On Monday, December 20, 2021 at 3:20:53 PM UTC-4, WM wrote:

> no natnumber is remaining in the intersection [S=I(E(1),E(2),E(3),...)]. That means all are deleted.

Nope. there are no numbers "deleted" from S. S is a set and does not change. The endsegments are sets and do not change.

--
William Hughes

On 12/20/2021 11:28 AM, WM wrote:
> Chris M. Thomasson schrieb am Montag, 20. Dezember 2021 um 09:48:17 UTC+1:
>> On 12/19/2021 1:52 AM, WM wrote:
>>
>> It goes on forever. There is no end.
>
> There is nothing beyond omega. And all natnumbers before omega can be exhausted in bijections if set theory is correct.

LOL! Are you smoking something strange?

infinity is REAL! Got it?

Wow!

måndag 20 december 2021 kl. 20:20:53 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Montag, 20. Dezember 2021 um 06:53:48 UTC+1:
> > fredag 17 december 2021 kl. 23:10:59 UTC+1 skrev WM:
> > > zelos...@gmail.com schrieb am Freitag, 17. Dezember 2021 um 13:27:08 UTC+1:
> > > > fredag 17 december 2021 kl. 11:06:25 UTC+1 skrev WM:
> > >
> > > >
> > > > But with that, congrats, you have L=N :)
> > > >
> > > > because n ∉ E(n+1) is true for ALL n
> > > What contain the non-empty endsegments?
> > > > >But you said that all endsegments are infinite. What is the contents of the infinite endsegments which have lost all natural numbers in ℕ?
> > > > Who says any endsegment is empty?
> > > What do non-empty endsegments contain when all n are gone?
> > what do you mean "when all n are gone"?
> Your proof is based on the idea that by
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> no natnumber is remaining in the intersection. That means all are deleted.. But this applies as well to the endsegments.
>
> Regards, WM

Nope, none was deleted or the like. It just means that for every natural number, there is at least 1 endsegment not having it.

Which is seen from your definition

> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}

for all k, E(k+1) does not contain it, ergo, is not in the intersection, ergo it is empty

On 12/20/2021 12:48 AM, Chris M. Thomasson wrote:
> On 12/19/2021 1:52 AM, WM wrote:
>> Chris M. Thomasson schrieb am Sonntag, 19. Dezember 2021 um 02:15:09
>> UTC+1:
>>> On 12/18/2021 7:29 AM, WM wrote:
>>
>>>> If all numbers were definable and all endsegments were infinite,
>>>> they would not have lost all their numbers. S would be an infinite set.
>>> [...]
>>>
>>> were infinite? They are infinite,
>>
>> Then they are not empty but contain natural numbers in common with all
>> other endsegments. Their intersetion is not empty.
>
> [0]            0
>               / \
>              /   \
>             /     \
>            /       \
>           /         \
> [1]      1           2
>         / \         / \
>        /   \       /   \
> [2]   3     4     5     6
>      / \   / \   / \   / \
> [3] 7   8 9  10 11 12 13  14
> ................................
>
> It goes on forever. There is no end. There are an infinite number of
> levels, however, no leafs in sight... ;^)
>
> Want to get to the number 12, we go: (0, 2, 5, 12), or, starting at the
> root, go (R, L, R) RLR... a finite path to 12. Want to get to a big
> number, go all rights... (0, 2, 6, 14, ...). Notice a pattern? You can
> never reach infinity, even though the tree is infinite in and of itself...
>
> starting at root, 9 = LRL, and 10 = LRR
[...]

Want to get to the children of a node? Say 5...

left = 2*5+1 = 11
right = 2*5+2 = 12

Say, 2...

left = 2*2+1 = 5
right = 2*2+2 = 6

Perhaps 1...

left = 2*1+1 = 3
right = 2*1+2 = 4

Oh shit how about zero:

left = 2*0+1 = 1
right = 2*0+2 = 2

Three:

left = 2*3+1 = 7
right = 2*3+2 = 8

On and on... On and on....

;^)

William schrieb am Montag, 20. Dezember 2021 um 21:18:12 UTC+1:

> Let S=I(E(1), E(2), E(3),...).
> From the facts that every E(i) is a set of natural numbers and for every natural number, m, there is
> an element of (E(1), E(2), E(3),...) that does note contain m we can conclude that S is empty.
> (note there is no need to be able to write m down).

From the fact that beyond every natural number nl, and therefore in every endsegment E(n), there remain ℵo natural numbers (∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo) we can conclude that S is not empty but your conclusion is disproved. Why do you try to repress this fact in a Freudian way?
>
> > Knowing that it is impossible to subtract individually definable natural numbers from ℕ such that ℕ gets exhausted
> > but that it is possible to subtract collectively natural numbers such that ℕ gets exhausted,

> "subtract individually" means subtract a finite set, and "subtract collectively" means subtract an infinite set.

So it is. Infinite sets contain numbers which cannot be treated individually. But in mappings all elements have to be individuals. Therefore there are no mappings between infinite sets.

> Since every natural number (even numbers you cannot write down) is an element of a finite set and an infinite set, this triviality does allow you to split the natural numbers into two parts.

∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo proves: Not all natural numbers can be removed by FISONs. In every case almost all ℵo numbers remain. They are dark.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 21. Dezember 2021 um 06:12:47 UTC+1:
> It just means that for every natural number, there is at least 1 endsegment not having it.
>
> Which is seen from your definition
> > ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> for all k, E(k+1) does not contain it, ergo, is not in the intersection, ergo it is empty

The last conclusion is wrong. From the fact that beyond every natural number n, and therefore in every endsegment E(n), there remain ℵo natural numbers (∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo) we can conclude that the intersection is not empty but your conclusion is disproved. You never talk about it. You seem to be unable to accept it but also you are unable to deny it. You try to repress this fact in a Freudian way. That's not how mathematics works.

Regards, WM

tisdag 21 december 2021 kl. 12:02:38 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 21. Dezember 2021 um 06:12:47 UTC+1:
> > It just means that for every natural number, there is at least 1 endsegment not having it.
> >
> > Which is seen from your definition
> > > ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> > for all k, E(k+1) does not contain it, ergo, is not in the intersection, ergo it is empty
> The last conclusion is wrong. From the fact that beyond every natural number n, and therefore in every endsegment E(n), there remain ℵo natural numbers (∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo) we can conclude that the intersection is not empty but your conclusion is disproved. You never talk about it. You seem to be unable to accept it but also you are unable to deny it. You try to repress this fact in a Freudian way. That's not how mathematics works.
>
> Regards, WM

>The last conclusion is wrong.

Nope, it is correct. If no k e N is in the intersection, it is empty.

>From the fact that beyond every natural number n

This changes nothing.

>and therefore in every endsegment E(n), there remain ℵo natural numbers (∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo)

Means nothing.

>we can conclude that the intersection is not empty

False, none of those matters because all that matters is
∀k ∈ ℕ: E(k+1) = E(k) \ {k}
meaning there is an endsegment where k is not in, therefore k is not in the intersection, that makes the intersection empty.

>but your conclusion is disproved.

Citing irrelevant stuff does not disprove anything.

>You never talk about it. You seem to be unable to accept it but also you are unable to deny it.

There is nothign to accept, you are factually wrong and stating unrelated irrelevant facts do not change it.

>You try to repress this fact in a Freudian way. That's not how mathematics works.

Projection much? You're the one trying to state faulty things and then think it proves your case.

The definition of the intersection is simple

It is the set of elements in all sets that are being intersected.

Or the largest common subset to all sets being intersected.

There is no such largest subset, other than empty

there is no element in all endsegments, making the intersection empty.

That is all that matters.

On 12/21/2021 4:57 AM, WM wrote:
> William schrieb am Montag, 20. Dezember 2021 um 21:18:12 UTC+1:
>
>> Let S=I(E(1), E(2), E(3),...).
>> From the facts that every E(i) is a set of natural numbers and for every natural number, m, there is
>> an element of (E(1), E(2), E(3),...) that does note contain m we can conclude that S is empty.
>> (note there is no need to be able to write m down).
>
> From the fact that beyond every natural number nl, and therefore in every endsegment E(n), there remain ℵo natural numbers (∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo) we can conclude that S is not empty but your conclusion is disproved. Why do you try to repress this fact in a Freudian way?

your conclusion is wrong.

>>
>>> Knowing that it is impossible to subtract individually definable natural numbers from ℕ such that ℕ gets exhausted
>>> but that it is possible to subtract collectively natural numbers such that ℕ gets exhausted,
>
>> "subtract individually" means subtract a finite set, and "subtract collectively" means subtract an infinite set.
>
> So it is. Infinite sets contain numbers which cannot be treated individually. But in mappings all elements have to be individuals. Therefore there are no mappings between infinite sets.

obviously totally wrong. ℕ and 2*ℕ

>
>> Since every natural number (even numbers you cannot write down) is an element of a finite set and an infinite set, this triviality does allow you to split the natural numbers into two parts.
>
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo proves: Not all natural numbers can be removed by FISONs. In every case almost all ℵo numbers remain. They are dark.

wrong again, sets are not variables they are fixed and do not change.

>
> Regards, WM

so far, every one of your posts has wrong information, therefore you intentionally mislead people.

On Tuesday, December 21, 2021 at 6:57:44 AM UTC-4, WM wrote:

∀n ∈ ℕ_P |ℕ_P \ {1, 2, 3, ..., n}| = ℵo proves
that there is no element of the set of FISONs which is not finite (do not stop the presses)

> Not all

"all FISONs", the set of FISONs, is infinite. The union of all elements of the set of FISONs is N_P, the Peano set of natural numbers

--
William Hughes

On 12/21/2021 5:02 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 21. Dezember 2021 um 06:12:47 UTC+1:
>> It just means that for every natural number, there is at least 1 endsegment not having it.
>>
>> Which is seen from your definition
>>> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
>> for all k, E(k+1) does not contain it, ergo, is not in the intersection, ergo it is empty
>
> The last conclusion is wrong. From the fact that beyond every natural number n, and therefore in every endsegment E(n), there remain ℵo natural numbers (∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo) we can conclude that the intersection is not empty but your conclusion is disproved. You never talk about it. You seem to be unable to accept it but also you are unable to deny it. You try to repress this fact in a Freudian way. That's not how mathematics works.
>
> Regards, WM

wrong again. Inspite of a simple proof that there is no number in the intersection of all endsegments, WM imagines there is.

the proof is true: "for all k, E(k+1) does not contain it, ergo, is not in the intersection, ergo it is empty."

yes and WM is projecting...

"You are unable to accept it, but also you are unable to deny it.
You try to repress this fact in a Freudian way.
That's not how mathematics works."

zelos...@gmail.com schrieb am Dienstag, 21. Dezember 2021 um 13:42:20 UTC+1:

> If no k e N is in the intersection, it is empty.

If infinitely many natnumbers are in all endsegments their intersection is infinite.

> >From the fact that beyond every natural number n
> This changes nothing.

says the victim of Freudian repression.

Regards, WM

William schrieb am Dienstag, 21. Dezember 2021 um 15:16:35 UTC+1:
> On Tuesday, December 21, 2021 at 6:57:44 AM UTC-4, WM wrote:
>
> ∀n ∈ ℕ_P |ℕ_P \ {1, 2, 3, ..., n}| = ℵo proves
> that there is no element of the set of FISONs which is not finite

Their finiteness is asily proved by the pigeonhole principle.

If each of the first n positive integers has a unary representation in form of a string, like ooooo, that is shorter than n then, by the pigeonhole principle, there must be two different positive integers defined by the same unary representation. Clearly this is absurd.

Same holds in case of ℵo finite strings. ℵo is a fixed quantity such that n < ℵo. If each one of all ℵo positive integers has a unary representation in form of a string that is shorter than ℵo then, by the pigeonhole principle, there must be two different positive integers defined by the same unary representation. Clearly this is absurd too.

Same holds for the Finite Initial Segments Of Natural numbers (FISONs). Since in (1, 2, 3, ..., n) the numbers do not change their positions, (1, 2, 3, 4, 5) has same information content as ooooo.
..
> "all FISONs", the set of FISONs, is infinite.

This is disprovede by the pigeonhole principle.

> The union of all elements of the set of FISONs is N_P, the Peano set of natural numbers

This is disproved by the impossibility to find any natnumber that is in the union but not in a FISON.

Regards, WM

On Tuesday, December 21, 2021 at 5:33:42 PM UTC-4, WM wrote:

Each element, n, of the set of natural numbers if finite. There is no largest n. The set of natural numbers is infinite.
Each string of n o's is finite, There is a string of n o's for every natural number. There is no longest string.. The set of strings is infinite

> ℵo is a fixed quantity

ℵo is not finite. The Pigeonhole principle applies to finite sets.

--
William Hughes

On 12/21/2021 3:27 PM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 21. Dezember 2021 um 13:42:20 UTC+1:
>
>> If no k e N is in the intersection, it is empty.
>
> If infinitely many natnumbers are in all endsegments their intersection is infinite.

wrong.

>
>> >From the fact that beyond every natural number n
>> This changes nothing.
>
> says the victim of Freudian repression.

says WM, with an error in EVERY post.

>
> Regards, WM

tisdag 21 december 2021 kl. 22:27:57 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 21. Dezember 2021 um 13:42:20 UTC+1:
>
> > If no k e N is in the intersection, it is empty.
> If infinitely many natnumbers are in all endsegments their intersection is infinite.
> > >From the fact that beyond every natural number n
> > This changes nothing.
> says the victim of Freudian repression.
>
> Regards, WM

>If infinitely many natnumbers are in all endsegments their intersection is infinite.

False, that is a complete non-sequitor. The cardinality of the sets have no effect on the intersection.

>says the victim of Freudian repression.

Says the person that knows logic and mathematics better than you :)

William schrieb am Dienstag, 21. Dezember 2021 um 23:19:57 UTC+1:
> On Tuesday, December 21, 2021 at 5:33:42 PM UTC-4, WM wrote:
>
> Each element, n, of the set of natural numbers if finite. There is no largest n. The set of natural numbers is infinite.

Potentially infinite.

> Each string of n o's is finite,

That means if you subtract it from the infinite string, infinitely man o's will remain.

> There is a string of n o's for every natural number. There is no longest string.. The set of strings is infinite

If you subtract any finite string from the infinite string, then infinitely man o's will remain. That holds for every string of the infinite set of finite strings. There is no way to cover all o's.

> > ℵo is a fixed quantity
> ℵo is not finite. The Pigeonhole principle applies to finite sets..

The pigeonhole principle applies to all sets of strings the elements of which shall be distinct. Why should the strings of an infinite set no longer be distinct? But how should they be distinguished if not by their number of o's? Your claim shows that set theory requires inconsistent assumptions.

Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 22. Dezember 2021 um 06:27:54 UTC+1:
> tisdag 21 december 2021 kl. 22:27:57 UTC+1 skrev WM:

> >If infinitely many natnumbers are in all endsegments their intersection is infinite.
> False, that is a complete non-sequitor. The cardinality of the sets have no effect on the intersection.

Your level has dropped to Sergio's. EOD.

Regards, WM

It happens that WM formulated :

> EOD.

Stop lying.

On Wednesday, December 22, 2021 at 4:52:24 AM UTC-4, WM wrote:
> William schrieb am Dienstag, 21. Dezember 2021 um 23:19:57 UTC+1:
> > On Tuesday, December 21, 2021 at 5:33:42 PM UTC-4, WM wrote:
> >
> > Each element, n, of the set of natural numbers if finite. There is no largest n. The set of natural numbers is infinite.
> Potentially infinite

The natural numbers are a Peano set. "Potentially infinite set" is nonsense.

--
William Hughes

On Wednesday, December 22, 2021 at 6:28:17 AM UTC-8, William wrote:
> On Wednesday, December 22, 2021 at 4:52:24 AM UTC-4, WM wrote:
> > William schrieb am Dienstag, 21. Dezember 2021 um 23:19:57 UTC+1:
> > > On Tuesday, December 21, 2021 at 5:33:42 PM UTC-4, WM wrote:
> > >
> > > Each element, n, of the set of natural numbers if finite. There is no largest n. The set of natural numbers is infinite.
> > Potentially infinite
> The natural numbers are a Peano set. "Potentially infinite set" is nonsense.
>
> --
> William Hughes

Nah, "potential" and "actual" are formal terms classically.

"If you're not making sense of it, don't go making nonsense of it."

That infinite unbounded integers are infinite and more than unbounded,
for example as you mention with respect to N_F, some theory,
that there are infinite members of the naturals and at least one ,
is a usual "resolution of paradox" better than "such words aren't".

Anyways for the Null Axiom Theory and more-or-less the initiative,
good luck in 2021 and that more-or-less foundations for mathematics and physics,
and logic, is primarily temporal and continuous.

About writing measure theory in that....

Anyways a usual sputnik of quantification results an infinite member,
of an infinite set of ordinals or naturals.

In terms of the whole of them for a usual classical infinity and time
it's also called "1". (For "zero through one".)

Mathematics deserves mostly the usual classical notions as the super-classical,
the foundations.

Usual planks of finitism like "these numbers though unbounded always
have fixed and constant differences", are simply that, a framework for relation,
a formalism, but greater theory for foundations has that contained.

So, the non-standard these days, are included in foundations.
Which, by definition, foundations, won't omit all the result in logic.

Such Cantorian, Goedelian, post-Goedelian, post-Cantorian theories as
modern foundations, of course includes ordinary foundations.

So, if you'd care to advance the argument 100 years, it's been so.

William schrieb am Mittwoch, 22. Dezember 2021 um 15:28:17 UTC+1:
> On Wednesday, December 22, 2021 at 4:52:24 AM UTC-4, WM wrote:
> > William schrieb am Dienstag, 21. Dezember 2021 um 23:19:57 UTC+1:
> > > On Tuesday, December 21, 2021 at 5:33:42 PM UTC-4, WM wrote:
> > >
> > > Each element, n, of the set of natural numbers if finite. There is no largest n. The set of natural numbers is infinite.
> > Potentially infinite
> The natural numbers are a Peano set. "Potentially infinite set" is nonsense.
>
All infinitely many FISONs (1, 2, 3, ..., n), when subtracted from ℕ, leave almost all natural numbers there

∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo

What happens when subtracting the union U{1, 2, 3, ..., n}? Often we hear the argument that the union of all FISONs is larger than all FISONs. Then we face two alternatives:
- Either even the union is not sufficient to exhaust ℕ completely. Then numbers remain which cannot be removed individually. They are dark.
- Or no numbers remain. Then the infinite difference of natural numbers can only be removed collectively (by the union of FISONs) but not individually (by the sequence of FISONs). That means, they are dark.

Regards, WM

On Wednesday, December 22, 2021 at 2:01:30 PM UTC-4, WM wrote:
> William schrieb am Mittwoch, 22. Dezember 2021 um 15:28:17 UTC+1:
> > On Wednesday, December 22, 2021 at 4:52:24 AM UTC-4, WM wrote:
> > > William schrieb am Dienstag, 21. Dezember 2021 um 23:19:57 UTC+1:
> > > > On Tuesday, December 21, 2021 at 5:33:42 PM UTC-4, WM wrote:
> > > >
> > > > Each element, n, of the set of natural numbers if finite. There is no largest n. The set of natural numbers is infinite.
> > > Potentially infinite
> > The natural numbers are a Peano set. "Potentially infinite set" is nonsense.

You keep saying "Potentially infinite" when talking about a set. "Potentially infinite set" is nonsense. Yes or no?

--
William Hughes

On Wednesday, December 22, 2021 at 7:01:30 PM UTC+1, WM wrote:

> All infinitely many FISONs (1, 2, 3, ..., n)

Learn to use standard notation, idiot!

(1, 2, 3, ..., n) is *not* a FISON, but {1, 2, 3, ..., n} is [for any n e IN].

On 12/22/2021 2:54 AM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 22. Dezember 2021 um 06:27:54 UTC+1:
>> tisdag 21 december 2021 kl. 22:27:57 UTC+1 skrev WM:
>
>>> If infinitely many natnumbers are in all endsegments their intersection is infinite.
>> False, that is a complete non-sequitor. The cardinality of the sets have no effect on the intersection.
>
> Your level has dropped to Sergio's. EOD.
>
> Regards, WM

I ask you simple questions you fail to answer.
I provide simple proofs, that openly disprove your math.
Your level is primitive and pre-algebra. My level is post grad Vector Calculus, and a lot more.
Your intent is to show Math as failure, to mislead people, it's all you can do, since you do not know nor understand math.
The turds you drop in here only stink, like your book.

William schrieb am Mittwoch, 22. Dezember 2021 um 20:00:10 UTC+1:
> On Wednesday, December 22, 2021 at 2:01:30 PM UTC-4, WM wrote:

> > > > > Each element, n, of the set of natural numbers if finite. There is no largest n. The set of natural numbers is infinite.
> > > > Potentially infinite
> > > The natural numbers are a Peano set. "Potentially infinite set" is nonsense.
> You keep saying "Potentially infinite" when talking about a set. "Potentially infinite set" is nonsense. Yes or no?

From the standpoint of set theory sets must be actually infinite. From the standpoint of German DIN-norm and even of the EU, zero is a natural number. I dare to tell my students that I do not accept zero as a natural number because it is one of the most unnatural numbers and I recommend that they should join me. And I dare to talk about potentially infinite sets because actually infinite sets are mostly dark. Sue me.

Regards, WM