap between 0 and the finite numbers where NUF(x) increases from 0 to ℵo.
So, NUF can jump from 0 to 1 at 1/ℵo which
if you consider ℵo to be a number that represents a count, then 1/ℵo is close enough to a unit
fraction.
Am Sat, 20 Jul 2024 12:19:26 +0000 schrieb WM:
Le 19/07/2024 à 16:18, Moebius a écrit :
Hint: For each and every natural number n: n is not in the (infinite)
end.segment {n+1, n+2, n+3, ...}.
Of course. But that does not concern infinite endsegments.
Quoted for contradiction.
Le 20/07/2024 à 16:00, Richard Damon a écrit :
On 7/20/24 9:37 AM, WM wrote:
Le 20/07/2024 à 15:07, Richard Damon a écrit :
On 7/20/24 8:37 AM, WM wrote:
Le 19/07/2024 à 19:10, Richard Damon a écrit :
Which means your logic includes transfinite numbers,But aleph_0 isn't a finite value, which are the only sorts of
numbers that WMs system can use.
No, NUF(x) = ℵo for every definable x > 0.
That is my premise.
And thus in includes all the natural numbers as defined.
No. It includes all, because we can manipulate them.
|ℕ \ {1, 2, 3, ...}| = 0
But not all are defined, bevause when we manipulate a defined one, most remain not manipulated
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
I would not call it 1/ℵo because in my system ℵo means infinitely
many with respect to all infinites, but it is not forbidden. The dark
number x where NUF(x) = 1 is certainly in the range of 1/ℵo.
But the point is that the number isn't in the range of the finite
numbers, like the conventional unit fractions.
All unit fractions are in the range of finite numbers.
Your "Dark numbers" are just the transfinite numbers,
No, all numbers less than ω are finite numbers.
ALL the members of the Natural, Rational, and Real numbers are definable.
Subtract all numbers from ℕ. Nothig remains.
Subtract the greatest number that can be defined from ℕ. ℵo numbers remain.
Regards, WM
since 1/ℵo is sort of a unit fraction
Subtract the greatest number that can be defined from ℕ. ℵo numbers
remain.
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM
<wolfgang.mueckenheim@tha.de>:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is all you need. [WM]
Which means that there can not be a smallest unit fraction, as if you
think think that 1/n is the smallest unit fraction, it shows that
1/(n+1) will be smaller [...]
It is not guarantedd that n+1 exists for every n. [WM]
I am left speechless.
Am 19.07.2024 um 16:18 schrieb Moebius:
Am 19.07.2024 um 16:07 schrieb WM:
Le 18/07/2024 à 23:51, Jim Burns a écrit :Nope.
It is insufficient to be in all _predecessor_ end.segments.
Infinitely many numbers are in all predecessors and in all *infinite* successors.
Le 19/07/2024 à 17:51, Richard Damon a écrit :
On 7/19/24 9:49 AM, WM wrote:
Try to excplain how NUF(x) can increase from 0 to infinity without
violating ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Then you will understand the inconsistency.
Because NUF(x) doesn't have a finite value of ANY finite x > 0,
That is your desired result, not an argment.
and thus is just UNDEFINED,
Not at all. NUF(x) is the number of unit fractions between 0 and x.
because its verbal definition makes in incorrect assumption that there
*IS* a smallest unit fraction,
No, there is no assumption but only a result from mathematics.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
is all you need. Do you believe it without deriving it from axioms?
Which means that there can not be a smallest unit fraction, as if you
think think that 1/n is the smallest unit fraction, it shows that
1/(n+1) will be smaller, but still greater than 0.
It is not guarantedd that n+1 exists for every n.
Try to understand and repeat the above. Then you will understand.
And you logic only works if you assume that there must be a largest
natural number, even if you don't know what it is.
I do not assume it but derive it from mathematics.
Your "Darkness" is just your inability to handle the unboundedness of
the infinite set of Natural Numbers.
You can't handle it either. NUF(x) is well defined. Your denial is wrong.
Regards, WM
ROFL!!!! WM's racist numbers, of course he calls them dark! lol.
Le 20/07/2024 à 15:28, joes a écrit :No number n is included in all segments, in particular no E(m) for m>n.
Am Sat, 20 Jul 2024 12:19:26 +0000 schrieb WM:
But infinitely many of them are the same natural numbers for allBut infinitely many are in all those endsegments which containThey are not the same ones.
infinitely many natnumbers.
infinite endsegments, because they stem from E(1) and have not been lost
as long as the endsegments are infinite.
Am Sat, 20 Jul 2024 13:48:44 +0000 schrieb WM:
But infinitely many of them [bla, bla, bla] for all infinite endsegments [blu blu blu]
No number n is included in all segments, in particular no[t in] E(m) for m > n.
Am 20.07.2024 um 15:35 schrieb joes:
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is all you need. [WM]
Which means that there can not be a smallest unit fraction, as if you
think think that 1/n is the smallest unit fraction, it shows that
1/(n+1) will be smaller [...]
Ja, denn aus der oben VON WM angeführten Formel ergibt sich sofort:
∀n ∈ ℕ: 1/n > 1/(n+1).
It is not guaranteed that n+1 exists for every n. [WM]
Doch das ist (im gegenwärtigen Kontext)
garantiert. WENN dem NICHT so WÄRE, dann dürftest Du Deine Formel erst
gar nicht verwenden. Denn dann könnte es ein n_0 e IN geben, so dass
"n_0 + 1" nicht definiert ist; aus Deiner Formel lässt sich aber durch Spezialisierung ableiten:
1/n_0 - 1/(n_0 + 1) > 0 .
Daher ergibt sich aus Deiner Formel (in diesem Kontext) in der Tat, (mit
der Def. des Begriffs /Stammbruchs/ und dem Umstand, dass mit n e IN
auch n+1 e IN ist [->Pean0 und Def. von +]) dass es zu jedem Stammbruch
einen kleineren Stammbruch gibt.
No number is "in all predecessors and in all *infinite* successors".
On 7/20/24 11:05 AM, WM wrote:
All unit fractions are in the range of finite numbers.
And NUF is undefined because it presumes doing something that can not be done.
Since there is not finite number x such that NUF(x) is 1, it shows that either its definition is just internally inconsistant, or it includes,
by necessisty, a broader class of values as "unit fractions" then you
are thinking of.
The lowest value of x where NUF(x) == 1 must be a "unit fraction" and it
can not be a "unit fraction of a natural number", as you agree that for
all of those NUF(x) is ℵo,
since 1/ℵo is sort of a unit fraction,
Subtract all numbers from ℕ. Nothig remains.
Subtract the greatest number that can be defined from ℕ. ℵo numbers remain.
Only because you have a broken definition of "defined" that can't handle
the rest of your system.
On 7/20/24 8:35 AM, WM wrote:
Le 19/07/2024 à 17:51, Richard Damon a écrit :
On 7/19/24 9:49 AM, WM wrote:
Try to excplain how NUF(x) can increase from 0 to infinity without
violating ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Then you will understand the inconsistency.
Because NUF(x) doesn't have a finite value of ANY finite x > 0,
That is your desired result, not an argment.
No, it is just a provable fact.
and thus is just UNDEFINED,
Not at all. NUF(x) is the number of unit fractions between 0 and x.
Which is something that can't be defined as to have any value other than
0 or ℵo since there is no finite value that has a finite non-zero number
of unit fractions below it.
Am Sat, 20 Jul 2024 13:48:44 +0000 schrieb WM:
Le 20/07/2024 à 15:28, joes a écrit :No number n is included in all segments, in particular no E(m) for m>n.
Am Sat, 20 Jul 2024 12:19:26 +0000 schrieb WM:But infinitely many of them are the same natural numbers for all
But infinitely many are in all those endsegments which containThey are not the same ones.
infinitely many natnumbers.
infinite endsegments, because they stem from E(1) and have not been lost
as long as the endsegments are infinite.
An e IN: E(n) is infinite.
In other words, for all X: X is an infinite endsegment iff X is an endsegment.
Le 20/07/2024 à 22:56, joes a écrit :I don't see it. What does your first sentence mean?
Am Sat, 20 Jul 2024 13:48:44 +0000 schrieb WM:But infinitely many numbers are included in all infinite endsegments.
Le 20/07/2024 à 15:28, joes a écrit :No number n is included in all segments, in particular no E(m) for m>n.
Am Sat, 20 Jul 2024 12:19:26 +0000 schrieb WM:But infinitely many of them are the same natural numbers for all
But infinitely many are in all those endsegments which containThey are not the same ones.
infinitely many natnumbers.
infinite endsegments, because they stem from E(1) and have not been
lost as long as the endsegments are infinite.
All endsegments are claimed to be infinite. Contradiction.
Am Sun, 21 Jul 2024 14:15:45 +0000 schrieb WM:
Le 20/07/2024 à 22:56, joes a écrit :I don't see it. What does your first sentence mean?
Am Sat, 20 Jul 2024 13:48:44 +0000 schrieb WM:But infinitely many numbers are included in all infinite endsegments.
Le 20/07/2024 à 15:28, joes a écrit :No number n is included in all segments, in particular no E(m) for m>n.
Am Sat, 20 Jul 2024 12:19:26 +0000 schrieb WM:But infinitely many of them are the same natural numbers for all
But infinitely many are in all those endsegments which containThey are not the same ones.
infinitely many natnumbers.
infinite endsegments, because they stem from E(1) and have not been
lost as long as the endsegments are infinite.
All endsegments are claimed to be infinite. Contradiction.
No number is a part of all segments.
Am Sun, 21 Jul 2024 14:02:43 +0000 schrieb WM:
Le 20/07/2024 à 18:53, Moebius a écrit :
Am 20.07.2024 um 15:35 schrieb joes:
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM
Was würde das heißen, dass ein Stammbruch nicht "existiert"?Die Formel garantiert nicht die Existenz aller Stammbrüche.Doch das ist (im gegenwärtigen Kontext)It is not guaranteed that n+1 exists for every n. [WM]
garantiert. WENN dem NICHT so WÄRE, dann dürftest Du Deine Formel erst >>> gar nicht verwenden. Denn dann könnte es ein n_0 e IN geben, so dass
"n_0 + 1" nicht definiert ist; aus Deiner Formel lässt sich aber durch
Spezialisierung ableiten:
1/n_0 - 1/(n_0 + 1) > 0 .
Le 20/07/2024 à 18:53, Moebius a écrit :Was würde das heißen, dass ein Stammbruch nicht "existiert"?
Am 20.07.2024 um 15:35 schrieb joes:
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM
Die Formel garantiert nicht die Existenz aller Stammbrüche.Doch das ist (im gegenwärtigen Kontext)It is not guaranteed that n+1 exists for every n. [WM]
garantiert. WENN dem NICHT so WÄRE, dann dürftest Du Deine Formel erst
gar nicht verwenden. Denn dann könnte es ein n_0 e IN geben, so dass
"n_0 + 1" nicht definiert ist; aus Deiner Formel lässt sich aber durch
Spezialisierung ableiten:
1/n_0 - 1/(n_0 + 1) > 0 .
Am Sun, 21 Jul 2024 14:15:45 +0000 schrieb WM:
Le 20/07/2024 à 22:56, joes a écrit :
No number n is included in all segments, in particular no[t in] E(m) for m > n. [joes]
But infinitely many numbers are included in all infinite endsegments. [...]
I don't see it. What does your first sentence mean?
No number is a part of all segments.
Am Sun, 21 Jul 2024 14:15:45 +0000 schrieb WM:
Le 20/07/2024 à 22:56, joes a écrit :
No number n is included in all segments, in particular no[t in] E(m) for m > n.
No number is a part of all segments.
Try this argument:
| For each end every natural number n: the endsegment {m e IN : m > n}
| is *infinite*, but n !e {m e IN : m > n}.
Hence for each and every natural number there is an *infinite*
endsegment it's not in.
Disproving WM's claim "infinitely many numbers are included in all
infinite endsegments" (since NOT EVEN ONE natural number is included in
all infinite endsegments).
on 7/21/2024, Moebius supposed :
Am 21.07.2024 um 16:48 schrieb joes:
Am Sun, 21 Jul 2024 14:15:45 +0000 schrieb WM:
Le 20/07/2024 à 22:56, joes a écrit :
No number n is included in all segments, in particular no[t in] E(m) for >>>>> m > n. [joes]
But infinitely many numbers are included in all infinite endsegments.
[...]
I don't see it. What does your first sentence mean?
It means that "infinitely many numbers are included in all *infinite*
endsegments".
Hint: IN WMs world there are finite endsegments.
No number is a part of all segments.
Right. That's because (due to WM) there are *finite* endsegments they are not
in.
Right, due to his not having an n+1 for each and every n guaranteed.
Kind of a wishy-washy logic.
Am 21.07.2024 um 16:48 schrieb joes:
Am Sun, 21 Jul 2024 14:15:45 +0000 schrieb WM:
Le 20/07/2024 à 22:56, joes a écrit :
No number n is included in all segments, in particular no[t in] E(m) for m > n.
[joes]
But infinitely many numbers are included in all infinite endsegments. [...]
I don't see it. What does your first sentence mean?
It means that "infinitely many numbers are included in all *infinite* endsegments".
Hence for each and every natural number there is an *infinite*
endsegment it's not in.
Disproving WM's claim "infinitely many numbers are included in all
infinite endsegments" (since NOT EVEN ONE natural number is included in
all infinite endsegments).
Am 21.07.2024 um 17:20 schrieb Moebius:
Hence for each and every natural number there is an *infinite*
endsegment it's not in.
Disproving WM's claim "infinitely many numbers are included in all
infinite endsegments" (since NOT EVEN ONE natural number is included
in all infinite endsegments).
But why does he still claim this nonsense?
Weil er nicht zwischen Ax Ey und Ey Ax unterscheiden kann.
Offenbar stimmt:
A IE e SET_OF_INFINITE_ENDSEGMENTS: E^oo n e IN: n e IE
also stimmt "in Mückenheims Welt" auch
E^oo n e IN: A IE e SET_OF_INFINITE_ENDSEGMENTS: n e IE .
Quantifier dyslexia.
Le 21/07/2024 à 17:20, Moebius a écrit :I have no idea how you got that from the above.
| For each end every natural number n: the endsegment {m e IN : m > n}Hereby you confess that infinitely many numbers cannot be used in
| is *infinite*, but n !e {m e IN : m > n}.
bijections.
Infinitely many, even.Hence for each and every natural number there is an *infinite*
endsegment it's not in.
But these infinite endsegments are much larger than the finite initial segments (1, 2, 3, ..., n). If they are in every endsegment, then theyWhat is in every segment?
yield an infinite intersection.
What? They are countable.Disproving WM's claim "infinitely many numbers are included in allHow can all endsegments remain infinite if all natnumbers leave?
infinite endsegments" (since NOT EVEN ONE natural number is included in
all infinite endsegments).
Le 21/07/2024 à 17:13, FromTheRafters a écrit :Only the finite intersections are not empty.
on 7/21/2024, Moebius supposed :
Am 21.07.2024 um 16:48 schrieb joes:
Am Sun, 21 Jul 2024 14:15:45 +0000 schrieb WM:It means that "infinitely many numbers are included in all *infinite*
Le 20/07/2024 à 22:56, joes a écrit :
I don't see it. What does your first sentence mean?No number n is included in all segments, in particular not inBut infinitely many numbers are included in all infinite
E(m) for m > n.
endsegments.
endsegments". Hint: IN WMs world there are finite endsegments.
All infinite endsegments E(n) and E(k) and E(m) and so on have anRight, due to his not having an n+1 for each and every n guaranteed.No number is a part of all segments.Right. That's because (due to WM) there are *finite* endsegments they
are not in.
Kind of a wishy-washy logic.
infinite intersection, whatever natnumbers n, k, j are. How can infinite endsegments have no natnumber in common?
Le 21/07/2024 à 16:44, joes a écrit :Nein, ich meine, wie kann man von etwas nicht Existentem sprechen?
Am Sun, 21 Jul 2024 14:02:43 +0000 schrieb WM:Das würde heißen, dass es einen letzten gibt, den man selbstverständlich nicht auffinden kann. Aber seine Existenz kann man beweisen unter den folgenden Prämissen:
Le 20/07/2024 à 18:53, Moebius a écrit :Was würde das heißen, dass ein Stammbruch nicht "existiert"?
Am 20.07.2024 um 15:35 schrieb joes:
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM
Die Formel garantiert nicht die Existenz aller Stammbrüche.Doch das ist (im gegenwärtigen Kontext)It is not guaranteed that n+1 exists for every n. [WM]
garantiert. WENN dem NICHT so WÄRE, dann dürftest Du Deine Formel
erst gar nicht verwenden. Denn dann könnte es ein n_0 e IN geben, so
dass "n_0 + 1" nicht definiert ist; aus Deiner Formel lässt sich aber >>>> durch Spezialisierung ableiten:
1/n_0 - 1/(n_0 + 1) > 0 .
1) Alle Stammbrüche existieren aktual (also nicht nur potentiell, woNur, wenn die Abstände konstant wären.
immer weitere erschaffen werden können).
2) Der Bereich, in dem Stammbrüche befinden können ist durch 0 und 1 begrenzt.
Daraus (lineares Problem) folgt, dass zu beiden Seiten ein erster oder mehrere erste existieren.
Le 21/07/2024 à 16:48, joes a écrit :But not the same ones.
Am Sun, 21 Jul 2024 14:15:45 +0000 schrieb WM:Every infinite endsegment has infinitely many elements from E(1) = ℕ.
Le 20/07/2024 à 22:56, joes a écrit :I don't see it. What does your first sentence mean?
Am Sat, 20 Jul 2024 13:48:44 +0000 schrieb WM:But infinitely many numbers are included in all infinite endsegments.
Le 20/07/2024 à 15:28, joes a écrit :No number n is included in all segments, in particular no E(m) for
Am Sat, 20 Jul 2024 12:19:26 +0000 schrieb WM:But infinitely many of them are the same natural numbers for all
But infinitely many are in all those endsegments which containThey are not the same ones.
infinitely many natnumbers.
infinite endsegments, because they stem from E(1) and have not been
lost as long as the endsegments are infinite.
n.
All endsegments are claimed to be infinite. Contradiction.
What do you mean "filling"?No number is a part of all segments.What numbers are filling the endsegments?
How can all endsegments remain infinite if all natnumbers leave?
What?
Am Sun, 21 Jul 2024 15:00:01 +0000 schrieb WM:
2) Der Bereich, in dem Stammbrüche befinden können ist durch 0 und 1
begrenzt.
Daraus (lineares Problem) folgt, dass [...] ein erster oder
mehrere erste existieren.
*) Wenn s ein Stammbruch ist, dann gibt es (per definitionem) eine natürliche Zahl, so dass s = 1/n ist. Sei n_0 eine natürliche Zahl mit s
= 1/n_0. Dann gibt es außerhalb der Mückenmatik (also z. B. im Kontext
der klassischen Mathematik) eine natürliche Zahl, die größer ist als
n_0, z. B. n_0 + 1 (->Peano und Def. der Addition sowie der Zahl 1).
Damit ist dann aber 1/(n_0 + 1) ein Stammbruch, der kleiner ist als s = 1/n_0. qed
On 7/20/2024 9:01 AM, WM wrote:
As long as endsegments have
["]lost["] only few numbers and
["]kept["] infinitely many,
they are infinite
and have an infinite intersection with
endsegments of this kind.
Le 20/07/2024 à 14:40, Jim Burns a écrit :
On 7/19/2024 10:07 AM, WM wrote:
Infinitely many numbers are
in all predecessors and
in all *infinite* successors.
Infinitely many of them are the same
because of inclusion monotony.
Infinite endsegments with empty intersection
would require an complete exchange of the numbers.
But that is excluded by inclusion montony.
You are wrong.
The natural numbers ℕⁿᵒᵗᐧᵂᴹ are
well.ordered with step.up and non.0.step.down.
If you say I am wrongᵂᴹ about that,
what you (WM) are calling "the natural numbers"
aren't the natural numbers.
What you call the natural numbers
does not exist
(as a set, but only as a collection).
Infinite endsegments without
infinite sets of natural numbers are impossible.
Exchanging natural numbers is impossible
for endsegments.
Infinitely many of them are the same
because of inclusion monotony.
Quantifier shift.
Call it as you like.
As long as endsegments have
lost only few numbers and
kept infinitely many,
they are infinite and have
an infinite intersection with
endsegments of this kind.
On 7/21/2024 7:13 AM, WM wrote:
If all are separated and if none is smaller than zero, then according
to logic there must be a first one or more than one first ones.
Huh? That "logic" makes NO sense whatsoever.
On 7/21/2024 7:13 AM, WM wrote:
If all are separated and if none is smaller than zero, then according
to logic there must be a first one or more than one first ones.
Huh? That "logic" makes NO sense whatsoever.
Le 20/07/2024 à 23:02, Moebius a écrit :
An e IN: E(n) is infinite.
In other words, for all X: X is an infinite endsegment iff X is an
endsegment.
Then infinitely many numbers from E(1) are contained in all endsegments.
Or what else makes them infinite?
Am 21.07.2024 um 16:17 schrieb WM:
Le 20/07/2024 à 23:02, Moebius a écrit :
An e IN: E(n) is infinite.
In other words, for all X: X is an infinite endsegment iff X is an
endsegment.
Then infinitely many numbers from E(1) are contained in all
endsegments. Or what else makes them infinite?
Mückenheim: Ax Ey <=/=> Ey Ax (in general).
Yes: "For all endsegments E there are infinitely many numbers n in E(1)
such that n e E"
No: "There are infinitely many numbers n in E(1) such that for all endsegments E: n e E"
On 7/21/2024 1:22 PM, Moebius wrote:
Am 21.07.2024 um 21:52 schrieb Chris M. Thomasson:
On 7/21/2024 7:13 AM, WM wrote:
If all are separated and if none is smaller than zero, then
according to logic there must be a first one or more than one first
ones.
Huh? That "logic" makes NO sense whatsoever.
We may consider a line segment AB and the midpoints p_1, p_2, p_3, ...
Where p_1 is the midpoint between A and B, p_2 is the midpoint between
A and p_1, p_3 is the midpoint between A and p_2, and so an (ad
infinitum).
All midpoints are separated and for none the distance between A and
it is zero, then according to Mückenheim-logic there must be a first
one or more than one first ones. (And since more than one first one is
not possible, there just is one first one.)
Actually, that's the reason why Achilles will never reach the
tortoise! So Mückenheim could finally solve this long-standing problem!
Here is a quick little example of AB:
https://i.ibb.co/M2f37Md/ct-pov.png
The line is red is AB
On 7/21/2024 3:57 PM, Chris M. Thomasson wrote:
Also, from the unit circle, all circles and ellipses can be created.
The ellipse aspect is that of a circle rotated in 3d is an ellipse
when viewed on the projected plane. Check this out in VR:
https://skfb.ly/6RozT
Think of a circle in 3d. Looking at it dead on as a circle. Then,
change the camera view. It becomes an ellipse.
A screenshot:
https://i.ibb.co/BcX7yQW/image.png
From any circle we can view it in a 3d realm such that it can create
any ellipse? Fair enough?
On 7/21/2024 4:33 PM, Moebius wrote:
Am 22.07.2024 um 01:31 schrieb Chris M. Thomasson:
From any circle we can view it in a 3d realm such that it can create
any ellipse? Fair enough?
Right. Depends on "the point of view". (lol)
Exactly. It's always a circle in the 3d realm, but from different points
of view, it can take the form of any ellipse... Fun to me! :^)
Think of a circle in 3d. Looking at it dead on as a circle. Then, change
the camera view. It becomes an ellipse.
Am 21.07.2024 um 17:20 schrieb Moebius:
Hence for each and every natural number there is an *infinite*
endsegment it's not in.
Disproving WM's claim "infinitely many numbers are included in all
infinite endsegments" (since NOT EVEN ONE natural number is included in
all infinite endsegments).
Offenbar stimmt:
A IE e SET_OF_INFINITE_ENDSEGMENTS: E^oo n e IN: n e IE
also stimmt "in Mückenheims Welt" auch
E^oo n e IN: A IE e SET_OF_INFINITE_ENDSEGMENTS: n e IE .
Quantifier dyslexia.
WM submitted this idea :
All infinite endsegments E(n) and E(k) and E(m) and so on have an infinite >> intersection, whatever natnumbers n, k, j are. How can infinite endsegments >> have no natnumber in common?
Infinity has some wierdness, you are still thinking finite where there
is a largest element every endsegment would have in common.
Am Sun, 21 Jul 2024 15:50:10 +0000 schrieb WM:
Le 21/07/2024 à 17:20, Moebius a écrit :I have no idea how you got that from the above.
| For each end every natural number n: the endsegment {m e IN : m > n}Hereby you confess that infinitely many numbers cannot be used in
| is *infinite*, but n !e {m e IN : m > n}.
bijections.
Infinitely many, even.Hence for each and every natural number there is an *infinite*
endsegment it's not in.
But these infinite endsegments are much larger than the finite initialWhat is in every segment?
segments (1, 2, 3, ..., n). If they are in every endsegment, then they
yield an infinite intersection.
Am Sun, 21 Jul 2024 15:02:30 +0000 schrieb WM:
Every infinite endsegment has infinitely many elements from E(1) = ℕ.But not the same ones.
What do you mean "filling"?No number is a part of all segments.What numbers are filling the endsegments?
Am Sun, 21 Jul 2024 15:00:01 +0000 schrieb WM:
Le 21/07/2024 à 16:44, joes a écrit :Nein, ich meine, wie kann man von etwas nicht Existentem sprechen?
Am Sun, 21 Jul 2024 14:02:43 +0000 schrieb WM:Das würde heißen, dass es einen letzten gibt, den man selbstverständlich >> nicht auffinden kann. Aber seine Existenz kann man beweisen unter den
Le 20/07/2024 à 18:53, Moebius a écrit :Was würde das heißen, dass ein Stammbruch nicht "existiert"?
Am 20.07.2024 um 15:35 schrieb joes:
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM
Die Formel garantiert nicht die Existenz aller Stammbrüche.Doch das ist (im gegenwärtigen Kontext)It is not guaranteed that n+1 exists for every n. [WM]
garantiert. WENN dem NICHT so WÄRE, dann dürftest Du Deine Formel
erst gar nicht verwenden. Denn dann könnte es ein n_0 e IN geben, so >>>>> dass "n_0 + 1" nicht definiert ist; aus Deiner Formel lässt sich aber >>>>> durch Spezialisierung ableiten:
1/n_0 - 1/(n_0 + 1) > 0 .
folgenden Prämissen:
1) Alle Stammbrüche existieren aktual (also nicht nur potentiell, woNur, wenn die Abstände konstant wären.
immer weitere erschaffen werden können).
2) Der Bereich, in dem Stammbrüche befinden können ist durch 0 und 1
begrenzt.
Daraus (lineares Problem) folgt, dass zu beiden Seiten ein erster oder
mehrere erste existieren.
Le 21/07/2024 à 20:24, joes a écrit :Every number is "lost", like Moebius said at the top.
Am Sun, 21 Jul 2024 15:50:10 +0000 schrieb WM:If all endsegments keep infinitely man natnumbers then these natnumbers
Le 21/07/2024 à 17:20, Moebius a écrit :I have no idea how you got that from the above.
| For each end every natural number n:Hereby you confess that infinitely many numbers cannot be used in
| the endsegment {m e IN : m > n}
| is *infinite*, but n !e {m e IN : m > n}.
bijections.
can never be lost and never be used anyway.
They are not the fucking same.Infinitely many, even.Hence for each and every natural number there is an *infinite*
endsegment it's not in.
But these infinite endsegments are much larger than the finite initial
segments (1, 2, 3, ..., n). If they are in every endsegment, then they
yield an infinite intersection.
Nevermind.What is in every segment?Nothing. But in every infinite endsegment there are infinitely many natnumbers.
Simple logic derived from inclusion monotony shows endsegments lose natnumbers one by one. As long as infinitely many natnumbers remain,I don't know what you wanted to say here.
they have not been lost. This is not contradicted by the fact that
quantifier cannot always exchanged.
Le 21/07/2024 à 19:19, Moebius a écrit :For every number there are even infinitely many segments that don't
Am 21.07.2024 um 17:20 schrieb Moebius:
Hence for each and every natural number there is an *infinite*
endsegment it's not in.
What is in the infinite endsegment? Infinitely many natnumbers not yet
lost.
The sentence starting with E^oo is false.Disproving WM's claim "infinitely many numbers are included in allOffenbar stimmt:
infinite endsegments" (since NOT EVEN ONE natural number is included
in all infinite endsegments).
A IE e SET_OF_INFINITE_ENDSEGMENTS: E^oo n e IN: n e IE
also stimmt "in Mückenheims Welt" auch
E^oo n e IN: A IE e SET_OF_INFINITE_ENDSEGMENTS: n e IE .
Quantifier dyslexia.
No, simple logic derived from inclusion monotony. Endsegments lose
natnumbers one by one. As long as infinitely many remain, they have not
been lost. This is not contradicted by the fact that quantifier cannot
always exchanged.
Le 21/07/2024 à 20:29, joes a écrit :Every SINGLE segment has infinitely many in common with N.
Am Sun, 21 Jul 2024 15:02:30 +0000 schrieb WM:
By inclusion monotony infinitely many are the same.Every infinite endsegment has infinitely many elements from E(1) = ℕ.But not the same ones.
[which = the numbers?]What numbers are the elements of infinite endsegments which are not inWhat do you mean "filling"?No number is a part of all segments.What numbers are filling the endsegments?
all infinite endsegments?
Note that infinite endsegments can have lost only finitely manyThere are infinitely many of them.
natnumbers.
Am Mon, 22 Jul 2024 10:03:59 +0000 schrieb WM:
Le 21/07/2024 à 19:19, Moebius a écrit :For every number there are even infinitely many segments that don't
Am 21.07.2024 um 17:20 schrieb Moebius:
Hence for each and every natural number there is an *infinite*
endsegment it's not in.
contain it (=numbers that are larger).
What is this "inclusion monotony"?
Am Mon, 22 Jul 2024 10:13:06 +0000 schrieb WM:
If all endsegments keep infinitely man natnumbers then these natnumbersEvery number is "lost", like Moebius said at the top.
can never be lost and never be used anyway.
They are not the fucking same.Infinitely many, even.Hence for each and every natural number there is an *infinite*
endsegment it's not in.
But these infinite endsegments are much larger than the finite initial >>>> segments (1, 2, 3, ..., n). If they are in every endsegment, then they >>>> yield an infinite intersection.
Am Mon, 22 Jul 2024 10:32:47 +0000 schrieb WM:
Every SINGLE segment has infinitely many in common with N.By inclusion monotony infinitely many are the same.Every infinite endsegment has infinitely many elements from E(1) = ℕ. >>> But not the same ones.
Every pair of segments has an infinite intersection.
There are infinitely many of them.
On 7/20/2024 9:01 AM, WM wrote:
Infinite endsegments without
infinite sets of natural numbers are impossible.
The intersection of the SET.of.ALL.end.segments of
well.ordered with step.up and non.min.step.down
is empty.
Exchanging natural numbers is impossible
for endsegments.
⎛ The intersection of a set.of.sets holds
⎜ exactly those elements which are common
⎜ to all those sets.
Infinitely many of them are the same
because of inclusion monotony.
Quantifier shift.
Call it as you like.
Zero of them are the same.
Le 22/07/2024 à 13:34, joes a écrit :But the intersection of all infinitely many segments is empty.
Am Mon, 22 Jul 2024 10:32:47 +0000 schrieb WM:
Every finite set of endsegments has an infinite intersection.Every SINGLE segment has infinitely many in common with N.By inclusion monotony infinitely many are the same.Every infinite endsegment has infinitely many elements from E(1) =But not the same ones.
ℕ.
Every pair of segments has an infinite intersection.
The fuck? All segments are "missing" only finitely many numbers.There are infinitely many of them.There are only finitely many infinite endsegments. Infinitely many endsegments E(n) have lost infinitely many numers n. They can contain
only finitely many numbers.
Le 22/07/2024 à 13:24, joes a écrit :All segments are infinite.
Am Mon, 22 Jul 2024 10:03:59 +0000 schrieb WM:But not infinite endsegments. Endsegments that contain infinitely many numbers have lost olny finitely many and belong to a finite set.
Le 21/07/2024 à 19:19, Moebius a écrit :For every number there are even infinitely many segments that don't
Am 21.07.2024 um 17:20 schrieb Moebius:
Hence for each and every natural number there is an *infinite*
endsegment it's not in.
contain it (=numbers that are larger).
Intersection with what?What is this "inclusion monotony"?E(2) c E(1), E(3) c E(2), and so on. That proves that all infinite endsegments are subsets of their predecessors and therefore have an
infinite intersection with them. All endsegments are subsets of their predecessors. All non-empty endsegments have a non-empty intersection.
Am Mon, 22 Jul 2024 13:18:42 +0000 schrieb WM:
Le 22/07/2024 à 13:34, joes a écrit :But the intersection of all infinitely many segments is empty.
Am Mon, 22 Jul 2024 10:32:47 +0000 schrieb WM:Every finite set of endsegments has an infinite intersection.
Every SINGLE segment has infinitely many in common with N.By inclusion monotony infinitely many are the same.Every infinite endsegment has infinitely many elements from E(1) = ℕ. [WM]But not the same ones.[joes]
Every pair of segments has an infinite intersection.
There are infinitely many of them. [joes]
There are only finitely many infinite endsegments. Infinitely many
endsegments E(n) have lost infinitely many numers n. They can contain
only finitely many numbers.
The fuck? All segments are "missing" only finitely many numbers.
There is no finite end segment, precisely because every natural number
in E(1) has a corresponding segment.
Am 22.07.2024 um 15:30 schrieb joes:
Am Mon, 22 Jul 2024 13:18:42 +0000 schrieb WM:
There are only finitely many infinite endsegments. Infinitely many
endsegments E(n) have lost infinitely many numers n. They can contain
only finitely many numbers.
I've already told you that, man: "Hint: IN WM's world there are finite endsegments."
On 7/22/2024 6:25 AM, WM wrote:
Le 21/07/2024 à 21:20, Jim Burns a écrit :[...]
On 7/20/2024 9:01 AM, WM wrote:
Infinite endsegments without
infinite sets of natural numbers are impossible.
The intersection of the SET.of.ALL.end.segments of
well.ordered with step.up and non.min.step.down
is empty.
Infinitely many endsegments must have lost infinitely many numbers.
There cannot remain infinitely many numbers in all of them.
Lost? Where did they go?
Le 21/07/2024 à 21:20, Jim Burns a écrit :
On 7/20/2024 9:01 AM, WM wrote:
Infinite endsegments without
infinite sets of natural numbers are impossible.
The intersection of the SET.of.ALL.end.segments of
well.ordered with step.up and non.min.step.down
is empty.
Infinitely many endsegments must have lost infinitely many numbers.
There cannot remain infinitely many numbers in all of them.
Exchanging natural numbers is impossible
for endsegments.
⎛ The intersection of a set.of.sets holds
⎜ exactly those elements which are common
⎜ to all those sets.
Inclusion monotony proves that all infinite endsegments have not yet
lost infinitely many numbers but have them in common.
Infinitely many of them are the same
because of inclusion monotony.
Quantifier shift.
Proved correct here by inclusion monotony.
Call it as you like.
Zero of them are the same.
They only lose. They do not acquire .
Inclusion momotony.
Le 21/07/2024 à 21:20, Jim Burns a écrit :
On 7/20/2024 9:01 AM, WM wrote:
Infinite endsegments without
infinite sets of natural numbers are impossible.
The intersection of the SET.of.ALL.end.segments of
well.ordered with step.up and non.min.step.down
is empty.
Infinitely many endsegments must have lost
infinitely many numbers.
There cannot remain
infinitely many numbers in all of them.
Momotony?
On 7/22/2024 9:25 AM, WM wrote:
Infinitely many endsegments must have lost
infinitely many numbers.
Each end.segment has lost only finitely.many numbers.
No end.segment has lost infinitely.many numbers.
There cannot remain
infinitely many numbers in all of them.
Each end.segment of ℕⁿᵒᵗᐧᵂᴹ is infinite.
Le 23/07/2024 à 02:32, Jim Burns a écrit :
On 7/22/2024 9:25 AM, WM wrote:
Infinitely many endsegments must have lost
infinitely many numbers.
Each end.segment has lost only finitely.many numbers.
No end.segment has lost infinitely.many numbers.
Infinitely many numbers cannot be lost individually
because they cannot be named as individuals.
The intersection over all non-empty terms is non-empty.
On 7/23/2024 1:01 PM, WM wrote:
The intersection over all non-empty terms is non-empty.
The intersection over all nonempty terms is
the set of elements which
each nonempty term holds in common.
In the natural numbers
(well.ordered with step.up and non.min.step.down)
there are no elements which
each nonempty term holds in common.
The intersection over all nonempty terms is empty.
Le 23/07/2024 à 20:17, Jim Burns a écrit :"Each"
On 7/23/2024 1:01 PM, WM wrote:
Of two endsegments A and B either A c B or B c A.The intersection over all non-empty terms is non-empty.The intersection over all nonempty terms is the set of elements which
each nonempty term holds in common.
In the natural numbers (well.ordered with step.up andName two non-empty endsegments without common numbers.
non.min.step.down) there are no elements which each nonempty term holds
in common.
What does this mean?The intersection over all nonempty terms is empty.The intersection over finitely many endsgements is infinite.
Infinited segments have reserved infinitely many numbers for their
contents.
Only finitely many are available as indices.And this?
1 The first segment is infinite.There are only finitely many infinite endsements.
On 7/24/2024 4:04 PM, WM wrote:
There are only finitely many infinite endsements. [WM]
IN \ F).
Le 23/07/2024 à 20:17, Jim Burns a écrit :
The intersection over all nonempty terms is empty.
The intersection over finitely many endsgements
is infinite.
Infinited segments have reserved
infinitely many numbers for their contents.
Only finitely many are available as indices.
There are only finitely many infinite endsements.
In the natural numbers
(well.ordered with step.up and non.min.step.down)
there are no elements which
each nonempty term holds in common.
Name two non-empty endsegments without common numbers.
The intersection over all non-empty terms is non-empty.
The intersection over all nonempty terms is
the set of elements which
each nonempty term holds in common.
Of two endsegments A and B either A c B or B c A.
Am Wed, 24 Jul 2024 20:04:03 +0000 schrieb WM:
Name two non-empty endsegments without common numbers."Each"
What does this mean?The intersection over all nonempty terms is empty.The intersection over finitely many endsgements is infinite.
Infinited segments have reserved infinitely many numbers for their
contents.
Only finitely many are available as indices.And this?
1 The first segment is infinite.There are only finitely many infinite endsements.
Am 25.07.2024 um 03:25 schrieb Jim Burns:
On 7/24/2024 4:04 PM, WM wrote:
There are only finitely many infinite endsements. [WM]
No, there are infinitely many infinite endsements.
Proof: IN is infinite.
a simple bijection from IN onto the set of FISONs: n |-> {1, ..., n}).
On 7/24/2024 4:04 PM, WM wrote:
Name two non-empty endsegments without common numbers.
There are more than two end.segments.
In the natural numbers
each element is not held in common by
all (which are more than two) end.segments.
Of two endsegments A and B either A c B or B c A.
For any two end.segments A and B
the elements in A\B aren't in
the intersection of all end segments.
Le 25/07/2024 à 03:25, Jim Burns a écrit :
On 7/24/2024 4:04 PM, WM wrote:
Name two non-empty endsegments without common numbers.
There are more than two end.segments.
In the natural numbers
each element is not held in common by
all (which are more than two) end.segments.
If infinite endsegments have an empty intersection,
then no number must be in all endsegments,
but all must contain infinitely many numbers.
If infinite endsegments have an empty intersection,
then no number must be in all endsegments,
but all must contain infinitely many numbers.
That is possible only
when endegments do not only lose numbers
but alos acquire numbers.
That however is impossible.
Of two endsegments A and B either A c B or B c A.
For any two end.segments A and B
the elements in A\B aren't in
the intersection of all end segments.
As long as numbers remain in endsegments,
the intersection is not empty.
Inclusion monotony.
On 7/25/2024 11:18 AM, WM wrote:
If infinite endsegments have an empty intersection,
then no number [is] in all endsegments,
but all [have] infinitely many numbers.
That is possible only
when endegments do not only lose numbers
but also acquire numbers.
No.
That however is impossible.
Am 25.07.2024 um 19:37 schrieb Jim Burns:
On 7/25/2024 11:18 AM, WM wrote:
If infinite endsegments have an empty intersection,
then no number [is] in all [infinite] endsegments,
but all [have] infinitely many numbers.
Right.
That is possible only
when endegments do not only lose numbers
but also acquire numbers.
Huh?! Nope.
No.
Right. :-)
That however is impossible.
Indeed!
It's just your [WM] claim that is wrong: "That is possible only when endegments do not only lose numbers but also acquire numbers."
Le 25/07/2024 à 03:25, Jim Burns a écrit :Correct.
On 7/24/2024 4:04 PM, WM wrote:
If infinite endsegments have an empty intersection, then no number mustName two non-empty endsegments without common numbers.There are more than two end.segments.
In the natural numbers each element is not held in common by all (which
are more than two) end.segments.
be in all endsegments, but all must contain infinitely many numbers.
That is possible only when endegments do not only lose numbers but alos acquire numbers. That however is impossible.That is not the case. You just don't understand that infinities don't
"As long as", but not for the infinite intersection.As long as numbers remain in endsegments, the intersection is not empty. Inclusion monotony.Of two endsegments A and B either A c B or B c A.For any two end.segments A and B the elements in A\B aren't in the
intersection of all end segments.
Le 25/07/2024 à 03:51, Moebius a écrit :They aren't. I haven't understood yet why you think that.
Am 25.07.2024 um 03:25 schrieb Jim Burns:Nonsense. The natural numbers cannot be split into two ℵo-sets.
On 7/24/2024 4:04 PM, WM wrote:No, there are infinitely many infinite endsements.
There are only finitely many infinite endsements. [WM]
Do you disagree with the bijection?Proof: IN is infinite.It is an ℵo-set where the first finitely many numbers can be split off
as indices of endsegments.
There are infinitely many FISONs (since there isNo, all FISONs have ℵo successors. Hence there can be only finittely
a simple bijection from IN onto the set of FISONs: n |-> {1, ..., n}).
many.
Le 24/07/2024 à 23:13, joes a écrit :No, you name a number that is common to EACH segment.
Am Wed, 24 Jul 2024 20:04:03 +0000 schrieb WM:
Name two.Name two non-empty endsegments without common numbers."Each"
What do you mean by "reserved"?Le 23/07/2024 à 20:17, Jim Burns a écrit :
The intersection over all nonempty terms is the set of elements
which each nonempty term holds in common.
In the natural numbers (well.ordered with step.up and
non.min.step.down) there are no elements which each nonempty term
holds in common.
The intersection over all nonempty terms is empty.The intersection over finitely many endsgements is infinite. Infinited
segments have reserved infinitely many numbers for their contents.
How so?Only finitely many are available as indices.
Every endsegment E(n) has an index n. As long as infinitely many numbers1 The first segment is infinite.There are only finitely many infinite endsements.
are in all infinite endsegments, only finitely many numbers have been
used as indices.
joes formulated on Friday :
Am Thu, 25 Jul 2024 15:08:39 +0000 schrieb WM:
Le 25/07/2024 à 03:51, Moebius a écrit :They aren't. I haven't understood yet why you think that.
Am 25.07.2024 um 03:25 schrieb Jim Burns:Nonsense. The natural numbers cannot be split into two ℵo-sets.
On 7/24/2024 4:04 PM, WM wrote:No, there are infinitely many infinite endsements.
There are only finitely many infinite endsements. [WM]
The infinite set of all (infinite) segments of course includes
all successors.
Do you disagree with the bijection?Proof: IN is infinite.It is an ℵo-set where the first finitely many numbers can be split off >>> as indices of endsegments.
There are infinitely many FISONs (since there isNo, all FISONs have ℵo successors. Hence there can be only finittely
a simple bijection from IN onto the set of FISONs: n |-> {1, ..., n}).
many.
Yes, he does. His idea is that bijection works for non-dark numbers
only. He thinks it is 'matching' elements from one set to another and
such 'matching' can't be done with elements which aren't 'distinguished'
or whatever crazy wishy-washy term he decides to use.
On 7/25/2024 11:18 AM, WM wrote:
If infinite endsegments have an empty intersection,
then no number must be in all endsegments,
but all must contain infinitely many numbers.
That is possible only
when endegments do not only lose numbers
but also acquire numbers.
No.
It is possible for the numbers to be
well.ordered with step.up and non.min.step.down
As long as numbers remain in endsegments,
the intersection is not empty.
As long as the intersection holds numbers,
NOT ALL end.segments are intersected.
The intersection of ALL end.segments
holds no numbers.
It's just your [WM] claim that is wrong: "That is possible only when endegments do not only lose numbers but also acquire numbers."
You just don't understand that infinities don't
shrink; they stay infinite when removing an arbitrary finite number.
As long as numbers remain in endsegments, the intersection is not empty."As long as", but not for the infinite intersection.
Inclusion monotony.
Le 26/07/2024 à 10:59, joes a écrit :Partly right: there are infinitely many "removals" individually,
You just don't understand that infinities don't shrink; they stayThey get empty by infinitely many removings. They happen in the infinite sequence of endsegments.
infinite when removing an arbitrary finite number.
Yes, in the sense that every number is eventually "missing".Then no numbers remain. If infinitely many endsegments lose infinitelyAs long as numbers remain in endsegments, the intersection is not"As long as", but not for the infinite intersection.
empty.
Inclusion monotony.
many numbers, then none remaons.
Le 25/07/2024 à 22:58, Moebius a écrit :They are all subsets. Every number is eventually "lost".
It's just your [WM] claim that is wrong: "That is possible only whenThe intersection of infinite endsegments can only be empty when all
endegments do not only lose numbers but also acquire numbers."
numbers differ. As long as endsegments only lose numbers that is
impossible.
Le 25/07/2024 à 19:37, Jim Burns a écrit :... in all of them, yes. Every number is only in a finite amount of sets.
On 7/25/2024 11:18 AM, WM wrote:
Infinite sets with empty intersection must contain elements which areIf infinite endsegments have an empty intersection,It is possible for the numbers to be well.ordered with step.up and
then no number must be in all endsegments,
but all must contain infinitely many numbers.
That is possible only when endegments do not only lose numbers but
also acquire numbers.
non.min.step.down
not identical.
Finally.That is very true.As long as numbers remain in endsegments,As long as the intersection holds numbers,
the intersection is not empty.
NOT ALL end.segments are intersected.
The intersection of ALL end.segments holds no numbers.That is very true.
But all endsegments have lost all numbers. Otherwise they are not all endsegments. And why should they stop to lose numbers?Because there are infinitely many of them. If you consider them ALL
Le 25/07/2024 à 19:37, Jim Burns a écrit :
On 7/25/2024 11:18 AM, WM wrote:
If infinite endsegments have an empty intersection,
then no number must be in all endsegments,
but all must contain infinitely many numbers.
That is possible only
when endegments do not only lose numbers
but also acquire numbers.
No.
You are wrong.
It is possible for the numbers to be
well.ordered with step.up and non.min.step.down
Infinite sets with empty intersection
must contain elements which are not identical.
As long as numbers remain in endsegments,
the intersection is not empty.
As long as the intersection holds numbers,
NOT ALL end.segments are intersected.
That is very true.
The intersection of ALL end.segments
holds no numbers.
That is very true.
But all endsegments have lost all numbers.
Otherwise they are not all endsegments.
And why should they stop to lose numbers?
On 7/26/2024 1:02 PM, WM wrote:
Le 25/07/2024 à 19:37, Jim Burns a écrit :
On 7/25/2024 11:18 AM, WM wrote:
If infinite endsegments have an empty intersection,
then no number must be in all endsegments,
but all must contain infinitely many numbers.
That is possible only
when endegments do not only lose numbers
but also acquire numbers.
No.
You are wrong.
It is possible for the numbers to be
well.ordered with step.up and non.min.step.down
Infinite sets with empty intersection
must contain elements which are not identical.
For each two end.segments E′ E″
either there is a number in E′\E″
or there is a number in E″\E′
That's always the case for sets which are different.
That's extensionality.
For end.segments E′ E″
either there is NO number in E′\E″
or there is NO number in E″\E′
That's inclusion.monotonicity.
Each number in the difference.set E′\E″
is not.in the intersection of ALL end.segments.
Each number is in the difference.set of some two end.segments.
Each number is not.in the intersection of ALL end.segments.
The intersection of ALL end.segments is empty.
As long as numbers remain in endsegments,
the intersection is not empty.
As long as the intersection holds numbers,
NOT ALL end.segments are intersected.
That is very true.
The intersection of ALL end.segments
holds no numbers.
That is very true.
But all endsegments have lost all numbers.
Otherwise they are not all endsegments.
Each end.segment has lost only finitely.many numbers.
Each number is not.yet.lost from only finitely.many end.segments.
⎛ 'Finitely.many' means that
⎜ each non.{}.subset of lost numbers is 2.ended.
⎝ Finite doesn't need to be small.
There are more.than.finitely.many end.segments.
The end.segments aren't 2.ended.
Each number is in fewer.than.all end.segments.
And why should they stop to lose numbers?
No end.segment is last to lose numbers.
The end.segments aren't 2.ended.
The numbers aren't 2.ended.
Each ordinal j -- followed by j∪{j} -- isn't the second end.
Am 26.07.2024 23:41:56 Jim Burns schrieb:
On 7/26/2024 1:02 PM, WM wrote:
Le 25/07/2024 à 19:37, Jim Burns a écrit :
On 7/25/2024 11:18 AM, WM wrote:
If infinite endsegments have an empty intersection,
then no number must be in all endsegments,
but all must contain infinitely many numbers.
That is possible only
when endegments do not only lose numbers
but also acquire numbers.
No.
You are wrong.
It is possible for the numbers to be
well.ordered with step.up and non.min.step.down
Infinite sets with empty intersection
must contain elements which are not identical.
For each two end.segments E′ E″
either there is a number in E′\E″
or there is a number in E″\E′
That's always the case for sets which are different.
That's extensionality.
For end.segments E′ E″
either there is NO number in E′\E″
or there is NO number in E″\E′
That's inclusion.monotonicity.
Each number in the difference.set E′\E″
is not.in the intersection of ALL end.segments.
Each number is in the difference.set of some two end.segments.
Each number is not.in the intersection of ALL end.segments.
The intersection of ALL end.segments is empty.
As long as numbers remain in endsegments,
the intersection is not empty.
As long as the intersection holds numbers,
NOT ALL end.segments are intersected.
That is very true.
The intersection of ALL end.segments
holds no numbers.
That is very true.
But all endsegments have lost all numbers.
Otherwise they are not all endsegments.
Each end.segment has lost only finitely.many numbers.
Each number is not.yet.lost from only finitely.many end.segments.
⎛ 'Finitely.many' means that
⎜ each non.{}.subset of lost numbers is 2.ended.
⎝ Finite doesn't need to be small.
There are more.than.finitely.many end.segments.
The end.segments aren't 2.ended.
Each number is in fewer.than.all end.segments.
And why should they stop to lose numbers?
No end.segment is last to lose numbers.
The end.segments aren't 2.ended.
The numbers aren't 2.ended.
Each ordinal j -- followed by j∪{j} -- isn't the second end.
Very, very well put!
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