Le 18/12/2023 à 04:08, "Chris M. Thomasson" a écrit :

I am thinking of a number.

Better think of maths.

ZFC supplies for the function Number of Unit Fractions between 0 and x the result
NUF(x) = 0 for x =< 0 and NUF(x) = ℵo for x > 0.
So, according to ZFC the number of unit fractions grows between 0 and (0,
1] by more than 2.
This is impossible because between any two unit fractions there are "uncountably" many points which do not fit between 0 and (0, 1]. This
proves that ZFC does not deliver correct mathematics in this case, like in several others: https://www.researchgate.net/publication/376587391_The_seven_deadly_sins_of_set_theory

Regards, WM

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On 12/28/23 3:38 AM, WM wrote:

Le 18/12/2023 à 04:08, "Chris M. Thomasson" a écrit :

I am thinking of a number.

Better think of maths.

ZFC supplies for the function Number of Unit Fractions between 0 and x
the result
NUF(x) = 0 for x =< 0 and NUF(x) = ℵo for x > 0.
So, according to ZFC the number of unit fractions grows between 0 and
(0, 1] by more than 2.
This is impossible because between any two unit fractions there are "uncountably" many points which do not fit between 0 and (0, 1]. This
proves that ZFC does not deliver correct mathematics in this case, like
in several others: https://www.researchgate.net/publication/376587391_The_seven_deadly_sins_of_set_theory

Regards, WM

And between 0 and (0, 1] is NOT a pair of unit fractions, so the logic
doesn't fit, and there isn't a first unit fraction to work your argueent
that it goes from 0 to first unit fraction to something.

Of course ZFC doesn't deliver the corrct mathematics in this case, as
you are working in a mathematical system above ZFC. You don't seem to understand what ZFC IS, it isn't a theory about all mathematics, but a
theory of a simple mathematics (Not even reaching to the full power of
the Natural Numbers) that can have some useful properties to explore.

IT is well known that pushing ZFC too far, extending it beyond its
design, will tend to. make it "blow up", and you can't determine in the
system, if that has happened yet.

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Le 28/12/2023 à 14:07, Richard Damon a écrit :

Of course ZFC doesn't deliver the corrct mathematics in this case,

It claims to cover and correctly describe all unit fractions.

IT is well known that pushing ZFC too far, extending it beyond its
design, will tend to. make it "blow up", and you can't determine in the system, if that has happened yet.

In mathematics, I can determine that after every unit fraction NUF(x) is constant. Hence there is a first step of height 1.

Regards, WM

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On 12/30/23 6:27 AM, WM wrote:

Le 28/12/2023 à 14:07, Richard Damon a écrit :

Of course ZFC doesn't deliver the corrct mathematics in this case,

It claims to cover and correctly describe all unit fractions.

But never says that there is a smallest of them.

There is a "first" but the order is based on the order of the Natural
Number, so the first is 1/1. There is no Smallest.

That is the problem with your logic, you don't understand how unbounded
sets work.

IT is well known that pushing ZFC too far, extending it beyond its
design, will tend to. make it "blow up", and you can't determine in
the system, if that has happened yet.

In mathematics, I can determine that after every unit fraction NUF(x) is constant. Hence there is a first step of height 1.

Only by FALSELY assuming there IS a first step. Since there actually is
no first step, you can't get it down to one.

If there WAS a value of x that NUF(x) = 1, then that x must be the
smallest unit fraction, but we know that if x is a unit fraction, so is
x/2, which is smaller, so no smallest x exists for NUF(x) to be 1.

The resolution would need to be expanding your number set to include the infinitesimals, where there does exist a "smallest value x", but you
have explicitly stated that you are not using them.

Note also, ZFC can gernerate sets of other Number Systems, even
"exotics" ones, if you inject other axioms into the system, like your
assertion that there exist a smallest Unit Fraction. Since you have
shown a lack of understanding of exactly what you are doing, this is a
likely case for you, and your "dark numbers" may just be some exotics
that are not actually Natural Numbers, but numbers resulting from your
"extra" axioms, like there is a smallest unit fraction.

Regards, WM

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Le 30/12/2023 à 17:12, Richard Damon a écrit :

On 12/30/23 6:27 AM, WM wrote:

In mathematics, I can determine that after every unit fraction NUF(x) is
constant. Hence there is a first step of height 1.

Only by FALSELY assuming there IS a first step.

Wrong. This being constant proves that there is a first step.

If there WAS a value of x that NUF(x) = 1, then that x must be the
smallest unit fraction, but we know that if x is a unit fraction, so is
x/2, which is smaller, so no smallest x exists for NUF(x) to be 1.

One of these two claims is wrong:
1) After every unit fraction NUF(x) is constant.
2) For every n there is n+1.

The solution is dark numbers, which also appear in other contexts like the intersection of endsegments and the x-O-problem.

The resolution would need to be expanding your number set to include the infinitesimals, where there does exist a "smallest value x", but you
have explicitly stated that you are not using them.

There are only natural numbers and fractions concerned with these
problems.

Regards, WM

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On 12/30/23 11:40 AM, WM wrote:

Le 30/12/2023 à 17:12, Richard Damon a écrit :

On 12/30/23 6:27 AM, WM wrote:

In mathematics, I can determine that after every unit fraction NUF(x)
is constant. Hence there is a first step of height 1.

Only by FALSELY assuming there IS a first step.

Wrong. This being constant proves that there is a first step.

But it is constant for less than the distance x, so it can't be the
first step.

If there WAS a value of x that NUF(x) = 1, then that x must be the
smallest unit fraction, but we know that if x is a unit fraction, so
is x/2, which is smaller, so no smallest x exists for NUF(x) to be 1.

One of these two claims is wrong:
1) After every unit fraction NUF(x) is constant.
2) For every n there is n+1.

Why do one of these need to be wrong?

The solution is dark numbers, which also appear in other contexts like
the intersection of endsegments and the x-O-problem.

No, there seem to be a solution for a non-problem cause by bad thinking

The resolution would need to be expanding your number set to include
the infinitesimals, where there does exist a "smallest value x", but
you have explicitly stated that you are not using them.

There are only natural numbers and fractions concerned with these problems.

And with just Unit Fractions, NUF(x) is never 1, and you logic that says
so based on the FALSE assumption that there exists a Naturl Number n
that doesn't have a successor, so its unit fraction can be first.

In other words, your theory is a theory of Unicorns.

Regards, WM

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Fritz Feldhase schrieb am Samstag, 30. Dezember 2023 um 15:25:54 UTC+1:

NUF(x) ist KONSTANT auf (0, 1], und zwar gilt NUF(x) = aleph_0 für alle x in (0, 1].

That means you cannot distinguish ℵo unit fractions by choosing an x > 0
or by any other means.

Regards, WM

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On 12/30/23 12:23 PM, WM wrote:

Fritz Feldhase schrieb am Samstag, 30. Dezember 2023 um 15:25:54 UTC+1:

NUF(x) ist KONSTANT auf (0, 1], und zwar gilt NUF(x) = aleph_0 für
alle x in (0, 1].

That means you cannot distinguish ℵo unit fractions by choosing an x > 0
or by any other means.

Regards, WM

What are you smoking?

The fact that NUF(x) has the value of "aleph_0", means that NUF(x) isn't
a function on the finites (as aleph_0 isn't a finite number).

Note also, that it is a fact that aleph_0 + 1 is aleph_0, means that the concept "Constant" as used for finite numbers doesn't apply to it.

Thus we can still have aleph_0 unit fractions in the range of (0,1] that
are all distinguishable.

Your logic just doesn't work, because you don't understand the domain of requard for it and use it outside of its applicability.

It is a known fact that when you introduce infinites into math, you need
to change the rules, as finite math can not handle them.

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On 12/30/23 2:16 PM, WM wrote:

Le 30/12/2023 à 18:39, Richard Damon a écrit :

On 12/30/23 12:23 PM, WM wrote:

Fritz Feldhase schrieb am Samstag, 30. Dezember 2023 um 15:25:54 UTC+1:

NUF(x) ist KONSTANT auf (0, 1], und zwar gilt NUF(x) = aleph_0 für
alle x in (0, 1].

That means you cannot distinguish ℵo unit fractions by choosing an x

0 or by any other means.

Thus we can still have aleph_0 unit fractions in the range of (0,1]
that are all distinguishable.

Try it. Fail. Feldhase correctly has recognized that it is impossible by
the choice of an x to distinguish ℵo unit fractions which are at the left-hand side of every x > 0 that can be choosen.

Regards, WM

Why not?

Note, is isn't for EVERY, but for ANY.

You are using the wrong qualificaiton.

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Le 30/12/2023 à 18:39, Richard Damon a écrit :

On 12/30/23 12:23 PM, WM wrote:

Fritz Feldhase schrieb am Samstag, 30. Dezember 2023 um 15:25:54 UTC+1:

NUF(x) ist KONSTANT auf (0, 1], und zwar gilt NUF(x) = aleph_0 für
alle x in (0, 1].

That means you cannot distinguish ℵo unit fractions by choosing an x > 0 >> or by any other means.

Thus we can still have aleph_0 unit fractions in the range of (0,1] that
are all distinguishable.

Try it. Fail. Feldhase correctly has recognized that it is impossible by
the choice of an x to distinguish ℵo unit fractions which are at the left-hand side of every x > 0 that can be choosen.

Regards, WM

--- SoupGate-Win32 v1.05
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Richard Damon schrieb am Samstag, 30. Dezember 2023 um 20:29:45 UTC+1:

On 12/30/23 2:16 PM, WM wrote:

Thus we can still have aleph_0 unit fractions in the range of (0,1]
that are all distinguishable.

Try it. Fail. Feldhase correctly has recognized that it is impossible by the choice of an x to distinguish ℵo unit fractions which are at the left-hand side of every x > 0 that can be choosen.

Why not?

Try it. Then you will see it.

Note, is isn't for EVERY, but for ANY.

Any and every chosen unit fraction leaves ℵo smaller unit fractions undistinguished.

You are using the wrong qualificaiton.

You are lacking some qualification.

Regards, WM

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On 12/31/23 3:57 AM, WM wrote:

Richard Damon schrieb am Samstag, 30. Dezember 2023 um 20:29:45 UTC+1:

On 12/30/23 2:16 PM, WM wrote:

Thus we can still have aleph_0 unit fractions in the range of (0,1]
that are all distinguishable. >> Try it. Fail. Feldhase correctly has recognized that it is

impossible by>> the choice of an x to

distinguish ℵo unit fractions which are at the
left-hand side of
every x > 0 that can be choosen.

Why not?

Try it. Then you will see it.

I did. The fact you don't understand is your fault.

IF you want to insist that I write out the distinguishing for all, I
will reply do it for the full N_vis

"Distinguishing" must be for an individual, not a whole set, or your
logic is just incapable of handling infinite sets.

Note, is isn't for EVERY, but for ANY.

Any and every chosen unit fraction leaves ℵo smaller unit fractions undistinguished.

It leaves ℵo unit fractions that CAN BE distinguished.

You are using the wrong qualificaiton.

You are lacking some qualification.

Which one? or is it just "dark"

Regards, WM

Your inability to understand basic theory isn't a reason to presume extraordinary theory is needed.

Your logic is just bad.

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Fritz Feldhase schrieb am Sonntag, 31. Dezember 2023 um 22:01:00 UTC+1:

On Saturday, December 30, 2023 at 5:44:39 PM UTC+1, WM wrote:

the matrix

X O O O...
X O O O...
X O O O...
X O O O...
...
will never be covered by X.

Das hat vor allem etwas damit zu tun, dass sich eine "gegebene" Matrix nicht verändert,

But the covering X change their positions.

Nun kann man aber auch eine Matrix A = (a_n,m)_(n,m e IN) definieren, die statt
Xen z. B. die natürlichen Zahlen 1, 2, 3, ... in der ersten Spalte enthält (und
sonst überall 0):

1 0 0 ...
2 0 0 ...
3 0 0 ...
...

ODER aber auch eine Matrix B = (b_n,m)_(n,m e IN) , die die Zahlen 1, 2, 3, 4,
... VERTEILT über sämtliche "Positionen" der Matrix enthält, und zwar gemäß
der Definition b_n.m = <1000-mal gepostet>.

1 2 4 ...
3 5 8 ...
6 9 13 ...
...

If both matrices were of same size, then the covering would be possible.
What should hinder it? However it is impossible. That falsifies your
claim.

Regards, WM

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Le 31/12/2023 à 13:38, Richard Damon a écrit :

On 12/31/23 3:57 AM, WM wrote:

Richard Damon schrieb am Samstag, 30. Dezember 2023 um 20:29:45 UTC+1:

On 12/30/23 2:16 PM, WM wrote:

Thus we can still have aleph_0 unit fractions in the range of (0,1]
that are all distinguishable. >> Try it. Fail. Feldhase correctly has >>> recognized that it is

impossible by>> the choice of an x to

distinguish ℵo unit fractions which are at the
left-hand side of
every x > 0 that can be choosen.

Why not?

Try it. Then you will see it.

I did.

Which one is next to zero?

IF you want to insist that I write out the distinguishing for all, I
will reply do it for the full N_vis

N_vis is potentially infinite. I did never claim that I could nam

Any and every chosen unit fraction leaves ℵo smaller unit fractions
undistinguished.

It leaves ℵo unit fractions that CAN BE distinguished.

It leaves ℵo unit fractions which are left undistinguished.

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/1/24 9:31 AM, WM wrote:

Fritz Feldhase schrieb am Sonntag, 31. Dezember 2023 um 22:01:00 UTC+1:

On Saturday, December 30, 2023 at 5:44:39 PM UTC+1, WM wrote:

the matrix > > X O O O... > X O O O... > X O O O... > X O O O...
... > will never be covered by X.

Das hat vor allem etwas damit zu tun, dass sich eine "gegebene" Matrix
nicht verändert,

But the covering X change their positions.

And thus your arguement breaks.

Nun kann man aber auch eine Matrix A = (a_n,m)_(n,m e IN) definieren,
die statt Xen z. B. die natürlichen Zahlen 1, 2, 3, ... in der ersten
Spalte enthält (und sonst überall 0):
1 0 0 ... 2 0 0 ... 3 0 0 ... ...
ODER aber auch eine Matrix B = (b_n,m)_(n,m e IN) , die die Zahlen 1,
2, 3, 4, ... VERTEILT über sämtliche "Positionen" der Matrix enthält,
und zwar gemäß der Definition b_n.m = <1000-mal gepostet>.
1 2 4 ... 3 5 8 ... 6 9 13 ... ...

If both matrices were of same size, then the covering would be possible.
What should hinder it? However it is impossible. That falsifies your claim.

But they ARE of the sames size, both have alpha_0 rows and columns and elements.

Regards, WM

Also, you seem to have a very broken Usenet reader (or are jus being intentionally deceptive), as your references don't match the message you
are quoting. And the messages you are quoting don't seem to be from the
group you are posting in.

This doesn't say much about why people should trust what you say, since
you seem to naturally do "broken" methods.

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On 1/1/24 9:20 AM, WM wrote:

Le 31/12/2023 à 13:38, Richard Damon a écrit :

On 12/31/23 3:57 AM, WM wrote:

Richard Damon schrieb am Samstag, 30. Dezember 2023 um 20:29:45 UTC+1:

On 12/30/23 2:16 PM, WM wrote:

Thus we can still have aleph_0 unit fractions in the range of

(0,1] >> that are all distinguishable. >> Try it. Fail. Feldhase
correctly has recognized that it is

impossible by>> the choice of an x to

distinguish ℵo unit fractions which are at the left-hand side of
every x > 0 that can be choosen.

Why not?

Try it. Then you will see it.

I did.

Which one is next to zero?

None of them. Why does there need to be one?

IF you want to insist that I write out the distinguishing for all, I
will reply do it for the full N_vis

N_vis is potentially infinite. I did never claim that I could nam

No, N_vis IS infinite (as a set) because ALL Natural Numbers are infinite.

Any and every chosen unit fraction leaves ℵo smaller unit fractions
undistinguished.

It leaves ℵo unit fractions that CAN BE distinguished.

It leaves ℵo unit fractions which are left undistinguished.

So?

Just because we haven't written out a name for something doesn't mean we
can't.

It seems your "dark" is not a property of the number it self, but a
transit property of what we have done with them. A "dark number" loses
its darkness when we look at it and write out its name.

Thus, it isn't an interesting property for the number itself.

Regards, WM

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Le 01/01/2024 à 16:33, Richard Damon a écrit :

On 1/1/24 9:31 AM, WM wrote:

Fritz Feldhase schrieb am Sonntag, 31. Dezember 2023 um 22:01:00 UTC+1:

On Saturday, December 30, 2023 at 5:44:39 PM UTC+1, WM wrote:

the matrix > > X O O O... > X O O O... > X O O O... > X O O O...
... > will never be covered by X.

Das hat vor allem etwas damit zu tun, dass sich eine "gegebene" Matrix
nicht verändert,

But the covering X change their positions.

And thus your arguement breaks.

No.

If both matrices were of same size, then the covering would be possible.
What should hinder it? However it is impossible. That falsifies your claim.

But they ARE of the sames size, both have alpha_0 rows and columns and elements.

As my proof shows, ℵo is not a valid measure. (Note the words you are
trying to use are aleph_0 and quantifier.)

Also, you seem to have a very broken Usenet reader (or are jus being intentionally deceptive), as your references don't match the message you
are quoting. And the messages you are quoting don't seem to be from the
group you are posting in.

The reason is that Franz Fritsche aka Fritz Feldhase is so ill-mannered
that his writing are rejected by news readers. But as long as Google is
active, I can see them. Alassince Google requires many Captchas (probably triggered by the Korean spammers in sci.logic), I answer his contributions here. It may happen, that by manual addressing mistakes occur.

Regards, WM

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On 1/1/24 2:18 PM, WM wrote:

Le 01/01/2024 à 16:33, Richard Damon a écrit :

On 1/1/24 9:31 AM, WM wrote:

Fritz Feldhase schrieb am Sonntag, 31. Dezember 2023 um 22:01:00 UTC+1: >>>> On Saturday, December 30, 2023 at 5:44:39 PM UTC+1, WM wrote:

the matrix > > X O O O... > X O O O... > X O O O... > X O O O...
... > will never be covered by X.

Das hat vor allem etwas damit zu tun, dass sich eine "gegebene"
Matrix nicht verändert,

But the covering X change their positions.

And thus your arguement breaks.

No.

Yes, it does. Your stupidity to not understand it, doesn't change that
your logic is just broken.

If both matrices were of same size, then the covering would be
possible. What should hinder it? However it is impossible. That
falsifies your claim.

But they ARE of the sames size, both have alpha_0 rows and columns and
elements.

As my proof shows, ℵo is not a valid measure. (Note the words you are trying to use are aleph_0 and quantifier.)

WHy not? You don't seem to understand the math needed for "infinite
measures"

Also, you seem to have a very broken Usenet reader (or are jus being
intentionally deceptive), as your references don't match the message
you are quoting. And the messages you are quoting don't seem to be
from the group you are posting in.

The reason is that Franz Fritsche aka Fritz Feldhase is so ill-mannered
that his writing are rejected by news readers. But as long as Google is active, I can see them. Alassince Google requires many Captchas
(probably triggered by the Korean spammers in sci.logic), I answer his contributions here. It may happen, that by manual addressing mistakes
occur.

If someone is so ill-mannered as to have there messages being blocked at
the news-server leverl (which doesn't really sound to be a thing), intenntionally doing things to get around that is just also ill-mannered.

It seems you have the same sort of problem with usenet as you do with mathematics, not understanding the rules and thus just ignoring them.

The reference headers of a message are supposed to point to the message
you are responding to. Erasing the message and putting a different
message in the message, just shows you are ignorant of what you are
doing, just like using the wrong rules of mathematics says you don't
know what you are doing in it.

Perhaps his problem is that he is posting in German, in a group that is designated as an "English" language newsgroup (as most of sci.* is.)

Regards, WM

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Le 01/01/2024 à 20:52, Richard Damon a écrit :

On 1/1/24 2:18 PM, WM wrote:

As my proof shows, ℵo is not a valid measure. (Note the words you are
trying to use are aleph_0 and quantifier.)

WHy not?

Because the X cannot cover the matrix.

Regards, WM

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

On 1/1/24 9:20 AM, WM wrote:

Which one is next to zero?

None of them. Why does there need to be one?

Because we have a linear set with gaps between the members. That enforces
a first one.

It leaves ℵo unit fractions which are left undistinguished.

So?

Just because we haven't written out a name for something doesn't mean we can't.

But ℵo will remain unnamed forever.

It seems your "dark" is not a property of the number it self, but a
transit property of what we have done with them. A "dark number" loses
its darkness when we look at it and write out its name.

Yes. But for ℵo numbers the property cannot be changed.

Thus, it isn't an interesting property for the number itself.

It is interesting enough to disprove Cantor'stheory.

Regards, WM

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On 1/2/24 2:11 PM, WM wrote:

Le 01/01/2024 à 20:52, Richard Damon a écrit :

On 1/1/24 2:18 PM, WM wrote:

As my proof shows, ℵo is not a valid measure. (Note the words you are
trying to use are aleph_0 and quantifier.)

WHy not?

Because the X cannot cover the matrix.

Regards, WM

But the remapped ones do.

You are just pointing the gun the wrong way and shooting yourself.

--- SoupGate-Win32 v1.05
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Le 03/01/2024 à 00:55, Richard Damon a écrit :

On 1/2/24 2:11 PM, WM wrote:

Le 01/01/2024 à 20:52, Richard Damon a écrit :

On 1/1/24 2:18 PM, WM wrote:

As my proof shows, ℵo is not a valid measure. (Note the words you are >>>> trying to use are aleph_0 and quantifier.)

WHy not?

Because the X cannot cover the matrix.

But the remapped ones do.

It appears so:
XOOO... XXOO... XXOO... XXXO... ... XXXX...
XOOO... OOOO... XOOO... XOOO... ... XXXX...
XOOO... XOOO... OOOO... OOOO... ... XXXX...
XOOO... XOOO... XOOO... OOOO... ... XXXX... .............................................................................

But since never an O leaves, the apparent picture is not the whole
picture.

Le 28/12/2023 à 19:34, Richard Damon a écrit: Who expects that one
column of a matrix will ever fill the full area of the SAME matrix?


Regards, WM

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Le 02/01/2024 à 21:42, FromTheRafters a écrit :

After serious thinking WM wrote :

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

On 1/1/24 9:20 AM, WM wrote:

Which one is next to zero?

None of them. Why does there need to be one?

Because we have a linear set with gaps between the members.

There are no gaps between the members, those are commas.

∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
d_n is not a comma.

Regards, WM

--- SoupGate-Win32 v1.05
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Le 03/01/2024 à 00:55, Richard Damon a écrit :

On 1/2/24 2:17 PM, WM wrote:

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

On 1/1/24 9:20 AM, WM wrote:

Which one is next to zero?

None of them. Why does there need to be one?

Because we have a linear set with gaps between the members. That
enforces a first one.

Nope. You are using BOUNDED logic rules on an UNBOUNDED set.

I am using logic. You are using matheological belief.

It leaves ℵo unit fractions which are left undistinguished.

So?

Just because we haven't written out a name for something doesn't mean
we can't.

But ℵo will remain unnamed forever.

But all are nameable.

Not the last ℵo.

It seems your "dark" is not a property of the number it self, but a
transit property of what we have done with them. A "dark number" loses
its darkness when we look at it and write out its name.

Yes. But for ℵo numbers the property cannot be changed.

We can for any of them.

Not for the last ℵo.

Regards, WM

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Fritz Feldhase schrieb am Mittwoch, 3. Januar 2024 um 00:58:35 UTC+1:

On Tuesday, January 2, 2024 at 8:11:25 PM UTC+1, WM wrote:

the X cannot cover the matrix.

1. Instead "the X" we should use the infinitely many elements in IN,

namely: 1, 2, 3, 4, ...

THIS WAY we may express (or prescribe) which element occupies

which position in the considered matrices.

This can also be done with the X:

XOOO... XXOO... XXOO... XXXO... ... XXXX...
XOOO... OOOO... XOOO... XOOO... ... XXXX...
XOOO... XOOO... OOOO... OOOO... ... XXXX...
XOOO... XOOO... XOOO... OOOO... ... XXXX... ...........................................

2. Now the elements 1, 2, 3, 4, ... c a n (since they do) "cover" the

matrix B = (b_n,m)_(n,m e IN) defined with b_n,m = (m + n - 1)(m + n -
2)/2 + m for all n,m e IN.

3. On the other hand they do NOT "cover" the matrix A = (a_n,m)

defined with a_n,1 = n for all n e IN and a_n,m = 0 for all n e IN\{1},
m e IN.

Not even after having been arranged as matrix B. Therefore B, the matrix
of defined indices is much smaller than the matrix A of all positive
fractions.

By the way that is clear because ∀n ∈ ℕ: between n and n+1 there are
more than two fractions on the real line.

Regards, WM

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On 1/3/24 3:25 AM, WM wrote:

Le 03/01/2024 à 00:55, Richard Damon a écrit :

On 1/2/24 2:11 PM, WM wrote:

Le 01/01/2024 à 20:52, Richard Damon a écrit :

On 1/1/24 2:18 PM, WM wrote:

As my proof shows, ℵo is not a valid measure. (Note the words you
are trying to use are aleph_0 and quantifier.)

WHy not?

Because the X cannot cover the matrix.

But the remapped ones do.

It appears so:
XOOO... XXOO... XXOO... XXXO... ... XXXX...
XOOO... OOOO... XOOO... XOOO... ... XXXX...
XOOO... XOOO... OOOO... OOOO... ... XXXX...
XOOO... XOOO... XOOO... OOOO... ... XXXX... .............................................................................

But since never an O leaves, the apparent picture is not the whole picture.

Le 28/12/2023 à 19:34, Richard Damon a écrit: Who expects that one
column of a matrix will ever fill the full area of the SAME matrix?

Regards, WM

In other words, you don't understand what you are talking about.

That seems normal for you.

Using illogic, you can make anything seems "reasonable" until you
understand how things actually work.

--- SoupGate-Win32 v1.05
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Le 03/01/2024 à 12:58, Richard Damon a écrit :

On 1/3/24 3:28 AM, WM wrote:

Le 03/01/2024 à 00:55, Richard Damon a écrit :

It seems your "dark" is not a property of the number it self, but a
transit property of what we have done with them. A "dark number"
loses its darkness when we look at it and write out its name.

Yes. But for ℵo numbers the property cannot be changed.

We can for any of them.

Not for the last ℵo.

Yes, all of them.

"All" which have ℵo successors. But that are not all.

Just saying they aren't doesn't make it so.

Just saying they can be named does not make it so. After every named
number ℵo remain. That cannoit be changed.

Regards, WM

--- SoupGate-Win32 v1.05
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Le 03/01/2024 à 12:58, Richard Damon a écrit :

On 1/3/24 3:25 AM, WM wrote:

Le 03/01/2024 à 00:55, Richard Damon a écrit :

On 1/2/24 2:11 PM, WM wrote:

Le 01/01/2024 à 20:52, Richard Damon a écrit :

On 1/1/24 2:18 PM, WM wrote:

As my proof shows, ℵo is not a valid measure. (Note the words you >>>>>> are trying to use are aleph_0 and quantifier.)

WHy not?

Because the X cannot cover the matrix.

But the remapped ones do.

It appears so:
XOOO... XXOO... XXOO... XXXO... ... XXXX...
XOOO... OOOO... XOOO... XOOO... ... XXXX...
XOOO... XOOO... OOOO... OOOO... ... XXXX...
XOOO... XOOO... XOOO... OOOO... ... XXXX...
.............................................................................

But since never an O leaves, the apparent picture is not the whole picture. >>
Le 28/12/2023 à 19:34, Richard Damon a écrit: Who expects that one
column of a matrix will ever fill the full area of the SAME matrix?

In other words, you don't understand what you are talking about.

The sentence above was yours.

Regards, WM

--- SoupGate-Win32 v1.05
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Fritz Feldhase schrieb am Mittwoch, 3. Januar 2024 um 17:19:58 UTC+1:

On Wednesday, January 3, 2024 at 12:58:56 PM UTC+1, Richard Damon wrote:

On 1/3/24 3:28 AM, WM wrote:

Le 03/01/2024 à 00:55, Richard Damon a écrit :

On 1/2/24 2:17 PM, WM wrote:

We can for any of them.

Not for the last ℵo.

What does this even mean?!

It means that "all" natural numbers which have ℵo natural numbers
following are not all natural numbers.

It seems that WM considers "the last ℵo [natural numbers]" to be a fixed set
of natural numbers.

No, the visible numbers are potentially infinite. But ℵo numbers remain
dark forever. Otherwise only finitely many numbers would follow upon some visible number.

Regards, WM

--- SoupGate-Win32 v1.05
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Fritz Feldhase schrieb am Mittwoch, 3. Januar 2024 um 16:57:15 UTC+1:

On Wednesday, January 3, 2024 at 11:55:02 AM UTC+1, WM wrote:

∀n ∈ ℕ: between n and n+1 there are more than two fractions on the real

line.
WOW, das ist beinahe abelpreiswürdig,

It shows in a simple manner that for all intervals (n, n+1] there are more fractions than natural numbers. In mathematics this proves that all
indexing can only happen "in the limit", i.e., nowhere.

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/4/24 5:16 AM, WM wrote:

Le 03/01/2024 à 12:58, Richard Damon a écrit :

On 1/3/24 3:25 AM, WM wrote:

Le 03/01/2024 à 00:55, Richard Damon a écrit :

On 1/2/24 2:11 PM, WM wrote:

Le 01/01/2024 à 20:52, Richard Damon a écrit :

On 1/1/24 2:18 PM, WM wrote:

As my proof shows, ℵo is not a valid measure. (Note the words you >>>>>>> are trying to use are aleph_0 and quantifier.)

WHy not?

Because the X cannot cover the matrix.

But the remapped ones do.

It appears so:
XOOO... XXOO... XXOO... XXXO... ... XXXX...
XOOO... OOOO... XOOO... XOOO... ... XXXX...
XOOO... XOOO... OOOO... OOOO... ... XXXX...
XOOO... XOOO... XOOO... OOOO... ... XXXX...
.............................................................................

But since never an O leaves, the apparent picture is not the whole
picture.

Le 28/12/2023 à 19:34, Richard Damon a écrit: Who expects that one
column of a matrix will ever fill the full area of the SAME matrix?

In other words, you don't understand what you are talking about.

The sentence above was yours.

Right YOU DON'T UNDERERSTAND THE PROBLEM.

You are just proving yourself too stupid to discuss these things with.

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/4/24 5:13 AM, WM wrote:

Le 03/01/2024 à 12:58, Richard Damon a écrit :

On 1/3/24 3:28 AM, WM wrote:

Le 03/01/2024 à 00:55, Richard Damon a écrit :

It seems your "dark" is not a property of the number it self, but
a transit property of what we have done with them. A "dark number" >>>>>> loses its darkness when we look at it and write out its name.

Yes. But for ℵo numbers the property cannot be changed.

We can for any of them.

Not for the last ℵo.

Yes, all of them.

"All" which have ℵo successors. But that are not all.

Which ones don't?

Answer, Only the ones that are not themselves Natural Numbers!

Remember, a basic property of ALL Natural Numbers (by their definitions)
is that every Natural Number has a successor Natural Number, and every
Natural Number has a finite construction, and that finite construction
can be turned into a name.

Just saying they aren't doesn't make it so.

Just saying they can be named does not make it so. After every named
number ℵo remain. That cannoit be  changed.

But that doesn't mean they all cannot be named?

I have shown why they all can be named, and the fact that you don't
beleive it just shows you don't understand the actual mathematics of the
set.

Your problem seems to be that you are stuck in a logic system that just
can't handled unbounded sets.

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/4/24 5:20 AM, WM wrote:

Fritz Feldhase schrieb am Mittwoch, 3. Januar 2024 um 16:57:15 UTC+1:

On Wednesday, January 3, 2024 at 11:55:02 AM UTC+1, WM wrote:

∀n ∈ ℕ: between n and n+1 there are more than two fractions on the >> real line.

WOW, das ist beinahe abelpreiswürdig,

It shows in a simple manner that for all intervals (n, n+1] there are
more fractions than natural numbers. In mathematics this proves that all indexing can only happen "in the limit", i.e., nowhere.

Regards, WM

But that isn't the question/claim.

The claim is that in any interval, there are a countable infinite number
of rational numbers.

That means we can either form a bijection between them, or a pair of
injections (one each way) to show that neither set is fundamentally
bigger. (some attempts are allowed to fail, the question is does a
successful one exist).

THAT can be done, so we can show the sets (the rationals in an interval,
and the Natural Numbers) are the same size.

Your failure to understand that just shows your ignorance.

In particular, your inability to understand that there CAN be "failed"
mappings that don't work, shows your lack of understanding.

Yes, when you reject that idea, your mathematics just becomes contradictory.

--- SoupGate-Win32 v1.05
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On 1/4/24 5:24 AM, WM wrote:

Fritz Feldhase schrieb am Mittwoch, 3. Januar 2024 um 17:19:58 UTC+1:

On Wednesday, January 3, 2024 at 12:58:56 PM UTC+1, Richard Damon
wrote: > On 1/3/24 3:28 AM, WM wrote: > > Le 03/01/2024 à 00:55,
Richard Damon a écrit : > >> On 1/2/24 2:17 PM, WM wrote: > > > >

We can for any of them. > > > > > Not for the last ℵo.

What does this even mean?!

It means that "all" natural numbers which have ℵo natural numbers
following are not all natural numbers.

It seems that WM considers "the last ℵo [natural numbers]" to be a
fixed set of natural numbers.

No, the visible numbers are potentially infinite. But ℵo numbers remain dark forever. Otherwise only finitely many numbers would follow upon
some visible number.

Regards, WM

No, the "visible" numbers are ACTUALLY INFINITE in number (but always
finite in value). There is no "Potentially" here by the normal definition.

You logic is just based on bad expectations.

--- SoupGate-Win32 v1.05
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Le 04/01/2024 à 13:13, Richard Damon a écrit :

On 1/4/24 5:13 AM, WM wrote:

Just saying they can be named does not make it so. After every named
number ℵo remain. That cannot be  changed.

I have shown why they all can be named,

If it is possible, then do it or let those do it whom you expect they
could do it. If it is impossible, then you are a liar.

Regards, WM

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

On 1/5/24 5:32 AM, WM wrote:

Le 04/01/2024 à 13:13, Richard Damon a écrit :

On 1/4/24 5:13 AM, WM wrote:

Just saying they can be named does not make it so. After every named
number ℵo remain. That cannot be  changed.

I have shown why they all can be named,

If it is possible, then do it or let those do it whom you expect they
could do it. If it is impossible, then you are a liar.

Regards, WM

So, you don't understand that the Natural Numbers are exactly the
OPPOSITE of your "dark" concept. They can all be used individually, but
they can not be "individually named" as the whole set.

May claim is that I can name "any" number I choose. Say, Ten Billion,
three hundred and seventy five Million, six hundred and thirty seven
thousand, four hundred and twenty two.

I can do that to any individual number I want.

I admit I can't individually name ALL of them at once.


Now, they can be formulaically named at once, in fact, the notation

1, 2, 3, 4, ... effectively does that.

But your logic can't handle that, as it can't handle infinite sets.

--- SoupGate-Win32 v1.05
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Le 05/01/2024 à 16:04, Richard Damon a écrit :

On 1/5/24 5:32 AM, WM wrote:

Le 04/01/2024 à 13:13, Richard Damon a écrit :

On 1/4/24 5:13 AM, WM wrote:

Just saying they can be named does not make it so. After every named
number ℵo remain. That cannot be  changed.

I have shown why they all can be named,

If it is possible, then do it or let those do it whom you expect they
could do it. If it is impossible, then you are a liar.

May claim is that I can name "any" number I choose. Say, Ten Billion,
three hundred and seventy five Million, six hundred and thirty seven thousand, four hundred and twenty two.

I can do that to any individual number I want.

But you cannot want it for every natural number because after every that
you want, ℵ are following.

I admit I can't individually name ALL of them at once.

You can name and subtract all at once collectively, such that no
successors remain. You cannot do so with in dividual numbers. There always
ℵ successors remain. This is a big difference.

Now, they can be formulaically named at once, in fact, the notation

1, 2, 3, 4, ... effectively does that.

Yes, but the "..." show that this means handling them collectively.

But your logic can't handle that, as it can't handle infinite sets.

I can handle them collectively only - just like you. But I have recognized
the difference. You not.

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/8/24 5:28 AM, WM wrote:

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

On 1/5/24 5:32 AM, WM wrote:

Le 04/01/2024 à 13:13, Richard Damon a écrit :

On 1/4/24 5:13 AM, WM wrote:

Just saying they can be named does not make it so. After every
named number ℵo remain. That cannot be  changed.

I have shown why they all can be named,

If it is possible, then do it or let those do it whom you expect they
could do it. If it is impossible, then you are a liar.

May claim is that I can name "any" number I choose. Say, Ten Billion,
three hundred and seventy five Million, six hundred and thirty seven
thousand, four hundred and twenty two.

I can do that to any individual number I want.

But you cannot want it for every natural number because after every that
you want, ℵ are following.

Why can't i "want" it for every natural number. I can want for a lot.

Your logic is just "dark".

I admit I can't individually name ALL of them at once.

You can name and subtract all at once collectively, such that no
successors remain. You cannot do so with in dividual numbers. There
always ℵ successors remain. This is a big difference.

Right, each number can be used individually.

All numbers must be used collectively.

That is the nature of unbounded sets,

Now, they can be formulaically named at once, in fact, the notation

1, 2, 3, 4, ... effectively does that.

Yes, but the "..." show that this means handling them collectively.

but each can be done individually by the pattern

But your logic can't handle that, as it can't handle infinite sets.

I can handle them collectively only - just like you. But I have
recognized the difference. You not.

Regards, WM

So, what is the collective set of just dark numbers.

If they can be used collectively, you should be able to show such a set.

--- SoupGate-Win32 v1.05
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Le 08/01/2024 à 13:21, Richard Damon a écrit :

On 1/8/24 5:28 AM, WM wrote:

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

On 1/5/24 5:32 AM, WM wrote:

Le 04/01/2024 à 13:13, Richard Damon a écrit :

On 1/4/24 5:13 AM, WM wrote:

Just saying they can be named does not make it so. After every
named number ℵo remain. That cannot be  changed.

I have shown why they all can be named,

If it is possible, then do it or let those do it whom you expect they
could do it. If it is impossible, then you are a liar.

May claim is that I can name "any" number I choose. Say, Ten Billion,
three hundred and seventy five Million, six hundred and thirty seven
thousand, four hundred and twenty two.

I can do that to any individual number I want.

But you cannot want it for every natural number because after every that
you want, ℵ are following.

Why can't i "want" it for every natural number.

Because every wanted number has ℵ successors. You cannot remove them individually but only collectively. The latter shows that all can be
removed.

Your logic is just "dark".

The blind sees only darkness.

I admit I can't individually name ALL of them at once.

You can name and subtract all at once collectively, such that no
successors remain. You cannot do so with in dividual numbers. There
always ℵ successors remain. This is a big difference.

Right, each number can be used individually.

All numbers must be used collectively.

That is the nature of unbounded sets,

Yes. That shows that not all numbers can be addressed individually. That
is what I call dark.

Regards, WM

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

On 1/9/24 12:10 PM, WM wrote:

Le 08/01/2024 à 13:21, Richard Damon a écrit :

On 1/8/24 5:28 AM, WM wrote:

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

On 1/5/24 5:32 AM, WM wrote:

Le 04/01/2024 à 13:13, Richard Damon a écrit :

On 1/4/24 5:13 AM, WM wrote:

Just saying they can be named does not make it so. After every
named number ℵo remain. That cannot be  changed.

I have shown why they all can be named,

If it is possible, then do it or let those do it whom you expect
they could do it. If it is impossible, then you are a liar.

May claim is that I can name "any" number I choose. Say, Ten
Billion, three hundred and seventy five Million, six hundred and
thirty seven thousand, four hundred and twenty two.

I can do that to any individual number I want.

But you cannot want it for every natural number because after every
that you want, ℵ are following.

Why can't i "want" it for every natural number.

Because every wanted number has ℵ successors. You cannot remove them individually but only collectively. The latter shows that all can be
removed.

So?

Your logic is just flawed if you expect to be able to create an infinite
set with a finite set.

Your logic is just "dark".

The blind sees only darkness.

I admit I can't individually name ALL of them at once.

You can name and subtract all at once collectively, such that no
successors remain. You cannot do so with in dividual numbers. There
always ℵ successors remain. This is a big difference.

Right, each number can be used individually.

All numbers must be used collectively.

That is the nature of unbounded sets,

Yes. That shows that not all numbers can be addressed individually. That
is what I call dark.

Regards, WM

Nope, where are the numbers that can't be addressed individually. The
fact that there are always more, but also that there are more addresses available, doesn't show that you can't address them all.

There can not be an actual Natural Number that can not be indexed, in
fact, that is essentially a tautology, as the Natural Numbers ARE the
indexes, so every Natural Number can be indexed by itself.

Now, your "dark" numbers could be those numbers past the finite numbers,
what are called the "transfinite numbers", except you specifically say
they aren't but that they are parts of the finite numbers that just
can't be "named" (even though the definitions of the numbers provide a
way to name any of them).

Also, the Transfinite numbers CAN be "named" individually, so they can't
be your "dark" numbers.

All your "dark" numbers are, is an artifact of trying to use logic that
can't handle infinite sets on infinite sets.

--- SoupGate-Win32 v1.05
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Le 10/01/2024 à 03:58, Richard Damon a écrit :

There can not be an actual Natural Number that can not be indexed, in
fact, that is essentially a tautology, as the Natural Numbers ARE the indexes, so every Natural Number can be indexed by itself.

But we cannot use these indices. Proof by the unit fractions 1/n. Since
all have distances, there is a first or smallest one, but we cannot
discern it.

Also, the Transfinite numbers CAN be "named" individually, so they can't
be your "dark" numbers.

Dark numbers like ω have no finite initial segments.

Regards, WM

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

On 1/10/24 1:49 PM, WM wrote:

Le 10/01/2024 à 03:58, Richard Damon a écrit :

There can not be an actual Natural Number that can not be indexed, in
fact, that is essentially a tautology, as the Natural Numbers ARE the
indexes, so every Natural Number can be indexed by itself.

But we cannot use these indices. Proof by the unit fractions 1/n. Since
all have distances, there is a first or smallest one, but we cannot
discern it.

You are just indexing from the wrong end.

1st is 1/1
2nd is 1/2

Also, the Transfinite numbers CAN be "named" individually, so they
can't be your "dark" numbers.

Dark numbers like ω have no finite initial segments.

But ω isn't a "Natural Number", it is never actually reached by the construction formula, only approached.

And ω is individually nameable, so isn' "dark" by your defiition.

Seems you have a bit of a definition problem.

Regards, WM

--- SoupGate-Win32 v1.05
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Le 11/01/2024 à 01:49, Richard Damon a écrit :

You are just indexing from the wrong end.

Points which are existing can be indexed from what we like.

But ω isn't a "Natural Number", it is never actually reached by the construction formula, only approached.

0 is reached by the cursor moving from 1 to 0.

And ω is individually nameable, so isn' "dark" by your defiition.

Seems you have a bit of a definition problem.

No, you are not able to read maths texts.

Definition: A natural number is "identified" or (individually) "defined"
or "instantiated" if it can be communicated such that sender and receiver understand the same and can link it by a finite initial segment to the
origin 0.

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/11/24 4:31 PM, WM wrote:

Le 11/01/2024 à 01:49, Richard Damon a écrit :

You are just indexing from the wrong end.

Points which are existing can be indexed from what we like.

So you say, and then you can't find the index.

But ω isn't a "Natural Number", it is never actually reached by the
construction formula, only approached.

0 is reached by the cursor moving from 1 to 0.

But 0 isn't ω. You seem to be having problems with definitions.

And ω is individually nameable, so isn' "dark" by your defiition.

Seems you have a bit of a definition problem.

No, you are not able to read maths texts.

Really, we could name the numbers you said couldn't be named.

Definition: A natural number is "identified" or (individually) "defined"
or "instantiated" if it can be communicated such that sender and
receiver understand the same and can link it by a finite initial segment
to the origin 0.
Regards, WM

And ω isn't a "Natural Number" as we can't reach it but the finite (but unbounded) application of the successor operator, which is what defines
the set of Natural Numbers.

So, ω not having a finite initial sequence doesn't make it "Dark", it
makes it not a Natural Number.

You don't seem to understand the difference.

--- SoupGate-Win32 v1.05
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Le 12/01/2024 à 04:07, Richard Damon a écrit :

On 1/11/24 4:31 PM, WM wrote:

Le 11/01/2024 à 01:49, Richard Damon a écrit :

You are just indexing from the wrong end.

Points which are existing can be indexed from what we like.

So you say, and then you can't find the index.

This proves the existence of dark points.

But ω isn't a "Natural Number", it is never actually reached by the
construction formula, only approached.

0 is reached by the cursor moving from 1 to 0.

But 0 isn't ω.

Both have equivalent properties in that they lie at the ends of the
sequences of natural numbers and of unit fractions.

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/12/24 8:54 AM, WM wrote:

Le 12/01/2024 à 04:07, Richard Damon a écrit :

On 1/11/24 4:31 PM, WM wrote:

Le 11/01/2024 à 01:49, Richard Damon a écrit :

You are just indexing from the wrong end.

Points which are existing can be indexed from what we like.

So you say, and then you can't find the index.

This proves the existence of dark points.

Nope, because all those numbers ARE indexed from the other end,

But ω isn't a "Natural Number", it is never actually reached by the
construction formula, only approached.

0 is reached by the cursor moving from 1 to 0.

But 0 isn't ω.

Both have equivalent properties in that they lie at the ends of the
sequences of natural numbers and of unit fractions.

And both of them are OUTSIDE the ends of the sequences, so not part of
the sets.

Regards, WM

--- SoupGate-Win32 v1.05
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Le 13/01/2024 à 00:46, Richard Damon a écrit :

On 1/12/24 8:54 AM, WM wrote:

Le 12/01/2024 à 04:07, Richard Damon a écrit :

On 1/11/24 4:31 PM, WM wrote:

Le 11/01/2024 à 01:49, Richard Damon a écrit :

You are just indexing from the wrong end.

Points which are existing can be indexed from what we like.

So you say, and then you can't find the index.

This proves the existence of dark points.

Nope, because all those numbers ARE indexed from the other end,

No. ℵ are missing.

But ω isn't a "Natural Number", it is never actually reached by the >>>>> construction formula, only approached.

0 is reached by the cursor moving from 1 to 0.

But 0 isn't ω.

Both have equivalent properties in that they lie at the ends of the
sequences of natural numbers and of unit fractions.

And both of them are OUTSIDE the ends of the sequences, so not part of
the sets.

Correct. But between all definable positive real numbers and these ends
there are ℵ dark natnumbers and ℵ dark unit fractions, respectively.

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/13/24 5:14 AM, WM wrote:

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

On 1/12/24 8:54 AM, WM wrote:

Le 12/01/2024 à 04:07, Richard Damon a écrit :

On 1/11/24 4:31 PM, WM wrote:

Le 11/01/2024 à 01:49, Richard Damon a écrit :

You are just indexing from the wrong end.

Points which are existing can be indexed from what we like.

So you say, and then you can't find the index.

This proves the existence of dark points.

Nope, because all those numbers ARE indexed from the other end,

No. ℵ are missing.

Which ones?

It seem your "Dark" numbers are just numbers that are not existing. We
can't "see" them because they don't exist.

But ω isn't a "Natural Number", it is never actually reached by
the construction formula, only approached.

0 is reached by the cursor moving from 1 to 0.

But 0 isn't ω.

Both have equivalent properties in that they lie at the ends of the
sequences of natural numbers and of unit fractions.

And both of them are OUTSIDE the ends of the sequences, so not part of
the sets.

Correct. But between all definable positive real numbers and these ends
there are ℵ dark natnumbers and ℵ dark unit fractions, respectively.

That is clearly false unless your "natnumbers" are not the Natural
Numbers and your unit fractions are not the actual Unit Fractions.

Between the Real Numbers 0.4 and 0.45 are NO Natural Numbers or Unit
Fractions, so you claim is just proved to be invalid.

If your "natnumbers" are something else, please define what that set of
numbers is, and how they are generated.

It almosts sounds like an intentional deception to confuse people.

As I have said, there ARE "transfinite" numbers, (that are NOT part of
the finite number sets) that have some of the properties you claim
(except for the not being namable or usable individually) but you have
seemed to try to make it clear that these are not what you are talking
about.

Regards, WM

--- SoupGate-Win32 v1.05
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Le 13/01/2024 à 14:22, Richard Damon a écrit :

On 1/13/24 5:14 AM, WM wrote:

Nope, because all those numbers ARE indexed from the other end,

No. ℵ are missing.

Which ones?

The dark numbers.

It seem your "Dark" numbers are just numbers that are not existing. We
can't "see" them because they don't exist.

But we know that there are ℵ unit fractions between zero and the
smallest named unit fraction.

between all definable positive real numbers and these ends
there are ℵ dark natnumbers and ℵ dark unit fractions, respectively.

Between the Real Numbers 0.4 and 0.45 are NO Natural Numbers or Unit Fractions, so you claim is just proved to be invalid.

Between the smallest visible unit fraction and zero there are ℵ dark
unit fractions.

As I have said, there ARE "transfinite" numbers, (that are NOT part of
the finite number sets) that have some of the properties you claim
(except for the not being namable or usable individually) but you have
seemed to try to make it clear that these are not what you are talking
about.

So it is. I am talking about natural numbers and the reciprocals.

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/14/24 7:27 AM, WM wrote:

Le 13/01/2024 à 14:22, Richard Damon a écrit :

On 1/13/24 5:14 AM, WM wrote:

Nope, because all those numbers ARE indexed from the other end,

No. ℵ are missing.

Which ones?

The dark numbers.

So you say, but cant show.

You are just chasing phantoms in your own mind.

It seem your "Dark" numbers are just numbers that are not existing. We
can't "see" them because they don't exist.

But we know that there are ℵ unit fractions between zero and the
smallest named unit fraction.

But there isn't a "smallest namable unit fraction".

"Named", as in someone has written the name, isn't a property of the
number, and thus not applicable here.

between all definable positive real numbers and these ends there are
ℵ dark natnumbers and ℵ dark unit fractions, respectively.

Between the Real Numbers 0.4 and 0.45 are NO Natural Numbers or Unit
Fractions, so you claim is just proved to be invalid.

Between the smallest visible unit fraction and zero there are ℵ dark
unit fractions.

But there is no "Smallest Visible Unit Fraction". If you think there is,
name it.

After all, it IS "visible" so namable.

As I have said, there ARE "transfinite" numbers, (that are NOT part of
the finite number sets) that have some of the properties you claim
(except for the not being namable or usable individually) but you have
seemed to try to make it clear that these are not what you are talking
about.

So it is. I am talking about natural numbers and the reciprocals.

And the numbers you are trying to talk about can't exist in those sets,
as all of the numbers in that set are "visible" and "namable" and Usable individually.

So, you are stuck in a contradiction.

You seem there is no actual boundary past which the visible numbers
can't reach, so all the numbers you want to call "dark" are visible, so
they aren't dark.

You are just lost in bad logic.

Regards, WM

--- SoupGate-Win32 v1.05
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On 1/14/2024 5:26 PM, Chris M. Thomasson wrote:

Some of my teachers wanted me to join GATE when, iirc, was in the 4'th
grade, but I did not want to. They even got to my parents.

What is GATE?

--- SoupGate-Win32 v1.05
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