1. Scrooge McDuck's bankrupt

Scrooge Mc Duck earns 1000 $ daily and spends only 1 $ per day. As a cartoon-figure he will live forever and his wealth will increase without
bound. But according to set theory he will get bankrupt if he spends the dollars in the same order as he receives them. Only if he always spends
them in another order, for instance every day the second dollar
received, he will get rich. These different results prove set theory to
be useless for all practical purposes.

The above story is only the story of Tristram Shandy in simplified
terms, which has been narrated by Fraenkel, one of the fathers of ZF set theory.

"Well known is the story of Tristram Shandy who undertakes to write his biography, in fact so pedantically, that the description of each day
takes him a full year. Of course he will never get ready if continuing
that way. But if he lived infinitely long (for instance a 'countable
infinity' of years [...]), then his biography would get 'ready',
because, expressed more precisely, every day of his life, how late ever, finally would get its description because the year scheduled for this
work would some time appear in his life." [A. Fraenkel: "Einleitung in
die Mengenlehre", 3rd ed., Springer, Berlin (1928) p. 24] "If he is
mortal he can never terminate; but did he live forever then no part of
his biography would remain unwritten, for to each day of his life a year devoted to that day's description would correspond." [A.A. Fraenkel, A.
Levy: "Abstract set theory", 4th ed., North Holland, Amsterdam (1976) p.
30]

2. Failed enumeration of the fractions

All natural numbers are said to be enough to index all positive
fractions. This can be disproved when the natural numbers are taken from
the first column of the matrix of all positive fractions

1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
.. .

To cover the whole matrix by the integer fractions amounts to the idea
that the letters X in

XOOO...
XOOO...
XOOO...
XOOO...
..

can be redistributed to cover all positions by exchanging them with the
letters O. (X and O must be exchanged because where an index has left,
there is no index remaining.) But where should the O remain if not
within the matrix at positions not covered by X?

3. Violation of translation invariance

Translation invariance is fundamental to every scientific theory. With n
m ∈ ℕ and q ∈ {ℚ ∩ (0, 1]} there is precisely the same number of rational points n + q in (n, n+1] as of rational points m + q in (m,
m+1] . However, half of all positive rational numbers of Cantor's
enumeration
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
5/1, ...
are of the form 0 + q and lie in the first unit interval between 0 and
1. There are less rational points in (1, 2] but more than in (2, 3] and
so on.

4. Violation of inclusion monotony

Every endsegment E(n) = {n, n+1, n+2, ...} of natural numbers has an
infinite intersection with all other infinite endsegments.
∀k ∈ ℕ_def: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀ .
Set theory however comes to the conclusion that there are only infinite endsegments and that their intersection is empty. This violates the
inclusion monotony of the endegments according to which, as long as only non-empty endsegments are concerned, their intersection is non-empty.

5. Actual infinity implies a smallest unit fraction

All unit fractions 1/n have finite distances from each other
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
Therefore the function Number of Unit Fractions between 0 and x, NUF(x),
cannot be infinite for all x > 0. The claim of set theory
∀x ∈ (0, 1]: NUF(x) = ℵo
is wrong. If every positive point has ℵo unit fractions at its left-hand side, then there is no positive point with less than ℵo unit fractions
at its left-hand side, then all positive points have ℵo unit fractions
at their left-hand side, then the interval (0, 1] has ℵo unit fractions
at its left-hand side, then ℵo unit fractions are negative.
Contradiction.

6. There are more path than nodes in the infinite Binary Tree

Since each of n paths in the complete infinite Binary Tree contains at
least one node differing from all other paths, there are not less nodes
than paths possible. Everything else would amount to having more houses
than bricks.

7. The diagonal does not define a number

An endless digit sequence without finite definition of the digits cannot
define a real number. After every known digit almost all digits will
follow.

Regards, WM

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On 1/4/24 10:59, WM wrote:

[snip]

what point are you trying to make? infinity is strange

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immibis schrieb am Donnerstag, 4. Januar 2024 um 22:53:35 UTC+1:

what point are you trying to make? infinity is strange

But it is based on logic. This logic is violated in the seven points I
made.

Regards, WM

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On 1/5/24 5:22 AM, WM wrote:

immibis schrieb am Donnerstag, 4. Januar 2024 um 22:53:35 UTC+1:

what point are you trying to make? infinity is strange

But it is based on logic. This logic is violated in the seven points I
made.

Regards, WM

Which are base on logic that doesn't handle infinity.

Yes, it is well know that infinite sets break some of the seemingly
obvious properties that hold for finite sets.

YOU just don't seem to understand and accept that fact, and keep on
making the ERROR of asuming it must.

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On 1/4/2024 4:59 AM, WM wrote:

7. The diagonal does not define a number

An endless digit sequence without
finite definition of the digits
cannot define a real number.
After every known digit
almost all digits will follow.

After every known and unknown digit,
almost all digits follow.

For each digit in the sequence,
the cardinal of digits.before is ⁺¹.able,
the cardinal of digits.after is
larger than each ⁺¹.able cardinal,
and thus is not.⁺¹.able.


It is sufficient that
an endless digit sequence without
finite definition of the digits
exist.

There is at most one real number
which is permitted by each
finite initial sub.sequence of digits.
That one real number is the one which
the endless digit sequence represents.

| Assume otherwise.
| Assume two points at a distance d > 0
| are permitted by each
| finite initial sub.sequence of digits.
|
| However,
| there is an initial n.digit sub.sequence which
| permits only points apart by 10-n < d
| Those two points cannot both be permitted
| by that n.digit sub.sequence or
| by each finite initial sub.sequence.
| Contradiction.

Therefore,
there is at most one real number
which is permitted by each
finite initial sub.sequence of digits.

Also,
there is at least one real number
which is permitted by each
finite initial sub.sequence of digits.

That follows from the requirement that
functions which jump
are discontinuous at some point.

For each endless digit sequence,
each digit preceded by ⁺¹.ably.many digits,
there is no more and no less than
one point.

For each point,
there is no less than one and
no more than one
(no more than two for trailing 0's and 9's)
endless digit sequence,
each digit preceded by ⁺¹.ably.many digits,

The representation of real points by digits
is obviously cousin to
the representation of rational points by digits.

However, the cousins are not the same.
The points _exist_
The digit.sequences _exist_
We know that by augmenting descriptive claims
with only not.first.false claims.
That isn't a calculation in the sense that
a rational point is calculated.

Nevertheless, we know they exist, uncalculated,
because we know that
a finite sequence of _claims_
if it has no first.false _claim_
has no false _claim_

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On 1/5/2024 5:58 PM, Ross Finlayson wrote:

On Friday, January 5, 2024
at 10:32:19 AM UTC-8, Jim Burns wrote:

On 1/4/2024 4:59 AM, WM wrote:

An endless digit sequence without
finite definition of the digits
cannot define a real number.
After every known digit
almost all digits will follow.

The representation of real points by digits
is obviously cousin to
the representation of rational points by digits.

However, the cousins are not the same.
The points _exist_
The digit.sequences _exist_
We know that by augmenting descriptive claims
with only not.first.false claims.
That isn't a calculation in the sense that
a rational point is calculated.

Nevertheless, we know they exist, uncalculated,
because we know that
a finite sequence of _claims_
if it has no first.false _claim_
has no false _claim_

If it's sufficient
to establish a model of arithmetic
then by the GIT's
you'll agree it's at best incomplete.

I'm not sure we're on the same page
with regard to incompleteness.

Incomplete == some things we can't know.
Were you (RF) seriously contemplating
not (some things we can't know)?

Gödel didn't surprise with the result itself.
Yes, some things we can't know. Whatever.
Gödel surprised with _proving_ it.
| On Formally Undecidable Propositions of
| Principia Mathematica and Related Systems
|
_Formally_ undecidable.

----
If a theory has a model,
then it's consistent.

If a theory can express
recursive definitions
non.recursively
(re.stated without the defined term)
(which is something arithmetic can do),
then
the theory can _quote_
recursive definitions of
what it is to be a formula and
what it is to be a proof.

By "quote", I mean: represent
an object of language (formula, proof) by
an object of study (number, set)
(Gödel.numbers et al)

If a theory can quote
what it is to be a formula and
what it is to be a proof,
then the theory can express G(x) such that
G("H(x)") ⟹ proof of H("H(x)") not.exists.
¬G("H(x)") ⟹ proof of contradiction exists.

If a theory can express G("H(x)")
then the theory can express G("G(x)")
G("G(x)") ⟹ proof of G("G(x)") not.exists.
¬G("G(x)") ⟹ proof of contradiction exists.

If a theory has a certain
very.attainable level of expressiveness,
then choose one: incomplete or inconsistent.

But there is a model. It's consistent.
So, it's incomplete.

... That it's false to say
it's, the "true", claim.
Only scientific and not falsified.

Platonism then sort of demands
"there are true numbers, so work it up".

"There is no 'but', only 'yet', ...."

We know that
each theory is true of
whatever that theory is true of.

Some theories are true of nothing.
Contradictions can be proved in those.

Some theories are true of something,
of what we intend or of something unintended.
Contradictions cannot be proved in those.

Some theories, numbers, for example,
if they can prove they're consistent,
would prove that they _aren't_ consistent.

There is no "yet" to not.proving consistency.
We know there is no such proof --
or, if (sadly) there is, it's meaningless,
about nothing.

However,
there are better options than
a theory proving itself consistent.

Gödel's work does not deny that
some other theory can prove
our theory consistent.

For example,
ZFC is a theorem of
ZFC+inaccessible.cardinal.exists

Some theories are so simple that,
even though we know there is no proof,
we don't take seriously the possibility
that they have no model.

For example,
the set.theory fragment,
-- {} exists
-- x∪{y} exists
-- same.element.sets x and y are equal
proves arithmetic and Gödel.incompleteness

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Le 05/01/2024 à 16:04, Richard Damon a écrit :

On 1/5/24 5:22 AM, WM wrote:

immibis schrieb am Donnerstag, 4. Januar 2024 um 22:53:35 UTC+1:

what point are you trying to make? infinity is strange

But it is based on logic. This logic is violated in the seven points I
made.

Which are base on logic that doesn't handle infinity.

This logic is indispensable.

Yes, it is well know that infinite sets break some of the seemingly
obvious properties that hold for finite sets.

But it has not yet been recognized that ZF breaks indispensable laws of
logic.

YOU just don't seem to understand and accept that fact, and keep on
making the ERROR of asuming it must.

That is not an error. It shows that ZF could never acquire any relevance
for reality where the basic laws of logic are referenced. It shows that ZF
is only a religion to be believed by its proponents and poor captured
students.

Regards, WM

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Le 05/01/2024 à 19:32, Jim Burns a écrit :

On 1/4/2024 4:59 AM, WM wrote:

7. The diagonal does not define a number

An endless digit sequence without
finite definition of the digits
cannot define a real number.
After every known digit
almost all digits will follow.

It is sufficient that
an endless digit sequence without
finite definition of the digits
exist.

No. Even and endless digit sequence does not describe an irrational
number. The irrational number is the limit only. Compare 0.999... which
does not contain 1 but only has the limit 1.

There is at most one real number
which is permitted by each
finite initial sub.sequence of digits.

There are infinitely many. Therefore it is impossible to decide which real number is described.

Therefore,
there is at most one real number
which is permitted by each
finite initial sub.sequence of digits.

By each defined digit sequence infinitely many numbers are permitted.
And the dark digits cannot describe any real number.

Regards, WM

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On 1/8/24 5:40 AM, WM wrote:

Le 05/01/2024 à 16:04, Richard Damon a écrit :

On 1/5/24 5:22 AM, WM wrote:

immibis schrieb am Donnerstag, 4. Januar 2024 um 22:53:35 UTC+1:

what point are you trying to make? infinity is strange

But it is based on logic. This logic is violated in the seven points
I made.

Which are base on logic that doesn't handle infinity.

This logic is indispensable.

Maybe for your small mind.

Yes, it is well know that infinite sets break some of the seemingly
obvious properties that hold for finite sets.

But it has not yet been recognized that ZF breaks indispensable laws of logic.

No, it shows that some supposed laws of logic can't handle unbounded sets.

YOU just don't seem to understand and accept that fact, and keep on
making the ERROR of asuming it must.

That is not an error. It shows that ZF could never acquire any relevance
for reality where the basic laws of logic are referenced. It shows that
ZF is only a religion to be believed by its proponents and poor captured students.

Regards, WM

In other words, you don't understand what it says so you ignore it.

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On 1/8/2024 7:18 AM, WM wrote:

Le 05/01/2024 à 19:32, Jim Burns a écrit :

An endless digit sequence without
finite definition of the digits
cannot define a real number.
After every known digit
almost all digits will follow.

There is at most one real number
which is permitted by each
finite initial sub.sequence of digits.

There are infinitely many.

Assume there are two permitted points.
Each power 10⁻ⁿ is larger than their distance,
a positive distance d > 0 apart.

β ≥ d > 0 is the least upper bound of
distances which each 10⁻ⁿ is larger than.

10β > β is not
a distance which each 10⁻ⁿ is larger than.
Thus b exists: 10⁻ᵇ < 10β

β/10 < β is
a distance which each 10⁻ⁿ is larger than.
Thus, in particular,
β/10 < 10⁻ᵇ⁻²

However,
(10⁻ᵇ)/100 < (10β)/100
10⁻ᵇ⁻² < β/10
Contradiction.

Therefore,
there _aren't_ two points which
are permitted by each
finite initial sub.sequence of
the endless digit sequence.

Therefore it is impossible to decide which
real number is described.

Whether or not we describe these points,
these points _exist_
no more than one point to each
endless digit sequence.

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Le 08/01/2024 à 19:46, Jim Burns a écrit :

On 1/8/2024 7:18 AM, WM wrote:

Therefore it is impossible to decide which
real number is described.

Whether or not we describe these points,
these points _exist_
no more than one point to each
endless digit sequence.

Dark numbers, dark points and dark parts of infinite sequences exist. But
we cannot distinguish and use them. And we can prove that there are not
more irrational numbers than rational numbers. Just today I showed some
proofs to my students. One is this: Between ***every*** pair of irrational numbers there is a rational number. Another one, which was very well
received, is the game Conquer the Binay Tree:

You start with one cent, buy a path in the Binary Tree and get one cent
for every covered node. Then you buy another path and get one cent for
every node not yet covered by the first path. You will never earn less
than one cent, because every path is distinct by at least one node from
every other path. Therefore you will not get bankrupt. But if there were
more paths than nodes, you would get bankrupt. Hence Cantor is defeated.

Regards, WM

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On 1/9/2024 12:40 PM, WM wrote:

Le 08/01/2024 à 19:46, Jim Burns a écrit :

On 1/8/2024 7:18 AM, WM wrote:

An endless digit sequence without
finite definition of the digits
cannot define a real number.
After every known digit
almost all digits will follow.

There is at most one real number
which is permitted by each
finite initial sub.sequence of digits.

There are infinitely many.

Therefore it is impossible to decide
which real number is described.

Whether or not we describe these points,
these points _exist_
no more than one point to each
endless digit sequence.

Dark numbers, dark points
and dark parts of infinite sequences
exist.

Elsewhere, you have told us that
darkᵂᴹ numbers and their ilk
are never one step away from
visibleᵂᴹ numbers and their ilk.

I'll clarify.
There is at most one point
which is permitted by each
finite initial sub.sequence of
visibleᵂᴹ digits.

Because,
if there are two permitted points,
there is a positive least upper bound β of
distances < each visibleᵂᴹ 10⁻ⁿ
β/10 < β < 10β
and visibleᵂᴹ 10⁻ᵇ < 10β
and visibleᵂᴹ 10⁻ᵇ⁻² > β/10
but also
10⁻ᵇ/100 < 10β/100
10⁻ᵇ⁻² < β/10
Contradiction.
Thus,
not two.

Just today I showed some proofs to my students.

If you are claiming your students for
some form of peer.review, then
you are accepting pre-calculus students as
_peers_

I won't object if you choose to do that.
I just want to make sure you understand that
they are who you are accepting as peers.

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On 1/9/2024 6:29 PM, Ross Finlayson wrote:

On Tuesday, January 9, 2024
at 11:34:19 AM UTC-8, Jim Burns wrote:

On 1/9/2024 12:40 PM, WM wrote:

Just today I showed some proofs to my students.

If you are claiming your students for
some form of peer.review, then
you are accepting pre-calculus students as
_peers_

Everybody knows
the entire point of pre-calculus is
to make it clear that the usual notion of
an Aristotelean continuum, [...]

I could have said that better.

In that post,
the only thing I meant by "pre-calculus"
was "having not yet been a calculus student".
I wasn't thinking of "being a student in
the course just before calculus".

Calculus and WM's dark.number game
make a poor fit together.
I expect WM to agree with that much,
even if he and I assign blame for
the poor fit differently.

I haven't actually heard from anyone that
WM's students are not.yet.calculus.
It only seems to me as though they must be.

Sorry for any confusion.


One sentence:

Everybody knows
the entire point of pre-calculus is
to make it clear that
the usual notion of
an Aristotelean continuum,
which is about
the most usual sort of intuition about
the continuum after constant motion,
and
the course-of-passage as through
points in a line,
is
gently shushing that down for
students who do have
such an intuitive notion of
analytical character,
and then
explaining for all
that the usual laws of
arithmetic and delta-epsilonics,
together,
make for defining infinite limit.

Gently shushing down notions which are
not the preferred notions
is the purpose for which we hold classes.

If it can be done,
I think that
giving the reasons that
some other notion is preferred
can be an excellent strategy for
effective gentle.shushing.down.

However,
in some instances,
we are, today, looking at preferences
resulting from resolution of a long controversy
between brilliant people
encyclopedically educated in their field.

Should we ask students,
upon their first contact with that field,
to reproduce
the best of that field?

If that is a reasonable ask,
that would be wonderful to see.

However,
it might well not.be a reasonable ask.

It might well be necessary for
students to _trust_ that
reasons exist for preferring
one set of notions over
another,
to trust, at least,
until they acquire their own
brilliance and
encyclopedic knowledge.

My own preference is
to share the beauty in
these deep notions,
with all and sundry,
with those with only
a momentary interest.

However,
I have to recognize that
I can't always get what I want.

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Le 09/01/2024 à 20:34, Jim Burns a écrit :

On 1/9/2024 12:40 PM, WM wrote:

Just today I showed some proofs to my students.

If you are claiming your students for
some form of peer.review, then
you are accepting pre-calculus students as
_peers_

Nonsense. There are 8th semester informatics and engineering studens. They
know mathematics very well (that which Cantor is not needed for, according
to Feferman) but they do not use matheology. Try to show how the player
gets bankrupt in the game "We conquer the Binary Tree".

Regards, WM

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Le 10/01/2024 à 06:49, Jim Burns a écrit :

Calculus and WM's dark.number game
make a poor fit together.

Of course. Calculus uses potential infinity only. Dark numbers can exist
only in actual infinity, the complement of potential infinity.
Feferman's book In the light of logic answers the question: Is Cantor
necessary for the maths of the real world? with a resounding no.

Regards, WM

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On 1/10/24 2:14 PM, WM wrote:

Le 10/01/2024 à 06:49, Jim Burns a écrit :

Calculus and WM's dark.number game
make a poor fit together.

Of course. Calculus uses potential infinity only. Dark numbers can exist
only in actual infinity, the complement of potential infinity.
Feferman's book In the light of logic answers the question: Is Cantor necessary for the maths of the real world? with a resounding no.

Regards, WM

But you claim them to be part of the sets that are only "Potential
Infinity", or are you retracting that claim. (All Natural Numbers are
only by your definition "Potentially Infinite")

All members of the Natural Numbers are Finite, so none of them are
"Actually Infinite".

The set of them has a SIZE that is infinite, but none of the members of
it ever are.

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On 1/10/24 8:10 PM, Ross Finlayson wrote:

On Wednesday, January 10, 2024 at 4:49:39 PM UTC-8, Richard Damon wrote:

On 1/10/24 2:14 PM, WM wrote:

Le 10/01/2024 à 06:49, Jim Burns a écrit :

Calculus and WM's dark.number game
make a poor fit together.

Of course. Calculus uses potential infinity only. Dark numbers can exist >>> only in actual infinity, the complement of potential infinity.
Feferman's book In the light of logic answers the question: Is Cantor
necessary for the maths of the real world? with a resounding no.

Regards, WM

But you claim them to be part of the sets that are only "Potential
Infinity", or are you retracting that claim. (All Natural Numbers are
only by your definition "Potentially Infinite")

All members of the Natural Numbers are Finite, so none of them are
"Actually Infinite".

The set of them has a SIZE that is infinite, but none of the members of
it ever are.

The order type of ordinals is an ordinal.

Finite ordinals as the classes of each those not containing themselves, according to simple quantification or "the set of all sets that don't
contain themselves", contains itself.

Such are simple reasons why the infinite integers have an infinite integer.

Yes I know that ZF two axioms of restriction of comprehension or "don't look" include one that is "there's an infinity that isn't thusly various".

So, you might find it interesting that in more naive theories,
it's those one-line proofs as above that do give that there are
true infinites courtesy the unbounded its quantification.

Then about MW's "I invoke the Absolute!", it's he's drunk
then also that potential and actual infinities do have differences
in real mathematics, one recalls "A Sober Mind Speaks".

So, RD's "there is only one theory and it ignores all my problems"
and MWs "there are at least two theories and they don't agree",
seems for a sort of "yet somehow, there's a theory".

... With laws, plural, of large numbers.

If you want a gentle introduction to set theory,
try reading Quine's Set Theory then after it introduces
the class-set distinction and x not in x and x not equals x,
and why neither of those is first-order in ZF, quit.

... in a theory, ....

I understand the multitude of different ways to define "Numbers", but
their are a core set of fundamental numbers with basically agreed on definitions, and reasonable people don't try calling something that
doesn't match the accepted definition as being one of those sets.

"Natural Numbers" is such a set, and while there may be several ways of expressing its "generation", they all are essentially, you start with a
base number, called Zero, and then you have an operation, called
something like Successor, and the set of the Natural Numbers is the set
of things you get by applying that operation an unlimited number of times.

Thus, I admit to other number theories, they just are not the "standard" systems of Natural Numbers, Integers, Rational Numbers, or Real Numbers.
(The Reals have a couple of different generational methods, but they
result in the same set of numbers with the same basic properties).

If you want to talk about other systems, feel free, just don't be
"deceptive" (or lying) and try to use the name for one of the standard sets.

And most of the "alternative" systems alternative have accepted names
for them, so that should be used.

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On 1/10/2024 2:04 PM, WM wrote:

Le 09/01/2024 à 20:34, Jim Burns a écrit :

On 1/9/2024 12:40 PM, WM wrote:

Just today I showed some proofs
to my students.

If you are claiming your students for
some form of peer.review, then
you are accepting pre-calculus students as
_peers_

Nonsense.
There are 8th semester informatics and
engineering studens.

Perhaps
the informatics and engineering students of
Wolfgang Mückenheim of Hochschule Augsburg
remember their arithmetic.

kᵢⱼ = i+(i+j-1)(i+j-2)/2

For each pair ⟨i,j⟩ of visibleᵂᴹ numbers
there is a unique visibleᵂᴹ number kᵢⱼ

sₖᵢⱼ = max{h| (h-1)(h-2)/2 < kᵢⱼ}
iₖᵢⱼ+jₖᵢⱼ = sₖᵢⱼ
iₖᵢⱼ = kᵢⱼ-(sₖᵢⱼ-1)(sₖᵢⱼ-2)/2
jₖᵢⱼ = sₖᵢⱼ(sₖᵢⱼ-1)/2-kᵢⱼ+1

For each visibleᵂᴹ number kᵢⱼ
there is a unique pair ⟨iₖᵢⱼ,jₖᵢⱼ⟩ of
visibleᵂᴹ numbers
⟨iₖᵢⱼ,jₖᵢⱼ⟩ = ⟨i,j⟩

There are as many visibleᵂᴹ numbers
as pairs of visibleᵂᴹ numbers.
Darkᵂᴹ numbers haven't entered the discussion.

----
For each non-empty split F,H of ⟨0,1,…,n⟩
some i‖i⁺¹ exists last‖first in F‖H
F‖H = ⟨0,1,…,i⟩‖⟨i⁺¹,…,n⟩

0‖n exists first‖last in ⟨0,1,…,n⟩

⟨0,1,…,n⟩ can't fit in ⟨1,…,n⟩
⟨0,1,…,n⟩ holds only visibleᵂᴹ numbers.

ℕ contains each ⟨0,1,…,n⟩ and nothing else.
ℕ\{0} contains each ⟨1,…,n⟩ and nothing else.

⟨0,1,…,n⟩ CAN fit in ⟨1,…,n,n⁺¹⟩
ℕ CAN fit in ℕ\{0}

If
darkᵂᴹ numbers are the reason that
ℕ CAN fit in ℕ\{0}
then
they must be
darkᵂᴹ visibleᵂᴹ in ⟨0,1,…,n⟩ in ℕ

In WM's scheme,
_sets_ inform _numbers_ that
the numbers are darkᵂᴹ
but we're not permitted to refer to
these darkᵂᴹ visibleᵂᴹ numbers,
only to the sets they're in.

Talking about
numbers which are darkᵂᴹ
is really nothing more than talking about
sets holding numbers which are darkᵂᴹ

The only use talking about darkᵂᴹ numbers has
is as a bit of techno.babble
to fill gaps in the discussion where,
otherwise, explanation would be expected.

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Le 11/01/2024 à 13:43, Jim Burns a écrit :

On 1/10/2024 2:04 PM, WM wrote:

Le 09/01/2024 à 20:34, Jim Burns a écrit :

On 1/9/2024 12:40 PM, WM wrote:

Just today I showed some proofs
to my students.

If you are claiming your students for
some form of peer.review, then
you are accepting pre-calculus students as
_peers_

Nonsense.
There are 8th semester informatics and
engineering studens.

Perhaps
the informatics and engineering students of
Wolfgang Mückenheim of Hochschule Augsburg
remember their arithmetic.

And now they know:

kᵢⱼ = i+(i+j-1)(i+j-2)/2

Every number defined like k does not belong to the domain covered by the smallest
ℵo unit fractions or by the largest ℵo natural numbers.

Regards, WM

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On 1/11/2024 11:49 AM, Ross Finlayson wrote:

On Thursday, January 11, 2024
at 4:43:40 AM UTC-8, Jim Burns wrote:

[...]

Then, the idea that there are
"descriptive dynamics",
has that
basically all theories
formally abstractly symbolically
result sorts of descriptive theories
in words, all one theory.

There is a Swiss.Army.knife theory, which
provides insight in many, many instances.

I'm thinking of the theory of
finite sequences of claims,
claims which might be about any of
many, many different domains of discussion.

We have a lot of confidence that,
in a finite sequence of claims,
if there is a false claim,
then there is a first.false claim.

Join that to our ability to _see_
in some sequences,
that some claims are not.first.false,
such as Q in ⟨ Q∨¬P P Q ⟩
and
we know we have, in some instances,
finite sequences of not.first.false claims,
which,
by our Swiss.Army.knife theory,
are finite sequences of not.false claims.

Even if
we can't see the objects of our claims,
we know they are true claims.
That seems magical to me --
"magical" perhaps not the best word for
mathematics, science, and engineering,
but there it is.

This Swiss.Army.knife theory of
finite sequences of claims
might be something like
what you (RF) mean by One Theory.
Or it might not be.

----
An essential step in using the Swiss.Army.knife
is description.
We decide to learn about widgets.
_We describe a widget_
We follow with not.first.false claims
about widgets, which,
because they are in that sequence,
we know are true about widgets.
Success.

That works because there are
pre.known claims about widgets mixed in.
That makes our post.known claims
specific to widgets _which we want_
since widgets are different from non.widgets.

But the pre.known claims are widget theory,
not the One Theory.
If we replace claims about widgets with
claims about anything,
following claims not.first.false about anything
shouldn't be telling us anything useful or
interesting, should it?
I don't see how it could.

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On 1/11/2024 4:38 PM, WM wrote:

Le 11/01/2024 à 13:43, Jim Burns a écrit :

On 1/10/2024 2:04 PM, WM wrote:

Le 09/01/2024 à 20:34, Jim Burns a écrit :

On 1/9/2024 12:40 PM, WM wrote:

Just today I showed some proofs
to my students.

If you are claiming your students for
some form of peer.review, then
you are accepting pre-calculus students as
_peers_

Nonsense.
There are 8th semester informatics and
engineering studens.

Perhaps
the informatics and engineering students of
Wolfgang Mückenheim of Hochschule Augsburg
remember their arithmetic.

And now they know:

kᵢⱼ = i+(i+j-1)(i+j-2)/2

Every number defined like k
does not belong to the domain covered by
the smallest ℵo unit fractions or by
the largest ℵo natural numbers.

Perhaps
the informatics and engineering students of
Wolfgang Mückenheim of Hochschule Augsburg
remember that,
in arithmetic,
the unit fractions and natural numbers
are closed under addition, subtraction,
multiplication, and division.

Since you (WM) have decided that
you are talking about _not.arithmetic_
this might be an especially apt time
for your students to remember that
a claim about _not.arithmetic_
even if it were _true_
doesn't contradict a claim about _arithmetic_

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On 1/11/2024 7:14 PM, FromTheRafters wrote:

Jim Burns has brought this to us :

On 1/11/2024 4:38 PM, WM wrote:

Le 11/01/2024 à 13:43, Jim Burns a écrit :

kᵢⱼ = i+(i+j-1)(i+j-2)/2

Every number defined like k
does not belong to the domain covered by
the smallest ℵo unit fractions or by
the largest ℵo natural numbers.

Perhaps
the informatics and engineering students of
Wolfgang Mückenheim of Hochschule Augsburg
remember that,
in arithmetic,
the unit fractions and natural numbers
are closed under addition, subtraction,
multiplication, and division.

Since you (WM) have decided that
you are talking about _not.arithmetic_
this might be an especially apt time
for your students to remember that
a claim about _not.arithmetic_
even if it were _true_
doesn't contradict a claim about _arithmetic_

Subtraction?

I may have been a little loose in
how I said that.

The operations are closed in _arithmetic_
except for division by 0.

There is
no single ℵ₀.set of smallest unit fractions and
no single ℵ₀.set of largest natural numbers.

There is,
for each pair ⟨i,j⟩ of visibleᵂᴹ numbers
a unique visibleᵂᴹ number kᵢⱼ
and,
for each visibleᵂᴹ number kᵢⱼ
a unique pair ⟨iₖᵢⱼ,jₖᵢⱼ⟩ of visibleᵂᴹ numbers ⟨iₖᵢⱼ,jₖᵢⱼ⟩ = ⟨i,j⟩

There are as many visibleᵂᴹ numbers
as pairs of visibleᵂᴹ numbers.
Darkᵂᴹ numbers haven't entered the discussion.

kᵢⱼ = i+(i+j-1)(i+j-2)/2

sₖᵢⱼ = max{h| (h-1)(h-2)/2 < kᵢⱼ}
iₖᵢⱼ+jₖᵢⱼ = sₖᵢⱼ
iₖᵢⱼ = kᵢⱼ-(sₖᵢⱼ-1)(sₖᵢⱼ-2)/2
jₖᵢⱼ = sₖᵢⱼ(sₖᵢⱼ-1)/2-kᵢⱼ+1

Because arithmetic.

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On 1/11/24 4:38 PM, WM wrote:

Le 11/01/2024 à 13:43, Jim Burns a écrit :

On 1/10/2024 2:04 PM, WM wrote:

Le 09/01/2024 à 20:34, Jim Burns a écrit :

On 1/9/2024 12:40 PM, WM wrote:

Just today I showed some proofs
to my students.

If you are claiming your students for
some form of peer.review, then
you are accepting pre-calculus students as
_peers_

Nonsense.
There are 8th semester informatics and
engineering studens.

Perhaps
the informatics and engineering students of
Wolfgang Mückenheim of Hochschule Augsburg
remember their arithmetic.

And now they know:

kᵢⱼ = i+(i+j-1)(i+j-2)/2

Every number defined like k does not belong to the domain covered by the smallest
ℵo unit fractions or by the largest ℵo natural numbers.

Regards, WM

Which of those largest natural numbers can k not get to?

What is the boundry that can not be passed?

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On 1/11/2024 8:54 PM, Ross Finlayson wrote:

On Thursday, January 11, 2024
at 4:56:46 PM UTC-8, Jim Burns wrote:

There are as many visibleᵂᴹ numbers
as pairs of visibleᵂᴹ numbers.
Darkᵂᴹ numbers haven't entered the discussion.
kᵢⱼ = i+(i+j-1)(i+j-2)/2
sₖᵢⱼ = max{h| (h-1)(h-2)/2 < kᵢⱼ}
iₖᵢⱼ+jₖᵢⱼ = sₖᵢⱼ
iₖᵢⱼ = kᵢⱼ-(sₖᵢⱼ-1)(sₖᵢⱼ-2)/2
jₖᵢⱼ = sₖᵢⱼ(sₖᵢⱼ-1)/2-kᵢⱼ+1
Because arithmetic.

What you can do is
define arithmetic just a little
differently than usual, that instead
of Peano/Presburger,
addition and multiplication,
and their inverses,
is that what you do is
start with increment and division.

Julia Robinson has a paper showing how
to define various arithmetical concepts
from each other.

Definability and decision problems in arithmetic

| In this paper, we are concerned with
| the arithmetical definability of certain notions
| of integers and rationals in terms of other notions.
| The results derived will be applied to obtain
| a negative solution of corresponding decision problems.
|
| In Section 1, we show that addition of
| positive integers can be defined arithmetically
| in terms of multiplication and the unary operation
| of successor S (where Sa = a + 1). Also, it is shown
| that both addition and multiplication can be defined
| arithmetically in terms of successor and
| the relation of divisibility |
| (where x|y means x divides y).
|
The Journal of Symbolic Logic,
Volume 14, Issue 2, 23 June 1949, pp. 98 - 114
DOI: https://doi.org/10.2307/2266510

That is also interesting because
some claims become easier or harder to prove,
starting from order and divisibility or
from addition and multiplication.

If I recall correctly,
unique prime factorization becomes easier
starting from order and divisibility.

Then there's as
increment addition
multiplication powers
exponent tetration

Once we have addition and multiplication,
however we get them,
the Chinese remainder theorem can be used
to encode an arbitrarily.long finite sequence,
and encoding finite sequences can be used
to re.phrase recursive definitions
non.recursively.

That makes exponentials, factorials,
tetration, and much, much more
Free At No Extra Charge.

Just saying.

--- SoupGate-Win32 v1.05
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Le 12/01/2024 à 00:34, Jim Burns a écrit :

Since you (WM) have decided that
you are talking about _not.arithmetic_
this might be an especially apt time
for your students to remember that
a claim about _not.arithmetic_
even if it were _true_
doesn't contradict a claim about _arithmetic_

That should not hinder an inquisitive student to learn that arithmetic
does not cover the domain of the smallest ℵo unit fractions and of the largest ℵo natural numbers.

Regards, WM

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Le 12/01/2024 à 04:09, Richard Damon a écrit :

Which of those largest natural numbers can k not get to?

If I could name it I had made it v isible.

What is the boundry that can not be passed?

That is the difficult point: There is no fixed threshold. Most can't
comprehend it. Potential infinity!

Regards, WM

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On 1/12/2024 9:01 AM, WM wrote:

Le 12/01/2024 à 00:34, Jim Burns a écrit :

Since you (WM) have decided that
you are talking about _not.arithmetic_
this might be an especially apt time
for your students to remember that
a claim about _not.arithmetic_
even if it were _true_
doesn't contradict a claim about _arithmetic_

That should not hinder an inquisitive student
to learn that
arithmetic does not cover the domain of
the smallest ℵo unit fractions and of
the largest ℵo natural numbers.

Do your students learn that it doesn't from
their instructor, from you?

If they don't,
that is what strikes me as
hindering the inquisitive student.

----
We can make a claim true of
each one of infinitely.many
which we know is true by knowing
_that is what we mean_ by
"natural number" or "unit fraction"
or something else.

Also,
we can make a claim not.first.false of
each one of the same infinitely.many
which we know is not.first.false by seeing
the sequence of claims we make.
For example, by seeing Q is after P and Q∨¬P

We can make a finite sequence of claims
in which each claim is not.first.false of
each one of the same infinitely.many.
This is more challenging, but the challenge
is the reason that mathematicians are
so famously well-paid.

A finite sequence in which
each claim is not.first.false about each
is finite sequence in which
each claim is not.false about each.

That is something we already know
_about finite sequences of claims_

By that knowledge,
we can increase our knowledge of other things:
natural numbers, unit fractions, and so on
_looking only at a sequence of claims_
and NOT looking at
natural numbers, unit fractions, and so on.

----
We can make a claim that
k can be counted to
and know that it is true of
each which can be counted to.

We can follow that with the claim that
k+1 can be counted to
and see that there is no way in which
k+1 can be counted to
can be first.false.

And, similarly, that
k isn't the largest which can be counted to.

That is how we know that
nothing exists which is the largest which
can be counted to,
even though we are finite beings,
even though we can't see the infinitely.many
which can be counted to.

We can see _the claims_
and recognize that they're not.first.false.

--- SoupGate-Win32 v1.05
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On 1/12/24 9:05 AM, WM wrote:

Le 12/01/2024 à 04:09, Richard Damon a écrit :

Which of those largest natural numbers can k not get to?

If I could name it I had made it v isible.

What is the boundry that can not be passed?

That is the difficult point: There is no fixed threshold. Most can't comprehend it. Potential infinity!

Regards, WM

So, you have two distinct sets with no boundry between them?

What makes them different?

If there isn't a line that keeps the describable numbers out of your
dark numbers, then aren't all your dark numbers describable?

If there is a line, then there must be a highest describable number, so
you can give it.

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Le 13/01/2024 à 00:46, Richard Damon a écrit :

On 1/12/24 9:05 AM, WM wrote:

Le 12/01/2024 à 04:09, Richard Damon a écrit :

Which of those largest natural numbers can k not get to?

If I could name it I had made it v isible.

What is the boundry that can not be passed?

That is the difficult point: There is no fixed threshold. Most can't
comprehend it. Potential infinity!

So, you have two distinct sets with no boundry between them?

The visible numbers are not a set but only a collection, because the
membership is not fixed.

What makes them different?

This: You cannot use real numbers of the domain containing the ℵ
smallest unit fractions and of the domain containing the ℵ largest
natural numbers. It is obviously impossible to come closer to zero. There
are always ℵ unit fractions between 0 and a number chosen by you. Same
for ω, but, contrary to zero, ω is a vague end of the scale.

If there isn't a line that keeps the describable numbers out of your
dark numbers, then aren't all your dark numbers describable?

Assume it. Then you can come close to zero such that nothing is between it
and your chosen number. But that is a contradiction.

If there is a line, then there must be a highest describable number, so
you can give it.

That is the point hard to swallow. There is no line.

Regards, WM

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Le 13/01/2024 à 00:46, Richard Damon a écrit :

On 1/12/24 9:01 AM, WM wrote:

arithmetic
does not cover the domain of the smallest ℵo unit fractions and of the
largest ℵo natural numbers.

But what arithmetic didn't cover them?

Addition, subtraction, multiplication, division, exponentiation,
tetration.

Regards, WM

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Le 12/01/2024 à 19:30, Jim Burns a écrit :

On 1/12/2024 9:01 AM, WM wrote:

Le 12/01/2024 à 00:34, Jim Burns a écrit :

Since you (WM) have decided that
you are talking about _not.arithmetic_
this might be an especially apt time
for your students to remember that
a claim about _not.arithmetic_
even if it were _true_
doesn't contradict a claim about _arithmetic_

That should not hinder an inquisitive student
to learn that
arithmetic does not cover the domain of
the smallest ℵo unit fractions and of
the largest ℵo natural numbers.

Do your students learn that it doesn't from
their instructor, from you?

They need not learn it, because it is clear to everybody, perhaps after a
short hint: You cannot use real numbers of the domain containing the ℵ smallest unit fractions and of the domain containing the ℵ largest
natural numbers.

It is obviously impossible to come closer to zero. There are always ℵ
unit fractions between 0 and a number chosen by you. Same for ω, but,
contrary to zero, ω is a vague end of the scale.

Regards, WM

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On 1/12/2024 10:18 PM, Ross Finlayson wrote:

On Friday, January 12, 2024
at 3:46:43 PM UTC-8, Richard Damon wrote:

[...]

Sorites and the Heap.

Make a claim which is true of each
and leap the Heap.

There are no standard models of integers.

The minimal inductive set is
a standard standard model of integers.

Are you currently considering
non.standard standard models of integers?

There are fragments,

A fragment isn't a model.

Granted, there are things other than
the minimal inductive set.
But, Shirley, that's not what you mean?

there are extensions,

An extension isn't standard.

Granted, there are things other than
the minimal inductive set.
But, Shirley, that's not what you mean?

the ordinary inductive set's
a non-logical constant.

The only place I remember seeing "non-logical"
(which doesn't mean "illogical") used
is with axioms _in addition to_ the logical axioms.

We could call this sense of non-logical
metalogical, after.logical, which makes it less
likely to be mistaken for an admission of failure.

If you (RF) don't mean metalogical, then
I don't know what you mean by non-logical.


The unformalizable goal of stating
metalogical axioms is that everyone can say
"Yes, that is what everyone means by X".
That wouldn't be a _logical_ result, but
it would be a fact about what everyone means.

For the more obscure X,
there isn't really an "everyone".
There is the author, and
there is what they mean by X

However, natural numbers aren't obscure.


For _what we mean by_ the non.negative integers,
the minimal inductive set is a standard model.

There are multiple models of integers.

Have we drifted away from (up.post)

There are no standard models of integers.

?

What I think we mean, what I know I mean
by a model of the non.negative integers
is that it contain at least
each standard non.negative integer,

by a standard model of the non.negative integers
is that it contains no extensions.

If any model of the non.negative integers exists,
then a standard model exists, provably.

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On 1/14/24 11:48 AM, WM wrote:

Le 13/01/2024 à 00:46, Richard Damon a écrit :

On 1/12/24 9:05 AM, WM wrote:

Le 12/01/2024 à 04:09, Richard Damon a écrit :

Which of those largest natural numbers can k not get to?

If I could name it I had made it v isible.

What is the boundry that can not be passed?

That is the difficult point: There is no fixed threshold. Most can't
comprehend it. Potential infinity!

So, you have two distinct sets with no boundry between them?

The visible numbers are not a set but only a collection, because the membership is not fixed.

What makes them different?

This: You cannot use real numbers of the domain containing the ℵ
smallest unit fractions and of the domain containing the ℵ largest
natural numbers. It is obviously impossible to come closer to zero.
There are always ℵ unit fractions between 0 and a number chosen by you. Same for ω, but, contrary to zero, ω is a vague end of the scale.

Really? You can't figure out how to use number like 1/10 or 12?

After all, those are part of the ℵ unit fractions between some unit
fraction and 0 and above some Natural Number.

There are no unit fractions smaller that ALL unit fractions or Natural
Numbers greater than ALL Natural Numbers, so the domain of numbers that
you can not use because of that condition is empty.

If there isn't a line that keeps the describable numbers out of your
dark numbers, then aren't all your dark numbers describable?

Assume it. Then you can come close to zero such that nothing is between
it and your chosen number. But that is a contradiction.

You can come as close as you want.

That doesn't means that you can reach the point that nothing is between.

That would require the existance of a Highest Natural Number.

That assumption is where the contradiction is, not that all Natural Numbers/Unit Fractions are visible.

If there is a line, then there must be a highest describable number,
so you can give it.

That is the point hard to swallow. There is no line.

Then all your "dark" numbers are also visible, and thus not Dark.

Without a line, what keeps the visible numbers out of the set of "Dark" numbers?

You are just hitting the flaw of naive set theory.

Regards, WM

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On 1/14/2024 11:43 AM, WM wrote:

Le 12/01/2024 à 19:30, Jim Burns a écrit :

On 1/12/2024 9:01 AM, WM wrote:

Le 12/01/2024 à 00:34, Jim Burns a écrit :

Since you (WM) have decided that
you are talking about _not.arithmetic_
this might be an especially apt time
for your students to remember that
a claim about _not.arithmetic_
even if it were _true_
doesn't contradict a claim about _arithmetic_

That should not hinder an inquisitive student
to learn that
arithmetic does not cover the domain of
the smallest ℵo unit fractions and of
the largest ℵo natural numbers.

Do your students learn that it doesn't from
their instructor, from you?

They need not learn it,

Which is to say:
No,
you,
Wolfgang Mückenheim of Hochschule Augsburg,
do not tell your students that,
when you talk about
the smallest ℵ₀ unit fractions and
the largest ℵ₀ natural numbers,
you aren't talking about
arithmetic.

because it is clear to everybody,
perhaps after a short hint:
You cannot use real numbers

Real numbers ℝ are the rationals ℚ and
enough points between non.empty splits of ℚ
that no function ℝ→ℝ which jumps is
continuous at each point in ℝ

Rationals ℚ are the integers ℤ and
enough points that each non.0 division in ℤ
has a solution in ℚ

Integers ℤ are the naturals ℕ and
enough points that each subtraction in ℕ
has a solution in ℤ

Naturals ℕ are each end of ordered ⟨0,…,k⟩
such that,
for each non.empty split F,H of ⟨0,…,k⟩
i‖i⁺¹ exists last‖first in F‖H,
and 0‖k exists first‖last in ⟨0,…,k⟩

i⁺¹ is non.0 non.doppelgänger non.final.

Also Known As arithmetic.

You cannot use real numbers of
the domain containing
the ℵ smallest unit fractions and of
the domain containing
the ℵ largest natural numbers.

Each element of ℝ is
not the smallest unit fraction and
not the largest natural number.

It is obviously impossible
to come closer to zero.

No,
it is obvious in arithmetic
that each x ≠ 0 can come closer.
To x/2 for example.

There are always ℵ unit fractions between
0 and a number [x] chosen by you.

For each x > 0
for each not-fitting-1-removed cardinality n
the cardinality of unit fractions ⊆ (0,x]
is not that cardinality.
|…,⅟mₓ⁺²,⅟mₓ⁺¹| ≥ |⅟mₓ⁺²,⅟mₓ⁺¹| > |⅟mₓ⁺¹|
|…,⅟mₓ⁺²,⅟mₓ⁺¹| ≥ |⅟mₓ⁺³,…,⅟mₓ⁺¹| > |⅟mₓ⁺²,⅟mₓ⁺¹|
|…,⅟mₓ⁺²,⅟mₓ⁺¹| ≥ |⅟mₓ⁺⁴,…,⅟mₓ⁺¹| > |⅟mₓ⁺³,…,⅟mₓ⁺¹|
|…,⅟mₓ⁺²,⅟mₓ⁺¹| ≥ |⅟mₓ⁺⁵,…,⅟mₓ⁺¹| > |⅟mₓ⁺⁴,…,⅟mₓ⁺¹|
|…,⅟mₓ⁺²,⅟mₓ⁺¹| ≥ |⅟mₓ⁺⁶,…,⅟mxₓ⁺¹| > |⅟mₓ⁺⁵,…,⅟mₓ⁺¹|
...

The cardinality of unit fractions ⊆ (0,x]
_is not_ any not.fitting.1.removed cardinality n

The cardinality of unit fractions ⊆ (0,x]
_is_ fitting.1.removed |…,⅟mₓ⁺⁴,⅟mₓ⁺³,⅟mₓ⁺²,⅟mₓ⁺¹| = |…,⅟mₓ⁺⁴,⅟mₓ⁺³,⅟mₓ⁺²| = |…,⅟mₓ⁺⁴,⅟mₓ⁺³| =
|…,⅟mₓ⁺⁴| =
... =
ℵ₀

Same for ω, but,
contrary to zero, ω is a vague end of the scale.

The cardinality of natural numbers ⊆ [x,∞)
_is not_ any not.fitting.1.removed cardinality n
|mₓ⁺¹,mₓ⁺²,…| ≥ |mₓ⁺¹,mₓ⁺²| > |mₓ⁺¹| |mₓ⁺¹,mₓ⁺²,…| ≥ |mₓ⁺¹,…,mₓ⁺³| > |mₓ⁺¹,mₓ⁺²| |mₓ⁺¹,mₓ⁺²,…| ≥ |mₓ⁺¹,…,mₓ⁺⁴| > |mₓ⁺¹,…,mₓ⁺³|
|mₓ⁺¹,mₓ⁺²,…| ≥ |mₓ⁺¹,…,mₓ⁺⁵| > |mₓ⁺¹,…,mₓ⁺⁴|
|mₓ⁺¹,mₓ⁺²,…| ≥ |mₓ⁺¹,…,mₓ⁺⁶| > |mₓ⁺¹,…,mₓ⁺⁵|
...

The cardinality of natural numbers ⊆ [x,∞)
_is_ fitting.1.removed
|mₓ⁺¹,mₓ⁺²,mₓ⁺³,mₓ⁺⁴,…| = |mₓ⁺²,mₓ⁺³,mₓ⁺⁴,…| =
|mₓ⁺³,mₓ⁺⁴,…| =
|mₓ⁺⁴,…| =
... =
ℵ₀ = |ω|

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Le 14/01/2024 à 19:42, Richard Damon a écrit :

There are no unit fractions smaller that ALL unit fractions

But there are ℵ unit fraction smaller than all you can name.

Assume it. Then you can come close to zero such that nothing is between
it and your chosen number. But that is a contradiction.

You can come as close as you want.

By naming an x you can come close as you want in the measure of distance
but not in the measure number of unit fractions between 0 and x.

∀eps > 0 ∀x ∈ (eps, 1]: NUF(x) = ℵo

That doesn't means that you can reach the point that nothing is between.

That would require the existance of a Highest Natural Number.

Nevertheless you can reach 0.

That assumption is where the contradiction is, not that all Natural Numbers/Unit Fractions are visible.

That is the same. Fact is that you cannot name almost all numbers.

Regards, WM

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Le 15/01/2024 à 01:23, Jim Burns a écrit :

On 1/14/2024 11:43 AM, WM wrote:

You cannot use real numbers of
the domain containing
the ℵ smallest unit fractions and of
the domain containing
the ℵ largest natural numbers.

Each element of ℝ is
not the smallest unit fraction and
not the largest natural number.

Each element of ℝ that can be named.

It is obviously impossible
to come closer to zero.

No,
it is obvious in arithmetic
that each x ≠ 0 can come closer.
To x/2 for example.

There is no boundary in the meaqsure of distance but there is a boundary
in the measure of remaining elements between 0 and the chosen eps > 0.

∀eps > 0, ∀x ∈ (eps, 1]: NUF(x) = ℵo

Regards, WM

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On 1/15/24 3:15 AM, WM wrote:

Le 14/01/2024 à 19:42, Richard Damon a écrit :

There are no unit fractions smaller that ALL unit fractions

But there are ℵ unit fraction smaller than all you can name.

No, there are not. As I have shown, I can name any of them.

Assume it. Then you can come close to zero such that nothing is
between it and your chosen number. But that is a contradiction.

You can come as close as you want.

By naming an x you can come close as you want in the measure of distance
but not in the measure number of unit fractions between 0 and x.

No, the number of unit fractions between 0 and x will always be Alpha_0,
the measure of Countable infinity, as there will ALWAYS be that many
unit fractions between 0 and x (unless x is 0, then the number is 0)

∀eps > 0 ∀x ∈ (eps, 1]: NUF(x) = ℵo

Right. There are always ℵo unit fractions below a positive number, so no number exists where NUF(x) == 1

That wasn't for all "visible" numbers, that was for all "Numbers" (from
one of the standard sets of numbers, like Rational or Reals)

That doesn't means that you can reach the point that nothing is between.

That would require the existance of a Highest Natural Number.

Nevertheless you can reach 0.

Only by leaving the set of unit fractions.

That assumption is where the contradiction is, not that all Natural
Numbers/Unit Fractions are visible.

That is the same. Fact is that you cannot name almost all numbers.

Except that you can.

Show a set that I can not name a member of it.

Your logic is just insufficient to handle unbounded sets.

Regards, WM

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On 1/15/2024 3:24 AM, WM wrote:

Le 15/01/2024 à 01:23, Jim Burns a écrit :

On 1/14/2024 11:43 AM, WM wrote:

It is obviously impossible
to come closer to zero.

No,
it is obvious in arithmetic
that each x ≠ 0 can come closer.
To x/2  for example.

There is no boundary in
the measure of distance
but
there is a boundary in
the measure of remaining elements
between 0 and the chosen eps > 0.

∀eps > 0, ∀x ∈ (eps, 1]: NUF(x) = ℵo

In arithmetic,
∀eps > 0
∀k ∈ ℕ
∃mᵉᵖˢ ∈ ℕ:
eps > ⅟mᵉᵖˢ > ⅟(mᵉᵖˢ+k+1) > 0
k < NUF(eps)

In arithmetic,
∀eps > 0
∀k ∈ ℕ
k < NUF(eps)
NUF(eps) ≮ NUF(eps)
NUF(eps) ∉ ℕ

In arithmetic,
∀eps > 0
NUF(eps) ∉ ℕ

In arithmetic,
¬∃mᵂᴹ ∈ ℕ:
∀eps > 0
∀k ∈ ℕ
eps > ⅟mᵂᴹ > ⅟(mᵂᴹ+k+1) > 0

there is a boundary in
the measure of remaining elements
between 0 and the chosen eps > 0.

In arithmetic,
there is a boundary in
the cardinality of remaining elements
between 0 and the chosen eps > 0.
but
it's not a positive boundary,
the boundary is 0

| Assume otherwise.
| Assume 0 < β/2 < β < 2β
| for remaining.element.card.boundary β
|
| β < 2β
| NUF(2β) ∉ ℕ
| ∀k ∈ ℕ
| ∃mₖ ∈ ℕ:
| 2β > ⅟mₖ > ⅟(mₖ+k+1) > 0
| [1]
| k < NUF(2β)
|
| β/2 < β
| NUF(β/2) ∈ ℕ
| ¬∀k ∈ ℕ
| ∃mₖ ∈ ℕ:
| β/2 > ⅟mₖ > ⅟(mₖ+k+1) > 0
|
| ∃k ∈ ℕ:
| ¬(k < NUF(β/2))
|
| ∃k ∈ ℕ:
| ∀m′ ∈ ℕ
| ¬(β/2 > ⅟m′ > ⅟(m′+k+1) > 0)
| β/2 ≤ ⅟m′ ∨ ¬(⅟m′ > ⅟(m′+k+1) > 0)
| ⅟m′ > ⅟(m′+k+1)) > 0
| β/2 ≤ ⅟m′
|
| ∀m′ ∈ ℕ
| β/2 ≤ ⅟m′
| [2]
|
| From [1]
| ∀k ∈ ℕ
| ∃mₖ ∈ ℕ:
| 2β > ⅟mₖ
| β/2 > ⅟(4mₖ)
|
| However, from [2]
| in the case of m′ = 4mₖ
| β/2 ≤ ⅟(4mₖ)
| Contradiction.

Therefore,
in arithmetic,
the remaining.element.measure.boundary β
isn't positive.

In arithmetic,
∀eps > 0:
NUF(eps) ∉ ℕ

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Le 15/01/2024 à 13:48, Richard Damon a écrit :

On 1/15/24 3:15 AM, WM wrote:

Le 14/01/2024 à 19:42, Richard Damon a écrit :

There are no unit fractions smaller that ALL unit fractions

But there are ℵ unit fraction smaller than all you can name.

No, there are not. As I have shown, I can name any of them.

Then show it. Name a unit fraction that has not ℵ smaller ones.

No, the number of unit fractions between 0 and x will always be Alpha_0,
the measure of Countable infinity, as there will ALWAYS be that many
unit fractions between 0 and x (unless x is 0, then the number is 0)

The term is aleph_0, not alpha_0.
The mathematics is this: if x is less than every positive number, then x
is less than (0, oo). That is impossible for positive x, let alone for
ℵo unit fractions.

∀eps > 0 ∀x ∈ (eps, 1]: NUF(x) = ℵo

Right. There are always ℵo unit fractions below a positive number, so no number exists where NUF(x) == 1

The statement says not below "a positive number" but below "every positive number", but that means below (0, oo) and is nonsense.

Your logic is just insufficient to handle unbounded sets.

My logic is based on mathematics. Your claims are not logic but dogmas of matheology in contradiction with mathematics.

Regards, WM

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On 1/16/24 6:16 AM, WM wrote:

Le 15/01/2024 à 20:57, Jim Burns a écrit :

In arithmetic,
∀eps > 0:
NUF(eps) ∉ ℕ

In arithmetic

∀x ∈ (0, 1]: y < x ==> y =< 0, i.e., y is not positive.

Regards, WM

Except that the statement does't hold if, say, y was x/2.

You also need to define the domain of x and y. If not specified it will
be presumed something like The Reals.

Your problem is you have the order of operations wrong in your logic,
the Qualifier is run first, so when we look at the inequality, we have
am x we can use.

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Le 15/01/2024 à 20:57, Jim Burns a écrit :

In arithmetic,
∀eps > 0:
NUF(eps) ∉ ℕ

In arithmetic

∀x ∈ (0, 1]: y < x ==> y =< 0, i.e., y is not positive.

Regards, WM

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On 1/16/24 6:12 AM, WM wrote:

Le 15/01/2024 à 13:48, Richard Damon a écrit :

On 1/15/24 3:15 AM, WM wrote:

Le 14/01/2024 à 19:42, Richard Damon a écrit :

There are no unit fractions smaller that ALL unit fractions

But there are ℵ unit fraction smaller than all you can name.

No, there are not. As I have shown, I can name any of them.

Then show it. Name a unit fraction that has not ℵ smaller ones.

So, you don't unddrstand English and just working with Strawman.

I never claimed there was a unit fraction that doesn't have aleph_0
smaller ones.

No, the number of unit fractions between 0 and x will always be
Alpha_0, the measure of Countable infinity, as there will ALWAYS be
that many unit fractions between 0 and x (unless x is 0, then the
number is 0)

The term is aleph_0, not alpha_0.
The mathematics is this: if x is less than every positive number, then x
is less than (0, oo). That is impossible for positive x, let alone for
ℵo unit fractions.

So, you just proved that there is no

∀eps > 0 ∀x ∈ (eps, 1]: NUF(x) = ℵo

Right. There are always ℵo unit fractions below a positive number, so
no number exists where NUF(x) == 1

The statement says not below "a positive number" but below "every
positive number", but that means below (0, oo) and is nonsense.

So, you agree that NUF(x) = ℵo for all positive x and is never a finite value.

Since for all eps > 0 NUF(x) == ℵo, there is no eps for which NUF(x) is smaller than that, not even a "dark" one.

Your logic is just insufficient to handle unbounded sets.

My logic is based on mathematics. Your claims are not logic but dogmas
of matheology in contradiction with mathematics.

Nope. YOUR claims are based on the dogma of WMism, not actual logic or mathematics.

You can't get to actual fundamental statements that show your claims, so
you are just shown to be a dogmatist.

Regards, WM

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On 1/16/2024 6:16 AM, WM wrote:

Le 15/01/2024 à 20:57, Jim Burns a écrit :

In arithmetic,
∀eps > 0:
NUF(eps) ∉ ℕ

In arithmetic
∀x ∈ (0, 1]: y < x ==> y =< 0,
i.e., y is not positive.

In arithmetic,
∀eps > 0: eps > 0
i.e., eps is positive.

There is an implicit ∀y

I don't know which you intend:

(i)
(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
true

if y is a lower bound of (0,1]
then y ≤ glb(0,1] = 0

0 ∉ (0,1]

(ii)
∀x ∈ (0,1]: (y < x ⇒ y ≤ 0)
false

∀y: ∀x ∈ (0,1]: (y < x ⇒ y ≤ 0)
if and only if
¬∃y: ∃x ∈ (0,1]: (y < x ∧ 0 < y)
if and only if
¬∃x ∈ (0,1]: ∃y ∈ (0,x)
false
Instead,
∃x ∈ (0,1]: ∃y ∈ (0,x)

x ∈ ⋃{ (eps,1] | eps ∈ (0,1] }
if and only if
x ∈ (0,1]


∀eps > 0: NUF(eps) ∉ ℕ
true

Regards, WM

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Richard Damon schrieb:

On 1/16/24 6:16 AM, WM wrote:

Le 15/01/2024 à 20:57, Jim Burns a écrit :

In arithmetic,
∀eps > 0:
NUF(eps) ∉ ℕ

In arithmetic

∀x ∈ (0, 1]: y < x ==> y =< 0, i.e., y is not positive.

Regards, WM

Except that the statement does't hold if, say, y was x/2.

That's right. But what WM means is the following:

for every y which is smaller than every x ∈ (0,1] : y <= 0.

One of his problems is that he doesn't understand quantifiers and their
proper application. Example: The correct statement about unit fractions
1/n (where n∈ℕ)

∀ x ∈ (0,1] ∃^ℵo 1/n : 1/n < x

he misunderstands in a way that is equivalent to the quantifier shift:

∃^ℵo 1/n ∀ x ∈ (0,1] : 1/n < x .

Then he concludes:

there are ℵo unit fractions left from zero.

Discussing with WM you should always keep in mind: WM doesn't write
about mathematics but his private system of ideas which are based on
his inability to understand infinite sets and set theory.

Dieter Heidorn

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On 1/15/2024 12:52 AM, Ross Finlayson wrote:

On Sunday, January 14, 2024
at 10:24:52 AM UTC-8, Jim Burns wrote:

[...]

Here are some things
you should be familiar with,
if you talking about models,
structures that embody all relations,
of integers,
here the natural or non-negative integers,
for that model-theory and proof-theory are
the same thing,
in terms of mathematical proofs,
insofar as that
a model of all relation and structurally,
is a proof,
and that
any proof has a corresponding model,
and vice-versa.

As I understand it,
the semantic (model.theory) and
the syntactic (proof.theory) points of view
are two sides of the same coin.

There are some very nice proofs showing
how the two are related, but
they aren't exactly the same thing.

For a certain theory (syntax),
there might or might not exist
a structure which it describes (semantics).
A structure which a theory describes
is a model of the theory.

Provably,
if no contradiction follows from a theory,
then a model of it exists.

The proof constructs (shows exists) a model from
the objects of the language of the theory.
That's why I note that, even if a theory has a model,
it still might not have the model you're thinking of.

Every structure that satisfies a theory is
a model of the theory.

Perhaps significantly different structures
satisfy the same theory (are models).

Provably,
if each model (semantics) which satisfies a theory
satisfies an additional formula,
then a proof (syntax) exists of that formula
starting from the theory (syntax).

----
Because there are formally undecidable formulas
in the natural numbers,
if any model exists,
then more than one model exists.

Also,
if any inductive set exists,
then a unique minimal inductive set exists,
subset to each inductive set, and
containing only each countable.to number.
By any other word, a standard model.

tl,;dr
The natural numbers either
are complete nonsense, top to bottom, or
have a standard model and a non.standard model.

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Le 16/01/2024 à 13:11, Richard Damon a écrit :

On 1/16/24 6:16 AM, WM wrote:

Le 15/01/2024 à 20:57, Jim Burns a écrit :

In arithmetic,
∀eps > 0:
NUF(eps) ∉ ℕ

In arithmetic

∀x ∈ (0, 1]: y < x ==> y =< 0, i.e., y is not positive.

Except that the statement does't hold if, say, y was x/2.

The statement concerns all real numbers.

You also need to define the domain of x and y. If not specified it will
be presumed something like The Reals.

Your problem is you have the order of operations wrong in your logic,
the Qualifier is run first, so when we look at the inequality, we have
am x we can use.

The "less than" relation is independent of quantifier-exchange. If for
every x ∈ (0, 1] there a smaller y, then there is an y smaller than
every x ∈ (0, 1] and hence smaller than (0, 1]. Of course for ℵ
elements y this is much clearer.

Regards, WM

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Le 16/01/2024 à 21:19, Dieter Heidorn a écrit :

The correct statement about unit fractions
1/n (where n∈ℕ)

∀ x ∈ (0,1] ∃^ℵo 1/n : 1/n < x

he misunderstands in a way that is equivalent to the quantifier shift:

∃^ℵo 1/n ∀ x ∈ (0,1] : 1/n < x .

Then he concludes:

there are ℵo unit fractions left from zero.

Would be required to make your statement correct.

For the less-than relation there is no quantifier magic.
NUF(x) cannot grow anywhere from 0 to ℵo without passing finite values.

Regards, WM

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Le 16/01/2024 à 18:11, Jim Burns a écrit :

(i)
(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
true

Yes, that is correct.

Regards, WM

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On 1/17/24 6:49 AM, WM wrote:

Le 16/01/2024 à 21:19, Dieter Heidorn a écrit :

The correct statement about unit fractions
1/n (where n∈ℕ)

    ∀ x ∈ (0,1]  ∃^ℵo 1/n : 1/n < x

he misunderstands in a way that is equivalent to the quantifier shift:

    ∃^ℵo 1/n  ∀ x ∈ (0,1] : 1/n < x .

Then he concludes:

    there are ℵo unit fractions left from zero.

Would be required to make your statement correct.

For the less-than relation there is no quantifier magic.
NUF(x) cannot grow anywhere from 0 to ℵo without passing finite values.

Regards, WM

Why not?

And what says those "values" that it happens at have to be in the set of
Unit Fractions/rational/real numbers?

If the "value" where NUF(x) == 1 isn't in the domain of Unit
factions/Rational Numbers/ Real Numbers, then there is no need for your
"dark" numbers, they are just visible numbers in some other system.

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Le 17/01/2024 à 13:55, Richard Damon a écrit :

On 1/17/24 6:49 AM, WM wrote:

NUF(x) cannot grow anywhere from 0 to ℵo without passing finite values.

Why not?

Because of mathemtics. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.

Regards, WM

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On 1/16/2024 11:39 PM, Ross Finlayson wrote:

On Tuesday, January 16, 2024
at 2:29:07 PM UTC-8, Ross Finlayson wrote:

[...]

In proof theory, it's provable, that
in classical logic,

proof.theory ≠ set.theory
proof ≠ set

logic < (non.logical) set.theory
set.theory = logic+set.description

set.description = infinity∨¬infinity
set.description = foundation∨¬foundation

different.description ⟺ different.models

one must always assert
Empty, Infinity, and well-foundedness
for the inference of the existence of any set,
for otherwise inference may arrive at
Russell's set via quantification.

Russell's ∃{x|x∉x} ⟸ unrestricted.comprehension

s/comprehension/specification+replacement

¬∃{x|x∉x} ⟸ classical
On the other hand,
revision.theory.of.truth+more?

So,
in classical logic,
ZF's axioms aren't independent.

Did you (RF) intend to make
that abrupt change of topic there?

That's not how I'd use "so"
Cha​cun à son goût.

Because models exist for
Infinity
¬Infinity
Foundation
¬Foundation
Choice
¬Choice
a (hypothetical) proof of
necessary.Infinity
necessary.Foundation
necessary.Choice
would be wrong.

tl;dr
Infinity, Foundation, Choice are
independent.

Of course, these days we know that
that's "quasi-modal" logic.

Uhm?
You mean?
https://philpapers.org/rec/GOLQEO-2
Quasi-Modal Equivalence of Canonical Structures

Yet, it's not the point here that
"classical" logic isn't monotonic.

Good to hear that it's not the point..
Classical is monotonic.

Yet, the idea that
"there's isn't a standard model of the integers",
and, also,
"there are multiple non-standard models of integers",
in set theory, and as well in number theory,
has a lot going on with
"there's an infinite integer in
some infinitudes of integers".

"has a lot going on with" = ?


Consider a finite set,
each element in its place.

A place is removed,
and not all elements have places.
We say:
the set's cardinality is finite.


Consider only all finite cardinals,
the ones which
cannot fit in a place removed.

For each finite cardinal,
inserting another element makes
a set which won't fit in its places,
one with a different cardinality.
Because finite.

Insert more and it still won't fit.
Insert all the cardinals and it still won't fit.

For each finite cardinal,
the cardinality of only all finite cardinals
is not its cardinality.

<drum.roll>
the cardinality of only all finite cardinals
is not a finite cardinality.
It is an infinite cardinality.
<cymbal.crash>

Check:
It isn't any cardinality which
cannot fit in a place removed.
It should fit in a place removed.

0@[1] 1@[2] 2@[3] 3@[4] ...
And check.

Weird as it is to fit in a place removed,
it must be so, or else
all the finite cardinals aren't
all the finite cardinals.

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On 1/17/2024 6:49 AM, WM wrote:

Le 16/01/2024 à 21:19, Dieter Heidorn a écrit :

[...]

For the less-than relation
there is no quantifier magic.

For anti.symmetric relations, such as less.than,
there is quantifier anti.magic.

| Assume P(x,y) ⇔ ¬P(y,x)
| Assume ∀x:∃y:P(x,y)
|
| ∀x:∃y:P(x,y)
| if and only if
| ¬∃x:∀y:¬P(x,y)
| if and only if
| ¬∃x:∀y:P(y,x)
| relabel 'x''z','y''x','z''y'
| ¬∃y:∀x:P(x,y)

Therefore,
P(x,y) ⇔ ¬P(y,x)
∀x:∃y:P(x,y)
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
¬∃y:∀x:P(x,y)
anti.magic

∀x:∃y≠x: x<y
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
¬∃y:∀y≠x: x<y
anti.magic

--- SoupGate-Win32 v1.05
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On 1/17/2024 7:55 AM, Richard Damon wrote:

On 1/17/24 6:53 AM, WM wrote:

Le 16/01/2024 à 18:11, Jim Burns a écrit :

(i)
(∀x ∈ (0,1]: y < x)  ⇒  y ≤ 0
true

Yes, that is correct.

Except we show it isn't for y = x/2

Well, for y = x/2, the antecedent is false,
thus the implication is true.
But WM probably isn't thinking of that.


(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
if and only if
0 < y ⇒ (∃x ∈ (0,1]: x ≤ y)

--- SoupGate-Win32 v1.05
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Fritz Feldhase schrieb am Mittwoch, 17. Januar 2024 um 20:15:47 UTC+1:

On Wednesday, January 17, 2024 at 12:49:48 PM UTC+1, WM wrote:

For the less-than relation there is no quantifier magic.

Was immer Deine "quantifier magic" auch sein soll; aber die (richtige) Reihenfolge der Quantoren ist wesentlich:

An e IN: Em e IN: n < m (true)

Em e IN: An e IN: n < m (false)

Both are true, but the average person cannot look into the infinite.
Therefore I have invented the simpler example with unit fractions.
If for every x ∈ (0, 1] there are ℵ smaller unit fractions, then ℵ
unit fractions are smaller than every x ∈ (0, 1] and lie left-hand side
of the whole interval. Otherwise a first one would appear within the
interval.

Please note: Without ℵ unit fractions left-hand side of (0, 1], the
above true statement cannot be true.

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/17/24 2:35 PM, Jim Burns wrote:

On 1/17/2024 7:55 AM, Richard Damon wrote:

On 1/17/24 6:53 AM, WM wrote:

Le 16/01/2024 à 18:11, Jim Burns a écrit :

(i)
(∀x ∈ (0,1]: y < x)  ⇒  y ≤ 0
true

Yes, that is correct.

Except we show it isn't for y = x/2

Well, for y = x/2, the antecedent is false,
thus the implication is true.
But WM probably isn't thinking of that.

What part of the antecedent (∀x ∈ (0,1]: y < x) is false?

x is such that x ∈ (0,1], and y is such that y < x

(∀x ∈ (0,1]: y < x)  ⇒  y ≤ 0
 if and only if
0 < y  ⇒  (∃x ∈ (0,1]: x ≤ y)

--- SoupGate-Win32 v1.05
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On 1/17/24 9:09 AM, WM wrote:

Le 17/01/2024 à 13:55, Richard Damon a écrit :

On 1/17/24 6:49 AM, WM wrote:

NUF(x) cannot grow anywhere from 0 to ℵo without passing finite values. >>

Why not?

Because of mathemtics. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
Regards, WM

And how does that say that NUF(x) can't grow to ℵo without passing
finite values.

That equation in fact proves that there can not be a smallest 1/n as the
'level gap' below 1/n is only 1/(n+1) of the distance between 0 and 1/n,
so there is room for at least n+1 more unit fractions below it.

You need to find a point where the gap is as big as 1/n, and it never is.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 1/17/2024 8:36 PM, Richard Damon wrote:

On 1/17/24 2:35 PM, Jim Burns wrote:

On 1/17/2024 7:55 AM, Richard Damon wrote:

On 1/17/24 6:53 AM, WM wrote:

Le 16/01/2024 à 18:11, Jim Burns a écrit :

(i)
(∀x ∈ (0,1]: y < x)  ⇒  y ≤ 0
true

Yes, that is correct.

Except we show it isn't for y = x/2

Well, for y = x/2, the antecedent is false,
thus the implication is true.
But WM probably isn't thinking of that.

What part of the antecedent
(∀x ∈ (0,1]: y < x) is false?

Excuse me, I was thinking of something else.

As I'm sure you know, if
x ∈ (0,1] ⇒ y < x
is false for any value of x,
then all of
(∀x ∈ (0,1]: y < x)
is false,
and all of
(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
is true.

For y > 0 and x = min{y/2,1}
x ∈ (0,1] ⇒ y < x
is false
(∀x ∈ (0,1]: y < x)
is false, and
(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
is true.

On the other hand,
for y ≤ 0
(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
is true,
whatever the case is for
∀x ∈ (0,1]: y < x

In sum,
(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
is true, but
its truth doesn't tell us about y


I confess that I don't know
what significance that formula has to WM.

It's possible that, for WM,
talking about
(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
replaces talking about
∀eps > 0: NUF(eps) ∉ ℕ

--- SoupGate-Win32 v1.05
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Le 18/01/2024 à 02:36, Richard Damon a écrit :

On 1/17/24 9:09 AM, WM wrote:

Le 17/01/2024 à 13:55, Richard Damon a écrit :

On 1/17/24 6:49 AM, WM wrote:

NUF(x) cannot grow anywhere from 0 to ℵo without passing finite values. >>>

Why not?

Because of mathemtics. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.

And how does that say that NUF(x) can't grow to ℵo without passing
finite values.

Because after every unit fraction the function NUF(x) is constant over d_n

0.

That equation in fact proves that there can not be a smallest 1/n as the 'level gap' below 1/n is only 1/(n+1) of the distance between 0 and 1/n,
so there is room for at least n+1 more unit fractions below it.

Nevertheless ***all*** unit fractions have gaps between each other. There
is no exception.

You need to find a point where the gap is as big as 1/n, and it never is.

We cannot investigate individuals within the dark domain.

Regards, WM

--- SoupGate-Win32 v1.05
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Le 18/01/2024 à 02:36, Richard Damon a écrit :

On 1/17/24 2:35 PM, Jim Burns wrote:

On 1/17/2024 7:55 AM, Richard Damon wrote:

On 1/17/24 6:53 AM, WM wrote:

Le 16/01/2024 à 18:11, Jim Burns a écrit :

(i)
(∀x ∈ (0,1]: y < x)  ⇒  y ≤ 0
true

Yes, that is correct.

Except we show it isn't for y = x/2

Well, for y = x/2, the antecedent is false,
thus the implication is true.
But WM probably isn't thinking of that.

What part of the antecedent (∀x ∈ (0,1]: y < x) is false?

x is such that x ∈ (0,1], and y is such that y < x

Only if y is less than all x ∈ (0,1], the implication holds. If the antecedent is violated, the implication is true nevertheless. But that is irrelevant.

(∀x ∈ (0,1]: y < x)  ⇒  y ≤ 0

Regards, WM

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Fritz Feldhase schrieb am Mittwoch, 17. Januar 2024 um 21:27:13 UTC+1:

On Wednesday, January 17, 2024 at 8:53:56 PM UTC+1, WM wrote:

Fritz Feldhase schrieb am Mittwoch, 17. Januar 2024 um 20:15:47 UTC+1:

die (richtige) Reihenfolge der Quantoren ist wesentlich:

An e IN: Em e IN: n < m (true)

Em e IN: An e IN: n < m (false)

Both are true,

Nope.

Yes, a mistake. Both are wrong. But the first statement is true for
*definable* numbers.

Regards, WM

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Le 18/01/2024 à 06:15, Jim Burns a écrit :

I confess that I don't know
what significance that formula has to WM.

If ∀x ∈ (0, 1]: NUF(x) = ℵo was true, it would prove negative unit fractions y.

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/18/24 4:09 AM, WM wrote:

Le 18/01/2024 à 02:36, Richard Damon a écrit :

On 1/17/24 9:09 AM, WM wrote:

Le 17/01/2024 à 13:55, Richard Damon a écrit :

On 1/17/24 6:49 AM, WM wrote:

NUF(x) cannot grow anywhere from 0 to ℵo without passing finite
values.

Why not?

Because of mathemtics. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.

And how does that say that NUF(x) can't grow to ℵo without passing
finite values.

Because after every unit fraction the function NUF(x) is constant over d_n

So? That doesn't mean that NUF(x) can't instantly grow to infinity
between 0 and the range (0,1]

0.

That equation in fact proves that there can not be a smallest 1/n as
the 'level gap' below 1/n is only 1/(n+1) of the distance between 0
and 1/n, so there is room for at least n+1 more unit fractions below it.

Nevertheless ***all*** unit fractions have gaps between each other.
There is no exception.

And again, we aren't talking BETWEEN unit fractions, but between 0 and
(0,1].

You need to find a point where the gap is as big as 1/n, and it never is.

We cannot investigate individuals within the dark domain.

You can not investigate the dark domain because it doesn't exist.

Regards, WM

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 1/18/2024 4:15 AM, WM wrote:

Le 18/01/2024 à 06:15, Jim Burns a écrit :

In sum,
(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
is true, but
its truth doesn't tell us about y

I confess that I don't know
what significance that formula has to WM.

If
∀x ∈ (0, 1]: NUF(x) = ℵo
was true,
it would prove negative unit fractions y.

How does
(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
and
∀x ∈ (0,1]: NUF(x) = ℵ₀
prove negative unit fractions y?

----
In arithmetic,
∀k <ℵ₀: ∀j <ℵ₀: j+k <ℵ₀

∀k <ℵ₀: ∀i <ℵ₀: i < |{j+k | j <ℵ₀}|

∀k <ℵ₀: ¬∃i <ℵ₀: i = |{j+k | j <ℵ₀}|

∀k <ℵ₀: ¬( |{j+k | j <ℵ₀}| <ℵ₀ )

----
In arithmetic,
∀k ∈ℕ₁: ∀j ∈ℕ: 0 < ⅟(j+k) ≤ 1

∀k ∈ℕ₁: {⅟(j+k) | j ∈ℕ} ⊆ (0,1] ∧
∀i ∈ℕ: i < |{⅟(j+k) | j ∈ℕ}|

∀k ∈ℕ₁: {⅟(j+k) | j ∈ℕ} ⊆ (0,1] ∧
¬∃i ∈ℕ: i = |{⅟(j+k) | j ∈ℕ}|

∀k ∈ℕ₁: {⅟(j+k) | j ∈ℕ} ⊆ (0,1] ∧
¬( |{⅟(j+k) | j ∈ℕ}| < |ℕ| )

At what formula, if any, have we stopped
using arithmetic?

Keep in mind that
_those formulas_ aren't dark.
You can see them on your screen.

--- SoupGate-Win32 v1.05
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Le 18/01/2024 à 13:44, FromTheRafters a écrit :

Yes, there are gaps in Q+ with respect to the positive reals fractional parts.

That means NUF(x) does not increase by more than 1 without stopping
afterwards. It starts with 0 and not with ℵ.

Regards, WM

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Le 18/01/2024 à 13:24, Jim Burns a écrit :

On 1/18/2024 4:15 AM, WM wrote:

Le 18/01/2024 à 06:15, Jim Burns a écrit :

In sum,
(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
is true, but
its truth doesn't tell us about y

I confess that I don't know
what significance that formula has to WM.

If
∀x ∈ (0, 1]: NUF(x) = ℵo
was true,
it would prove negative unit fractions y.

How does
(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
and
∀x ∈ (0,1]: NUF(x) = ℵ₀
prove negative unit fractions y?

For every point x > 0 there are ℵ smaller unit fractions.

There exists no point x > 0 without ℵ smaller unit fractions.

There exist ℵ negative unit fractions.

Regards, WM

--- SoupGate-Win32 v1.05
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Le 18/01/2024 à 13:51, Richard Damon a écrit :

On 1/18/24 4:09 AM, WM wrote:

Because after every unit fraction the function NUF(x) is constant over d_n

So? That doesn't mean that NUF(x) can't instantly grow to infinity
between 0 and the range (0,1]

It does mean exactly this. Otherwise:

If for every point x > 0 there are ℵ smaller unit fractions,
then there exists no point x > 0 without ℵ smaller unit fractions.
Then there exist ℵ negative unit fractions.

Regards, WM

--- SoupGate-Win32 v1.05
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Le 18/01/2024 à 18:02, FromTheRafters a écrit :

WM wrote :

Le 18/01/2024 à 13:44, FromTheRafters a écrit :

Yes, there are gaps in Q+ with respect to the positive reals fractional
parts.

That means NUF(x) does not increase by more than 1 without stopping
afterwards.

No, it doesn't.

In mathematics it does. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

It starts with 0 and not with ℵ.

It is not an action.

Hogwash. It is a function.

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/18/2024 11:45 AM, WM wrote:

Le 18/01/2024 à 13:24, Jim Burns a écrit :

On 1/18/2024 4:15 AM, WM wrote:

Le 18/01/2024 à 06:15, Jim Burns a écrit :

In sum,
(∀x ∈ (0,1]: y < x)  ⇒  y ≤ 0
is true, but
its truth doesn't tell us about y

I confess that I don't know
what significance that formula has to WM.

If
∀x ∈ (0, 1]: NUF(x) = ℵo
was true,
it would prove negative unit fractions y.

How does
(∀x ∈ (0,1]: y < x)  ⇒  y ≤ 0
and
∀x ∈ (0,1]: NUF(x) = ℵ₀
prove negative unit fractions y?

For every point x > 0
there are ℵ smaller unit fractions.

There exists no point x > 0 without
ℵ smaller unit fractions.

https://sneltraining.nl/then-a-miracle-occurs/

"then a miracle occurs"

"I think you should be more explicit
here in step two"

There exist ℵ negative unit fractions.

An ordinal after each final ordinal
is not any of the final ordinals.
Thus, it is non.final.

A non.final ordinal, by definition,
is followed by
an ordinal with the same cardinality.

--- SoupGate-Win32 v1.05
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On 1/18/2024 12:12 PM, Ross Finlayson wrote:

On Wednesday, January 17, 2024
at 1:07:02 PM UTC-8, Ross Finlayson wrote:

On Wednesday, January 17, 2024
at 8:29:31 AM UTC-8, Jim Burns wrote:

[...]

[...]

Here though,
the notion of the Aristotle's continuum, is that
the segment, is finite, but
is equi-partitioned with infinitely-many.
This is the most simple usual model of line-drawing or
putting pencil to paper, drawing, and lifting it,
relating exactly this model in space according to
any model in time, in exactly the course-of-passage,
in space through time.

My vague sense of these issues is that
Zeno argued very effectively that
that simplest theory of the line is too simple.

Our current theory of the line,
a theory stress.tested beyond the dreams of Zeno,
is that, in topological terms,
the line is _one component_ that it can't be
partitioned into two non.empty open sets.

Without that being true,
there can be curves y=f(x) which are continuous at
each point of the line, but which, nonetheless,
jump over lines between <x1,y1> and <x2,y2>

Since jumping over isn't the sort of behavior
which the things we consider curves get up to,
it seems unavoidable for us to require
_one component_

One effect of there being only one component is
that any countable series of points must have
a split with no series.point between sides.
But there must be a point.between,
or there will be more than one component.

One component implies a point.between, but
a point.between which is not in the series.
And so on, for each series of points.
_Each_ series not holding all points
implies that mere infinitely.small partitions
aren't enough for a good theory of the line.

--- SoupGate-Win32 v1.05
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On 1/18/24 11:51 AM, WM wrote:

Le 18/01/2024 à 13:51, Richard Damon a écrit :

On 1/18/24 4:09 AM, WM wrote:

Because after every unit fraction the function NUF(x) is constant
over d_n

So? That doesn't mean that NUF(x) can't instantly grow to infinity
between 0 and the range (0,1]

It does mean exactly this. Otherwise:

If for every point x > 0 there are ℵ smaller unit fractions,
then there exists no point x > 0 without ℵ smaller unit fractions.
Then there exist ℵ negative unit fractions.

Regards, WM

How do you get from

there exists no point x > 0 without ℵ smaller unit fractions.

to

Then there exist ℵ negative unit fractions.



You don't seem to understand how mathematics works with transfinite
numbers like ℵ

Likely because you are trying to treat it as a finite number, which it
isn't.

--- SoupGate-Win32 v1.05
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Le 18/01/2024 à 19:39, FromTheRafters a écrit :

WM formulated on Thursday :

Le 18/01/2024 à 18:02, FromTheRafters a écrit :

WM wrote :

Le 18/01/2024 à 13:44, FromTheRafters a écrit :

Yes, there are gaps in Q+ with respect to the positive reals fractional >>>>> parts.

That means NUF(x) does not increase by more than 1 without stopping
afterwards.

No, it doesn't.

In mathematics it does. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

It starts with 0 and not with ℵ.

It is not an action.

Hogwash. It is a function.

Then map it, done. No start here and finish there involved.

The start is NUF(x) = 0 for an arbitrary x =< 0.
Math proves ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
This excludes an increase from 0 to more than 1 at some point.
Finish is NUF(x) = ℵ for arbitrary x >= 1.

Regards, WM

Regards, WM

--- SoupGate-Win32 v1.05
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Le 19/01/2024 à 02:26, Richard Damon a écrit :

On 1/18/24 11:48 AM, WM wrote:

Le 18/01/2024 à 13:44, FromTheRafters a écrit :

Yes, there are gaps in Q+ with respect to the positive reals
fractional parts.

That means NUF(x) does not increase by more than 1 without stopping
afterwards. It starts with 0 and not with ℵ.

And then hits a gap in space between 0 and (0,1] where it jumps at not
unit fractions.

No. There is no gap in space. And NUF jumps only at unit fractions.

Regards, WM

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Le 19/01/2024 à 02:26, Richard Damon a écrit :

On 1/18/24 11:51 AM, WM wrote:

Le 18/01/2024 à 13:51, Richard Damon a écrit :

On 1/18/24 4:09 AM, WM wrote:

Because after every unit fraction the function NUF(x) is constant
over d_n

So? That doesn't mean that NUF(x) can't instantly grow to infinity
between 0 and the range (0,1]

It does mean exactly this. Otherwise:

If for every point x > 0 there are ℵ smaller unit fractions,
then there exists no point x > 0 without ℵ smaller unit fractions.
Then there exist ℵ negative unit fractions.

How do you get from

there exists no point x > 0 without ℵ smaller unit fractions.

to

Then there exist ℵ negative unit fractions.

There are ℵ unit fractions in total. No point of (0, 1] has ℵ at its right-hand side, because every point has ℵ at its left-hand side. The interval (0, 1] is nothing but all its points. Therefore ℵ unit
fractions are at the left-hand side of the interval (0, 1].


Regards, WM

--- SoupGate-Win32 v1.05
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Le 18/01/2024 à 19:55, Jim Burns a écrit :

On 1/18/2024 11:45 AM, WM wrote:

For every point x > 0
there are ℵ smaller unit fractions.

There exists no point x > 0 without
ℵ smaller unit fractions.

"I think you should be more explicit
here in step two"

No, this is simply a reformulation based on logic.
If all have property P, then none exists without property P.

There exist ℵ negative unit fractions.

An ordinal after each final ordinal
is not any of the final ordinals.
Thus, it is non.final.

A non.final ordinal, by definition,
is followed by
an ordinal with the same cardinality.

Waffle.

There are ℵ unit fractions in total. No point of (0, 1] has ℵ at its right-hand side, because every point has ℵ at its left-hand side. The interval (0, 1] is nothing but all its points. Therefore ℵ unit
fractions are at the left-hand side of the interval (0, 1].


Regards, WM

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Le 19/01/2024 à 02:26, Richard Damon a écrit :

On 1/18/24 4:12 AM, WM wrote:

Le 18/01/2024 à 02:36, Richard Damon a écrit :

On 1/17/24 2:35 PM, Jim Burns wrote:

(i)
(∀x ∈ (0,1]: y < x)  ⇒  y ≤ 0
true

Yes, that is correct.

Except we show it isn't for y = x/2

Well, for y = x/2, the antecedent is false,
thus the implication is true.
But WM probably isn't thinking of that.

What part of the antecedent (∀x ∈ (0,1]: y < x) is false?

x is such that x ∈ (0,1], and y is such that y < x

Only if y is less than all x ∈ (0,1], the implication holds. If the
antecedent is violated, the implication is true nevertheless. But that
is irrelevant.

I never claimed that y was less than ALL x, only any particular x.

But the antecedent concerns y smaller than all x.

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/19/2024 3:45 AM, WM wrote:

Le 18/01/2024 à 19:55, Jim Burns a écrit :

On 1/18/2024 11:45 AM, WM wrote:

For every point x > 0
there are ℵ smaller unit fractions.

There exists no point x > 0 without
ℵ smaller unit fractions.

"I think you should be more explicit
here in step two"

No, this is simply
a reformulation based on logic.
If all have property P,
then none exists without property P.

I think you should be more explicit
in the step from
| There exists no point x > 0 without
| ℵ smaller unit fractions.
to
| There exist ℵ negative unit fractions.

There exist ℵ negative unit fractions.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 1/19/24 3:50 AM, WM wrote:

Le 19/01/2024 à 02:26, Richard Damon a écrit :

On 1/18/24 4:12 AM, WM wrote:

Le 18/01/2024 à 02:36, Richard Damon a écrit :

On 1/17/24 2:35 PM, Jim Burns wrote:

(i)
(∀x ∈ (0,1]: y < x)  ⇒  y ≤ 0
true

Yes, that is correct.

Except we show it isn't for y = x/2

Well, for y = x/2, the antecedent is false,
thus the implication is true.
But WM probably isn't thinking of that.

What part of the antecedent (∀x ∈ (0,1]: y < x) is false?

x is such that x ∈ (0,1], and y is such that y < x

Only if y is less than all x ∈ (0,1], the implication holds. If the
antecedent is violated, the implication is true nevertheless. But
that is irrelevant.

I never claimed that y was less than ALL x, only any particular x.

But the antecedent concerns y smaller than all x.

Regards, WM

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Le 19/01/2024 à 16:54, Richard Damon a écrit :

On 1/19/24 3:48 AM, WM wrote:

Le 19/01/2024 à 02:26, Richard Damon a écrit :

On 1/18/24 11:48 AM, WM wrote:

Le 18/01/2024 à 13:44, FromTheRafters a écrit :

Yes, there are gaps in Q+ with respect to the positive reals
fractional parts.

That means NUF(x) does not increase by more than 1 without stopping
afterwards. It starts with 0 and not with ℵ.

And then hits a gap in space between 0 and (0,1] where it jumps at not
unit fractions.

No. There is no gap in space. And NUF jumps only at unit fractions.

So you say, but can't show how to do that.

I need not do that or show how to do that. NUF is well defined so.

As to "No Gap" it depends on your definition of "gap", as due to the
density of the Reals and Rationals, ssince between any two elements of
those sets, there are an infinite number of other elements, any
operation based on "counting" can see a gap.

Between 0 ad the real interval (0, 1] there is nothing.

Therefore I use the formulation: There is no x > 0 without ℵ smaller
unit fractions (following from set theory: Every x > 0 has ℵ smaller
unit fractions). This formulation covers all points of (0, 1] without any exception. Therefore it claims negative unit fractions. Contradiction.

Regards, WM

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Le 19/01/2024 à 16:54, Richard Damon a écrit :

On 1/19/24 3:31 AM, WM wrote:

Le 19/01/2024 à 02:26, Richard Damon a écrit :

On 1/18/24 11:51 AM, WM wrote:

Le 18/01/2024 à 13:51, Richard Damon a écrit :

On 1/18/24 4:09 AM, WM wrote:

Because after every unit fraction the function NUF(x) is constant
over d_n

So? That doesn't mean that NUF(x) can't instantly grow to infinity
between 0 and the range (0,1]

It does mean exactly this. Otherwise:

If for every point x > 0 there are ℵ smaller unit fractions,
then there exists no point x > 0 without ℵ smaller unit fractions.
Then there exist ℵ negative unit fractions.

How do you get from

there exists no point x > 0 without ℵ smaller unit fractions.

to

Then there exist ℵ negative unit fractions.

There are ℵ unit fractions in total. No point of (0, 1] has ℵ at its
right-hand side, because every point has ℵ at its left-hand side. The
interval (0, 1] is nothing but all its points. Therefore ℵ unit
fractions are at the left-hand side of the interval (0, 1].

You can't use a property of the individual members to the set.

That is a myth of set theory which sometimes is valid and sometimes is
not. But here we need no set theory at all. If this property holds for all points, then it holds for the interval.

Yes, for all x in the set, there are ℵ unit fractions to its left.

That doesn't apply to the "Set" (0,1], since the set isn't a number.

I use n interval and I use the geometrical property "is left-hand side
of".

Note, the claim that (0,1] is nothing but all its points you make
elsewhere isn't a valid claim, as a set is a different thing then its members, which is part of the logic behind a set can't contain itself.

That logic is irrelevant here, because I use geometry of points and
intervals without reference to sets, just correct mathematics, so to say.

Regards, WM

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On 1/19/24 12:38 PM, WM wrote:

Le 19/01/2024 à 16:54, Richard Damon a écrit :

On 1/19/24 3:48 AM, WM wrote:

Le 19/01/2024 à 02:26, Richard Damon a écrit :

On 1/18/24 11:48 AM, WM wrote:

Le 18/01/2024 à 13:44, FromTheRafters a écrit :

Yes, there are gaps in Q+ with respect to the positive reals
fractional parts.

That means NUF(x) does not increase by more than 1 without stopping
afterwards. It starts with 0 and not with ℵ.

And then hits a gap in space between 0 and (0,1] where it jumps at
not unit fractions.

No. There is no gap in space. And NUF jumps only at unit fractions.

So you say, but can't show how to do that.

I need not do that or show how to do that. NUF is well defined so.

But nothing in NUF(x) says that in needs to have a value of 1 at any
actual real number.

As to "No Gap" it depends on your definition of "gap", as due to the
density of the Reals and Rationals, ssince between any two elements of
those sets, there are an infinite number of other elements, any
operation based on "counting" can see a gap.

Between 0 ad the real interval (0, 1] there is nothing.

Only if your domain of interest are the finites.

Therefore I use the formulation: There is no x > 0 without ℵ smaller
unit fractions (following from set theory: Every x > 0 has ℵ smaller
unit fractions). This formulation covers all points of (0, 1] without
any exception. Therefore it claims negative unit fractions. Contradiction.

Nope.

Yes, EVERY x > 0 has an infinite number of smaller unit fractions.

all that proves is that there is no smallest unit fraction.

You can't seem to understand unbounded sets, likely because your brain
power is too limited.

Regards, WM

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Le 19/01/2024 à 16:54, Richard Damon a écrit :

On 1/19/24 3:35 AM, WM wrote:

Le 18/01/2024 à 19:39, FromTheRafters a écrit :

WM formulated on Thursday :

Le 18/01/2024 à 18:02, FromTheRafters a écrit :

WM wrote :

Le 18/01/2024 à 13:44, FromTheRafters a écrit :

Yes, there are gaps in Q+ with respect to the positive reals
fractional parts.

That means NUF(x) does not increase by more than 1 without stopping >>>>>> afterwards.

No, it doesn't.

In mathematics it does. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

It starts with 0 and not with ℵ.

It is not an action.

Hogwash. It is a function.

Then map it, done. No start here and finish there involved.

The start is NUF(x) = 0 for an arbitrary x =< 0. Math proves ∀n ∈ ℕ: 1/n
- 1/(n+1) > 0.
This excludes an increase from 0 to more than 1 at some point.

It excludes an increase from 0 to more than 1 at any finite point.

Yes, of course. Any.

Not at any point.

Unit fractions sit only at finite points. Only there NUF can grow.

Finish is NUF(x) = ℵ for arbitrary x >= 1.

No, for any arbitrary finite x > 0.

Wrong. For an arbitrary eps > 0.

Regards, WM

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On 1/19/24 12:30 PM, WM wrote:

Le 19/01/2024 à 16:54, Richard Damon a écrit :

On 1/19/24 3:31 AM, WM wrote:

Le 19/01/2024 à 02:26, Richard Damon a écrit :

On 1/18/24 11:51 AM, WM wrote:

Le 18/01/2024 à 13:51, Richard Damon a écrit :

On 1/18/24 4:09 AM, WM wrote:

Because after every unit fraction the function NUF(x) is constant >>>>>>> over d_n

So? That doesn't mean that NUF(x) can't instantly grow to infinity >>>>>> between 0 and the range (0,1]

It does mean exactly this. Otherwise:

If for every point x > 0 there are ℵ smaller unit fractions,
then there exists no point x > 0 without ℵ smaller unit fractions. >>>>> Then there exist ℵ negative unit fractions.

How do you get from

there exists no point x > 0 without ℵ smaller unit fractions.

to

Then there exist ℵ negative unit fractions.

There are ℵ unit fractions in total. No point of (0, 1] has ℵ at its >>> right-hand side, because every point has ℵ at its left-hand side. The
interval (0, 1] is nothing but all its points. Therefore ℵ unit
fractions are at the left-hand side of the interval (0, 1].

You can't use a property of the individual members to the set.

That is a myth of set theory which sometimes is valid and sometimes is
not. But here we need no set theory at all. If this property holds for
all points, then it holds for the interval.

Says what theory?

Maye you are using an incorrect theour that doesn't understand the
distinction?

Yes, for all x in the set, there are ℵ unit fractions to its left.

That doesn't apply to the "Set" (0,1], since the set isn't a number.

I use n interval and I use the geometrical property "is left-hand side of".

But what says that (0,1] HAS a "left hand side"?

Note, the claim that (0,1] is nothing but all its points you make
elsewhere isn't a valid claim, as a set is a different thing then its
members, which is part of the logic behind a set can't contain itself.

That logic is irrelevant here, because I use geometry of points and
intervals without reference to sets, just correct mathematics, so to say.

No, you use incorrect mathematics.

Mathematics (and Geometry) is based on logic.

You need to show what logic your theory is working from, and how it
supports your claims.

Regards, WM

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On 1/19/2024 3:45 AM, WM wrote:

Le 18/01/2024 à 19:55, Jim Burns a écrit :

An ordinal after each final ordinal
is not any of the final ordinals.
Thus, it is non.final.

A non.final ordinal, by definition,
is followed by
an ordinal with the same cardinality.

Waffle.

α+1 is the ordinal next.after α
The set of ordinals preceding α+1 is
the set of ordinals preceding α and α

Represent an ordinal α as
the set of ordinals preceding α
∀β: β ∈ α ⟺ β < α
0 = {}
1 = {0}
2 = (0,1}
...
ω = {0,1,2,…}
ω+1 = {0,1,2,…;ω}
...

α+1 = α∪{α}

If
no 1.to.1 map exists from
the set α+1 of ordinals preceding α+1 to
the set α of ordinals preceding α
¬(α+1 ⇉ α)
then
set α+1 has a larger cardinality than set α
|α+1| > |α|

If
set α+1 has a larger cardinality than set α
then
α is a final ordinal.
αᑉ⁺¹

Define ω as
the set of final ordinals
ω = {α| αᑉ⁺¹}

ω is not a final ordinal.

| Assume otherwise.
| Assume ω is a final ordinal.
|
| No 1.to.1 map exists from
| the set ω+1 of ordinals preceding ω+1 to
| the set ω of ordinals preceding ω
| ¬(ω+1 ⇉ ω)
| ¬({0,1,2,…;ω} ⇉ {0,1,2,…})
|
| However,
| ω+1 = {0,1,2,…;ω} only holds final ordinals,
| thus only elements of the set ω
| Identity α ⟼ α is 1.to.1 and
| it is from ω+1 to ω
| ω+1 ⇉ ω
| Contradiction.

Therefore,
ω is not a final ordinal.
{0,1,2,…;ω} ⇉ {0,1,2,…}

There are ℵ unit fractions in total.

Also, there are ℵ₀ unit fractions in part.
... = |⅟ℕ∩(0,⅟k)| = |⅟ℕ∩(0,⅟k⁺¹)| = ...

No point of (0, 1] has ℵ at its right-hand side,
because every point has ℵ at its left-hand side.

Each point of (0,1] has, to its right,
unit fractions with a finite ordinal, and
to its left, unit fractions with an ordinal
_after_ each final ordinal, thus non.final.

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On 1/19/24 1:02 PM, WM wrote:

Le 19/01/2024 à 16:54, Richard Damon a écrit :

On 1/19/24 3:35 AM, WM wrote:

Le 18/01/2024 à 19:39, FromTheRafters a écrit :

WM formulated on Thursday :

Le 18/01/2024 à 18:02, FromTheRafters a écrit :

WM wrote :

Le 18/01/2024 à 13:44, FromTheRafters a écrit :

Yes, there are gaps in Q+ with respect to the positive reals
fractional parts.

That means NUF(x) does not increase by more than 1 without
stopping afterwards.

No, it doesn't.

In mathematics it does. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

It starts with 0 and not with ℵ.

It is not an action.

Hogwash. It is a function.

Then map it, done. No start here and finish there involved.

The start is NUF(x) = 0 for an arbitrary x =< 0. Math proves ∀n ∈ ℕ: >>> 1/n - 1/(n+1) > 0.
This excludes an increase from 0 to more than 1 at some point.

It excludes an increase from 0 to more than 1 at any finite point.

Yes, of course. Any.

Not at any point.

Unit fractions sit only at finite points. Only there NUF can grow.

Nothing in its definition says that.

That assumes that unit fractions are indexable from their smallest value
end, which they are not.

Finish is NUF(x) = ℵ for arbitrary x >= 1.

No, for any arbitrary finite x > 0.

Wrong. For an arbitrary eps > 0.

Only FINITE eps.

if eps is not finite, but some transfinite small number, your
definitions don't apply. In the transfinitely small gap between 0 and
(0,1] NUF(x) isn't defined, and can jump.

Regards, WM

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Le 19/01/2024 à 19:26, Jim Burns a écrit :

On 1/19/2024 3:45 AM, WM wrote:

Le 18/01/2024 à 19:55, Jim Burns a écrit :

An ordinal after each final ordinal
is not any of the final ordinals.
Thus, it is non.final.

A non.final ordinal, by definition,
is followed by
an ordinal with the same cardinality.

Waffle.

α+1 is the ordinal next.after α
The set of ordinals preceding α+1 is
the set of ordinals preceding α and α

Represent an ordinal α as
the set of ordinals preceding α
∀β: β ∈ α ⟺ β < α
0 = {}
1 = {0}
2 = (0,1}
...
ω = {0,1,2,…}
ω+1 = {0,1,2,…;ω}

See? The last ones are not definable. They have no FISON.

No point of (0, 1] has ℵ at its right-hand side,
because every point has ℵ at its left-hand side.

Each point of (0,1] has, to its right,
unit fractions with a finite ordinal, and
to its left, unit fractions with an ordinal
_after_ each final ordinal, thus non.final.

The interval (0, 1] has in this respect the same property as all its
points. Geometry. No matheologial nonsense, no quantifier magic.

Regards, WM

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Le 19/01/2024 à 19:22, Richard Damon a écrit :

On 1/19/24 1:02 PM, WM wrote:

Unit fractions sit only at finite points. Only there NUF can grow.

Nothing in its definition says that.

Try again.

That assumes that unit fractions are indexable from their smallest value
end, which they are not.

It assumes that all unit fractions are existing and are indexed by their denominator.

Finish is NUF(x) = ℵ for arbitrary x >= 1.

No, for any arbitrary finite x > 0.

Wrong. For an arbitrary eps > 0.

Only FINITE eps.

if eps is not finite, but some transfinite small number, your
definitions don't apply. In the transfinitely small gap between 0 and
(0,1] NUF(x) isn't defined, and can jump.

No, there are no indexed unit fractions in this gap (and there is no such
gap).

Regards, WM

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On 1/19/2024 2:22 PM, WM wrote:

Le 19/01/2024 à 19:26, Jim Burns a écrit :

On 1/19/2024 3:45 AM, WM wrote:

Le 18/01/2024 à 19:55, Jim Burns a écrit :

An ordinal after each final ordinal
is not any of the final ordinals.
Thus, it is non.final.

A non.final ordinal, by definition,
is followed by
an ordinal with the same cardinality.

Waffle.

α+1 is the ordinal next.after α
The set of  ordinals preceding α+1  is
the set of  ordinals preceding α and α

Represent an ordinal α as
the set of ordinals preceding α
∀β: β ∈ α  ⟺  β < α
0 = {}
1 = {0}
2 = (0,1}
...
ω = {0,1,2,…}
ω+1 = {0,1,2,…;ω}

See?
The last ones are not definable.
They have no FISON.

ω is the set of final ordinals.
ω = {α ∈ Ord| | ¬(α+1 ⇉ α)}

The last final ordinal not.exists.

| Assume otherwise.
| Assume ¬(λ+1 ⇉ λ) ∧ λ+2 ⇉ λ+1
|
| Exists g: λ+2 ⇉ λ+1
|
| Define f: λ+1 ⇉ λ
| if g(α) ≠ λ+1
| then f(α) = g(α)
| else f(α) = g(λ+1)
|
| f is 1.to.1 from λ+1 to λ
|
| However, ¬(λ+1 ⇉ λ)
| No such map exists.
| Contradiction.

Therefore,
the last final ordinal not.exists.

The last ones are not definable.
They have no FISON.

They not.exist.
"They have no FISON" is true the same way
"The present king of France is bald" is true.
In the least useful way possible.

Am I right to think that
you (Wolfgang Mückenheim of Hochschule Augsburg)
don't object to reasoning from contradictions?

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On 1/18/2024 10:09 PM, Ross Finlayson wrote:

On Thursday, January 18, 2024
at 12:28:54 PM UTC-8, Jim Burns wrote:

Our current theory of the line,
a theory stress.tested beyond the dreams of Zeno,
is that, in topological terms,
the line is _one component_ that it can't be
partitioned into two non.empty open sets.

Well there are lots of topologies.

There is one meaning I intend for "component",
as used in topology.area.of.study.

One meaning across all
topologies.set.collections.designated.the.opens.

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On 1/19/24 2:26 PM, WM wrote:

Le 19/01/2024 à 19:22, Richard Damon a écrit :

On 1/19/24 1:02 PM, WM wrote:

Unit fractions sit only at finite points. Only there NUF can grow.

Nothing in its definition says that.

Try again.

You try again. Where does the definiton of NUF(x) say what you are claiming.

It only has values define at finite value. Nothing in the definiton of
NUF(x) talks about the difference of NUF(x) and a prior NUF(x)

That assumes that unit fractions are indexable from their smallest
value end, which they are not.

It assumes that all unit fractions are existing and are indexed by their denominator.

which indexes them from 1/1, which is the LARGEST value of x given to NUF(x)

Finish is NUF(x) = ℵ for arbitrary x >= 1.

No, for any arbitrary finite x > 0.

Wrong. For an arbitrary eps > 0.

Only FINITE eps.

if eps is not finite, but some transfinite small number, your
definitions don't apply. In the transfinitely small gap between 0 and
(0,1] NUF(x) isn't defined, and can jump.

No, there are no indexed unit fractions in this gap (and there is no
such gap).

and no reason NUF(x) can't jump to infinity for every unit fraction.

Regards, WM

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Le 19/01/2024 à 23:26, FromTheRafters a écrit :

WM wrote :

Le 19/01/2024 à 19:26, Jim Burns a écrit :

On 1/19/2024 3:45 AM, WM wrote:

Le 18/01/2024 à 19:55, Jim Burns a écrit :

An ordinal after each final ordinal
is not any of the final ordinals.
Thus, it is non.final.

A non.final ordinal, by definition,
is followed by
an ordinal with the same cardinality.

Waffle.

α+1 is the ordinal next.after α
The set of ordinals preceding α+1 is
the set of ordinals preceding α and α

Represent an ordinal α as
the set of ordinals preceding α
∀β: β ∈ α ⟺ β < α
0 = {}
1 = {0}
2 = (0,1}
...
ω = {0,1,2,…}
ω+1 = {0,1,2,…;ω}

See? The last ones are not definable. They have no FISON.

Last ones? These are ordinals, ω has an IISON.

Definabl numbers have FISONs.

Regards, WM

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Le 19/01/2024 à 21:33, Jim Burns a écrit :

"They have no FISON" is true the same way
"The present king of France is bald" is true.
In the least useful way possible.

Wrong. There is no present king but, assuming actual infinity, there is
ω.

Regards, WM

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Le 19/01/2024 à 22:48, Richard Damon a écrit :

On 1/19/24 2:26 PM, WM wrote:

Le 19/01/2024 à 19:22, Richard Damon a écrit :

On 1/19/24 1:02 PM, WM wrote:

Unit fractions sit only at finite points. Only there NUF can grow.

Nothing in its definition says that.

Try again.

You try again. Where does the definiton of NUF(x) say what you are claiming.

NUF(x) counts the unit fractions between 0 and x.

It only has values define at finite value. Nothing in the definiton of
NUF(x) talks about the difference of NUF(x) and a prior NUF(x)

This formula does: ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0

That assumes that unit fractions are indexable from their smallest
value end, which they are not.

It assumes that all unit fractions are existing and are indexed by their
denominator.

which indexes them

Yes.

from 1/1,

No, the index has no direction. It is but a number, namely the
denominator.

Regards, WM

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On 1/19/24 5:49 PM, WM wrote:

Le 19/01/2024 à 21:33, Jim Burns a écrit :

"They have no FISON" is true the same way
"The present king of France is bald" is true.
In the least useful way possible.

Wrong. There is no present king but, assuming actual infinity, there is ω.

Regards, WM

Which isn't a Natural Number.

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Ross Finlayson schrieb:

It's pretty simple that anything in proof theory has a model in model theory.

You are a complete moron. A proof theoretical statement
is an existential statement of the form:

T |- A <=>

∃j T |- j: A

T is syntactically derivable from A when there is a
judgement j, i.e. a deduction of A from T. A model
theoretical statement is an universal statement of the form

T |= A <=>

∀M (M[T]=1 => M[A]=1)

T is semantically derivable from A when every model
M that satisfies T also satisfies A.


There is no theorem anywhere in classical proof theory and
model theory that would say a proof is a model.

There is something in non-classical proof theory where
a judgegement indicates a model.

And then there is Gödels completness theorem, which says
that if the is no judgement then there is a counter model.

What are you talking about peanut brain?

--- SoupGate-Win32 v1.05
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peanut brain, you need some real spanking. You are talking
nonsense all day long all day in. What are you trying
the British Museum approach. Hope that you one day say s
something sensible , you monkey with a type writer.

Mild Shock schrieb:

Ross Finlayson schrieb:

It's pretty simple that anything in proof theory has a model in model
theory.

You are a complete moron. A proof theoretical statement
is an existential statement of the form:

T |- A <=>

∃j T |- j: A

T is syntactically derivable from A when there is a
judgement j, i.e. a deduction of A from T. A model
theoretical statement is an universal statement of the form

T |= A <=>

∀M (M[T]=1 => M[A]=1)

T is semantically derivable from A when every model
M that satisfies T also satisfies A.

There is no theorem anywhere in classical proof theory and
model theory that would say a proof is a model.

There is something in non-classical proof theory where
a judgegement indicates a model.

And then there is Gödels completness theorem, which says
that if the is no judgement then there is a counter model.

What are you talking about peanut brain?

--- SoupGate-Win32 v1.05
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Idiot brainless spam without any content.
Just random nonsense stringed together.

Rossy Boy the ape sitting in the Britsh Museum
in front of his typewriter.

Ross Finlayson schrieb:

On Friday, January 19, 2024 at 12:51:15 PM UTC-8, Jim Burns wrote:

On 1/18/2024 10:09 PM, Ross Finlayson wrote:

On Thursday, January 18, 2024
at 12:28:54 PM UTC-8, Jim Burns wrote:

Our current theory of the line,
a theory stress.tested beyond the dreams of Zeno,
is that, in topological terms,
the line is _one component_ that it can't be
partitioned into two non.empty open sets.

Well there are lots of topologies.

There is one meaning I intend for "component",
as used in topology.area.of.study.

One meaning across all
topologies.set.collections.designated.the.opens.

Why not?

I'm unfamiliar with this usage of, "component".
Can you detail its definition?

Of course you know that "open" and "closed" aren't
together a binary property, in topology.

Then just saying so, Dedekind cuts only follow
equivalence classes of sequences that are Cauchy,
and initial ordinals only follow equivalence classes
of sets that are cardinals, and furthermore vary
according to the statuses of CH and GCH. I.e.
any definition you provide has to be written in
terms of equivalence classes of sequences, or series
if you will that are Cauchy.

Anyways can you help me know where to find
a definition of "component" or whether it's yours.

A definition of "continuous domain" is easier to find these days.

I'm glad I sat a mathematics degree, it covered
many things expected to be known by all competent formalists.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

I'm glad I sat a mathematics degree

I didn't know that they give math degrees
in mexican border prisons.

Mild Shock schrieb:

Idiot brainless spam without any content.
Just random nonsense stringed together.

Rossy Boy the ape sitting in the Britsh Museum
in front of his typewriter.

Ross Finlayson schrieb:

On Friday, January 19, 2024 at 12:51:15 PM UTC-8, Jim Burns wrote:

On 1/18/2024 10:09 PM, Ross Finlayson wrote:

On Thursday, January 18, 2024
at 12:28:54 PM UTC-8, Jim Burns wrote:

Our current theory of the line,
a theory stress.tested beyond the dreams of Zeno,
is that, in topological terms,
the line is _one component_ that it can't be
partitioned into two non.empty open sets.

Well there are lots of topologies.

There is one meaning I intend for "component",
as used in topology.area.of.study.

One meaning across all
topologies.set.collections.designated.the.opens.

Why not?

I'm unfamiliar with this usage of, "component".
Can you detail its definition?

Of course you know that "open" and "closed" aren't
together a binary property, in topology.

Then just saying so, Dedekind cuts only follow
equivalence classes of sequences that are Cauchy,
and initial ordinals only follow equivalence classes
of sets that are cardinals, and furthermore vary
according to the statuses of CH and GCH.  I.e.
any definition you provide has to be written in
terms of equivalence classes of sequences, or series
if you will that are Cauchy.

Anyways can you help me know where to find
a definition of "component" or whether it's yours.

A definition of "continuous domain" is easier to find these days.


I'm glad I sat a mathematics degree, it covered
many things expected to be known by all competent formalists.

--- SoupGate-Win32 v1.05
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Le 20/01/2024 à 00:18, Richard Damon a écrit :

On 1/19/24 5:54 PM, WM wrote:

Le 19/01/2024 à 22:48, Richard Damon a écrit :

On 1/19/24 2:26 PM, WM wrote:

Le 19/01/2024 à 19:22, Richard Damon a écrit :

On 1/19/24 1:02 PM, WM wrote:

Unit fractions sit only at finite points. Only there NUF can grow.

Nothing in its definition says that.

Try again.

You try again. Where does the definiton of NUF(x) say what you are
claiming.

NUF(x) counts the unit fractions between 0 and x.

And why does that say it can have the value of 1?

That would require there to be a last unit fraction, and thus a highest Natural Number, which doesn't exist.

The sequence of unit fractions ends before zero. Therefore there is a last
one or more last ones. Since all have gaps, there is one last one.

No, the index has no direction. It is but a number, namely the denominator.

But it does, it indexes them from 1, and thus 1/1

Indexed DO have direction, since the first number is not the successor
of any number.

The first number has no direction, the last number has no direction, no
index has a direction. They are simply there or are not there. Actual
infinity says that all are there. That is presumed.

Regards, WM

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 1/20/24 6:00 AM, WM wrote:

Le 20/01/2024 à 00:18, Richard Damon a écrit :

On 1/19/24 5:54 PM, WM wrote:

Le 19/01/2024 à 22:48, Richard Damon a écrit :

On 1/19/24 2:26 PM, WM wrote:

Le 19/01/2024 à 19:22, Richard Damon a écrit :

On 1/19/24 1:02 PM, WM wrote:

Unit fractions sit only at finite points. Only there NUF can grow. >>>>>>

Nothing in its definition says that.

Try again.

You try again. Where does the definiton of NUF(x) say what you are
claiming.

NUF(x) counts the unit fractions between 0 and x.

And why does that say it can have the value of 1?

That would require there to be a last unit fraction, and thus a
highest Natural Number, which doesn't exist.

The sequence of unit fractions ends before zero. Therefore there is a
last one or more last ones. Since all have gaps, there is one last one.

Nope. You don't understand unbounded sets.

Note "More last ones" just means they continue.

No, the index has no direction. It is but a number, namely the
denominator.

But it does, it indexes them from 1, and thus 1/1

Indexed DO have direction, since the first number is not the successor
of any number.

The first number has no direction, the last number has no direction, no
index has a direction. They are simply there or are not there. Actual infinity says that all are there. That is presumed.

No, the first has a direction, the dirrection of the "next" operator.

First has a next but no previous, so it establishes a direction.

In fact, every number has a next that establishes the direction that
indexing runs in.

You don't seem to know the meaning of the words, or are just ignoring
the meaning because it is inconvient for your ideas (which makes you a
liar).

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/19/2024 6:29 PM, Ross Finlayson wrote:

On Friday, January 19, 2024
at 12:51:15 PM UTC-8, Jim Burns wrote:

On 1/18/2024 10:09 PM, Ross Finlayson wrote:

On Thursday, January 18, 2024
at 12:28:54 PM UTC-8, Jim Burns wrote:

Our current theory of the line,
a theory stress.tested beyond the dreams of Zeno,
is that, in topological terms,
the line is _one component_ that it can't be
partitioned into two non.empty open sets.

Well there are lots of topologies.

There is one meaning I intend for "component",
as used in topology.area.of.study.

One meaning across all
topologies.set.collections.designated.the.opens.

I'm unfamiliar with this usage of, "component".
Can you detail its definition?

We need "topological space" "disconnected" and
"connected component"

| A topological space is the most general type of
| a mathematical space that allows for the definition of
| limits, continuity, and connectedness. [...]
|
| A topology on a set X may be defined as
| a collection τ of subsets of X, called open sets
| and satisfying the following axioms:
|
| 1. The empty set and X itself belong to τ
| 2. Any arbitrary (finite or infinite) union of
| members of τ belongs to τ
| 3. The intersection of any finite number of
| members of τ belongs to τ
[1]

| A topological space X is said to be disconnected if
| it is the union of two disjoint non-empty open sets.
| Otherwise, X is said to be connected. [...]
|
| The connected component of a point x in X is
| the union of all connected subsets of X that contain x
| it is the unique largest (with respect to ⊆)
| connected subset of X that contains x
[2]

I said: ℝ has one component.
Better: ℝ is connected.

Delete ℼ from ℝ and ℝ\{ℼ} is disconnected.
ℝ\{ℼ} = (-∞,ℼ)∪(ℼ,+∞)

It's possible to map
(-∞,ℼ) ⟶ 0 and (ℼ,+∞) ⟶ 1
and be continuous at each point in ℝ\{ℼ}
But that's not what we want
when we ask for a continuous function.

In order to get what we want,
we must include ℼ in our discussion,
even if we never use ℼ
We must include uncountably.many points,
which even an endless process of naming can't use.

They must be there, used or unused, in order to
be able talk about continuous functions
with it meaning what we want.

[1]
https://en.wikipedia.org/wiki/Topological_space

[2]
https://en.wikipedia.org/wiki/Connected_space

Of course you know that "open" and "closed" aren't
together a binary property, in topology.

Point x is in the interior of set S if
some open set Oₓ which holds x is entirely in S

Point x is in the exterior of set S if
some open set O′ₓ which holds x is entirely not-in S

Point x is in the boundary of set S if
if it isn't in its interior or its exterior.

A closed set contains all of its boundary.
An open set contains none of its boundary.

And then, there are the sets containing
neither all nor none.


For some sets, their boundary is the empty set.
They contain all and none of the empty set.
They are closed and open.
They are called clopen sets,
which proves mathematicians have a sense of humor.

--- SoupGate-Win32 v1.05
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Richard Damon schrieb:

On 1/20/24 6:00 AM, WM wrote:

Le 20/01/2024 à 00:18, Richard Damon a écrit :

On 1/19/24 5:54 PM, WM wrote:

Le 19/01/2024 à 22:48, Richard Damon a écrit :

On 1/19/24 2:26 PM, WM wrote:

Le 19/01/2024 à 19:22, Richard Damon a écrit :

On 1/19/24 1:02 PM, WM wrote:

Unit fractions sit only at finite points. Only there NUF can grow. >>>>>>>

Nothing in its definition says that.

Try again.

You try again. Where does the definiton of NUF(x) say what you are
claiming.

NUF(x) counts the unit fractions between 0 and x.

And why does that say it can have the value of 1?

That would require there to be a last unit fraction, and thus a
highest Natural Number, which doesn't exist.

The sequence of unit fractions ends before zero. Therefore there is a
last one or more last ones. Since all have gaps, there is one last one.

Nope. You don't understand unbounded sets.

Note "More last ones" just means they continue.

No, the index has no direction. It is but a number, namely the
denominator.

But it does, it indexes them from 1, and thus 1/1

Indexed DO have direction, since the first number is not the
successor of any number.

The first number has no direction, the last number has no direction,
no index has a direction. They are simply there or are not there.
Actual infinity says that all are there. That is presumed.

No, the first has a direction, the dirrection of the "next" operator.

First has a next but no previous, so it establishes a direction.

In fact, every number has a next that establishes the direction that
indexing runs in.

Perhaps it would be better to use the term "ordered set" instead of "direction". See Cantor:

----------------------------------------------------------------------
"We call an aggregate M 'simply ordered' if a definite 'order of
precedence' (Rangordnung) rules over its elements m, so that, of
every two elements m_1 and m_2 one takes the 'lower' and the other
the 'higher' rank, and so that, if of three elements m_1, m_2 , and
m_3, m_1, say, is of lower rank than m_2, and m_2 is of lower rank
than m_3, then m_1 is of lower rank than m_3.
The relation of two elements m_1 and m_2 in which m_1 has the lower
rank in the given order of precedence and m_2 the higher, is expressed
by the formulae :

(1) m_1 <' m_2 , m_2 '> m_1 .

Thus, for example, every aggregate P of points defined on a straight
line is a simply ordered aggregate if, of every two points p_1 and p_2 belonging to it, that one whose co-ordinate (an origin and a positive
direction having been fixed upon) is the lesser is given the lower
rank."

(Georg Cantor: Contributions to the founding of the theory of
transfinite numbers. Dover Publications, 1915; p.110) ----------------------------------------------------------------------

Dieter Heidorn

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On 1/19/2024 5:49 PM, WM wrote:

Le 19/01/2024 à 21:33, Jim Burns a écrit :

"They have no FISON" is true the same way
"The present king of France is bald" is true.
In the least useful way possible.

Wrong.
There is no present king but,
assuming actual infinity,
there is ω.

ω is defined to be the set of final.ordinals
ω isn't a final.ordinal.
No final.ordinal has.no.FISON.
No element.of.ω has.no.FISON.
No final.ordinal is the last final.ordinal.
No element.of.ω is the last element.of.ω

ω isn't a final ordinal
|
| Assume otherwise.
| Assume ω is a final ordinal
|
| ω is the set of final.ordinals.
| ω+1 = {0,1,2,…;ω} ⊆ ω
| ≡: ω+1 ⇉ ω 1.to.1
| |ω| = |ω+1|
| ω isn't a final ordinal.
| Contradiction.

No final.ordinal is the last final.ordinal.
|
| Assume otherwise.
| Assume λ is the last final.ordinal.
| |λ| < |λ+1| = |λ+2|
|
| |λ+1| = |λ+2|
| exists g: λ+2 ⇉ λ+1 1.to.1
|
| Define f: λ+1 ⇉ λ
| such that
| if g(α) ≠ λ+1
| then f(α) = g(α)
| else f(α) = g(λ+1)
|
| exists f: λ+1 ⇉ λ 1.to.1
| |λ| = |λ+1|
|
| However,
| |λ| < |λ+1|
| not.exists f: λ+1 ⇉ λ 1.to.1
| Contradiction.

--- SoupGate-Win32 v1.05
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On 1/20/2024 1:05 PM, Ross Finlayson wrote:

On Saturday, January 20, 2024
at 4:47:48 AM UTC-8, Jim Burns wrote:

[...]

So, line-reals follow the integer continuum,
a unit magnitude of an integer,
for the integer part, and non-integer part,
of a real number, of the linear continuum.

Of course,
their establishment as continuous domains, involves
their relations, in function theory,
and topology.

What I consider essential:

Natural numbers are not too big to fail at
matching nearby natural numbers.
There are no 1.to.1 maps from 1.inserted.
¬(⟨0,…,k⟩ ⇇ ⟨0,…,k,k+1⟩) ⟺ k ∈ ℕ

Rational numbers do not have an inherent scale.
∀p ∈ ℚ: ∀n ∈ ℕ₁: p/n ∈ ℚ ∧ n⋅p ∈ ℚ

Real numbers do not have functions.which.jump
and are continuous everywhere.
For each non.empty split F,H of ℝ
exists x ∈ ℝ between F and H


These claims are sufficient to begin,
for our purposes, for _those_ purposes:
not.matching, scaling, not.jumping.

They can be augmented by
claims which are not.first.false

--- SoupGate-Win32 v1.05
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Le 20/01/2024 à 17:24, Jim Burns a écrit :

On 1/19/2024 5:49 PM, WM wrote:

ω is defined to be the set of final.ordinals
ω isn't a final.ordinal.

And many finite ordinals before ω are no final ordinals too. They have no FISONs. They are dark.

No final.ordinal has.no.FISON.

By definition.

No element.of.ω has.no.FISON.

Wrong. This would imply that no element of (0, 1] has less than ℵ LHS
unit fractions, but the interval has no LHS unit fractions. A beautiful contradiction in geometry.

Regards, WM

--- SoupGate-Win32 v1.05
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Le 20/01/2024 à 13:29, Richard Damon a écrit :

On 1/20/24 6:00 AM, WM wrote:

Le 20/01/2024 à 00:18, Richard Damon a écrit :

On 1/19/24 5:54 PM, WM wrote:

Le 19/01/2024 à 22:48, Richard Damon a écrit :

On 1/19/24 2:26 PM, WM wrote:

Le 19/01/2024 à 19:22, Richard Damon a écrit :

On 1/19/24 1:02 PM, WM wrote:

Unit fractions sit only at finite points. Only there NUF can grow. >>>>>>>

Nothing in its definition says that.

Try again.

You try again. Where does the definiton of NUF(x) say what you are
claiming.

NUF(x) counts the unit fractions between 0 and x.

And why does that say it can have the value of 1?

That would require there to be a last unit fraction, and thus a
highest Natural Number, which doesn't exist.

The sequence of unit fractions ends before zero. Therefore there is a
last one or more last ones. Since all have gaps, there is one last one.

Nope.

Note "More last ones" just means they continue.

They cannot continue beyond zero. More last ones would mean many between 0
and (0, 1]. Your infinitesimals.

The first number has no direction, the last number has no direction, no
index has a direction. They are simply there or are not there. Actual
infinity says that all are there. That is presumed.

No, the first has a direction, the dirrection of the "next" operator.

The operator does not belong to the number.

First has a next but no previous, so it establishes a direction.

The set has an order, may be called a direction, but not any number.

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/21/24 3:25 AM, WM wrote:

Le 20/01/2024 à 13:29, Richard Damon a écrit :

On 1/20/24 6:00 AM, WM wrote:

Le 20/01/2024 à 00:18, Richard Damon a écrit :

On 1/19/24 5:54 PM, WM wrote:

Le 19/01/2024 à 22:48, Richard Damon a écrit :

On 1/19/24 2:26 PM, WM wrote:

Le 19/01/2024 à 19:22, Richard Damon a écrit :

On 1/19/24 1:02 PM, WM wrote:

Unit fractions sit only at finite points. Only there NUF can grow. >>>>>>>>

Nothing in its definition says that.

Try again.

You try again. Where does the definiton of NUF(x) say what you are >>>>>> claiming.

NUF(x) counts the unit fractions between 0 and x.

And why does that say it can have the value of 1?

That would require there to be a last unit fraction, and thus a
highest Natural Number, which doesn't exist.

The sequence of unit fractions ends before zero. Therefore there is a
last one or more last ones. Since all have gaps, there is one last one.

Nope.

Note "More last ones" just means they continue.

They cannot continue beyond zero. More last ones would mean many between
0 and (0, 1]. Your infinitesimals.

But they do extend from what ever point you are at towards zero, and
there is an unbounded number of them, so you never get to the last one,
as such a thing doesn't exist.

The first number has no direction, the last number has no direction,
no index has a direction. They are simply there or are not there.
Actual infinity says that all are there. That is presumed.

No, the first has a direction, the dirrection of the "next" operator.

The operator does not belong to the number.

No, but it creates numbers, and establishes a direction of the number
sequence.

First has a next but no previous, so it establishes a direction.

The set has an order, may be called a direction, but not any number.

Then why did YOU try to establish a "forward" direction from the "last"
(you call it lowest) number?

NUF(x) == 1 requires that you can find a lowest x that is a unit
fraction that only has a forward towards larger unit fractions, (lower
natural numbers), but we know that every natural number has a higher
natural number, and thus every unit fraction has a lower unit fraction,
and thus no unit fraction is at the point where NUF(x) == 1.

Your whole logic is based on the idea that there must be a "last"
natural number, (to make a smallest unit fraction) but ZFC says such a
thing does not exist.

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/21/2024 3:34 AM, WM wrote:

Le 20/01/2024 à 17:24, Jim Burns a écrit :

On 1/19/2024 5:49 PM, WM wrote:

ω is defined to be the set of final.ordinals
ω isn't a final.ordinal.

And many finite ordinals before ω are
no final ordinals too.

Each ordinal is represented as
the set of ordinals preceding it.
α ∈ β ⟺ α < β

ω is defined to be the set of final.ordinals.
Each ordinal is represented as
the set of ordinals preceding it.

| ∀α:( α ∈ ω ⟺ ¬(α ⇇ α∪{α}) )
is the same as
| ∀α:( α < ω ⟺ ¬(α ⇇ α∪{α}) )

| α is in ω iff α is not.final
is the same as
| α is before ω iff α is not.final

¬∃α < ω: ¬¬(α ⇇ α∪{α})
No not.final.ordinal is before ω

They have no FISONs.
They are dark.

They are not in/before ω

I think that,
when you (WM) see "infinite"
you think "ridiculously.big",
a thought I think at least partly because,
when you (WM) try to give a counter.example to Cantor,
you describe some ridiculously.big
(final) number.which.inserting.another.does.not.fit.

That's not what "infinite" means.
Infinitely.many is more than each
(final) number.which.inserting.another.does.not.fit,
even ridiculously.big ones.

No final.ordinal has.no.FISON.

By definition.

No element.of.ω has.no.FISON.

Wrong.

And yet,
final.ordinal == element.of.ω

How can one be true and the other false?

This would imply that
no element of (0, 1] has less than
ℵ LHS unit fractions, but
the interval has no LHS unit fractions.

In (0,x]
for each
(final) number.which.inserting.another.does.not.fit,
there are more unit fractions than that.
|⅟ℕ₁∩(0,x]| ∉ ω

In (x,1]
there is some
(final) number.which.inserting.another.does.not.fit
number of unit fractions.
|⅟ℕ₁∩(x,1]| ∈ ω

In (-∞,0]
there are no unit fractions.
|⅟ℕ₁∩(-∞,0]| = 0

A beautiful contradiction in geometry.

Yes,
the set ω of final.ordinals is not a final.ordinal.

However,
that isn't a contradiction.
We get that by augmenting what we mean
with only not.first.false claims.

--- SoupGate-Win32 v1.05
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Le 21/01/2024 à 14:03, Richard Damon a écrit :

On 1/21/24 3:25 AM, WM wrote:

The sequence of unit fractions ends before zero. Therefore there is a
last one or more last ones. Since all have gaps, there is one last one. >>>

Nope.

Note "More last ones" just means they continue.

They cannot continue beyond zero. More last ones would mean many between
0 and (0, 1]. Your infinitesimals.

But they do extend from what ever point you are at towards zero, and
there is an unbounded number of them, so you never get to the last one,
as such a thing doesn't exist.

Anyhow you cannot name all, because "all" implies that none is missing.

The first number has no direction, the last number has no direction,
no index has a direction. They are simply there or are not there.
Actual infinity says that all are there. That is presumed.

No, the first has a direction, the dirrection of the "next" operator.

The operator does not belong to the number.

No, but it creates numbers, and establishes a direction of the number sequence.

Yes, a direction of the sequence but not of any number.

First has a next but no previous, so it establishes a direction.

The set has an order, may be called a direction, but not any number.

Then why did YOU try to establish a "forward" direction from the "last"
(you call it lowest) number?

Because the sequence has an order.

Your whole logic is based on the idea that there must be a "last"
natural number, (to make a smallest unit fraction) but ZFC says such a
thing does not exist.

My logic is based on mathematics:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
∀1/n ∈ ℝ: (∀x ∈ (0,1]: 1/n < x) ⇒ 1/n ≤ 0.
∀x ∈ (0, 1] ∃^ℵo 1/n < x

Regards, WM

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Le 21/01/2024 à 20:58, Jim Burns a écrit :

Infinitely.many is more than each
(final) number.

Infinity is more than can be expressed by finite numbers. But all visible numbers are measured by the last temporarily last final or visible number,
and therefore are far less than ω or ℵ.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵ.

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/22/2024 3:16 AM, WM wrote:

Le 21/01/2024 à 20:58, Jim Burns a écrit :

Infinitely.many is more than each
(final) number.

Infinity is more than
can be expressed by finite numbers.

ω is the set of finite (final) ordinals.
ω is an infinite (not.final) ordinal.
α ∈ ω ⟺ ¬(a ⇇ a+1)
ω ⇇ ω+1


ordinal β is the set of preceding ordinals
α ∈ β ⟺ α < β

α+1 means α∪{α}

α ⇇ β
means
exists 1.to.1 map to α from β
means
¬(|α| < |β|)

α ∈ ω ⟺ |α| < |α+1|
¬(|ω| < |ω+1|)

|α| < |β| ⊻ α ⇇ β

non.empty ordinal.set S holds a first
S ⊆ Ord ∧ S≠{} :⇒
∃α∈S ∀β∈S α≤β

But all visible numbers are measured by
the last temporarily last
final or visible number,

No final number n measures all final numbers.
final: |n| < |n+1| < |n+2|

and therefore are far less than ω or ℵ.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵ.

∀n ∈ ℕ:
ℕ\{1,…,n} ⇇ ℕ
ℕ\{1,…,n} ⇉ ℕ

∀n ∈ ℕ:
¬(|ℕ\{1….,n}| < |ℕ|)
¬(|ℕ\{1,…,n}| > |ℕ|)
|ℕ\{1,…,n}| = |ℕ|

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On 1/22/24 3:16 AM, WM wrote:

Le 21/01/2024 à 20:58, Jim Burns a écrit :

Infinitely.many is more than each
(final) number.

Infinity is more than can be expressed by finite numbers. But all
visible numbers are measured by the last temporarily last final or
visible number, and therefore are far less than ω or ℵ.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵ.

Regards, WM

But there is no "last" even temporarily.

All you are saying really is that any chosen finite initial sequence of
the Natural Numbers leaves most of the Natural Numbers out.

So?

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