> B is a counterexample.

>No. It is only an example of switching between actual infinity and potential infinity

Piffle. Every B(i) is an unchangeing entity. The sequence B is an unchanging entity. The words "Potential infinitu" are meaningless when applied to something that does not change. You are staring at a simple counterexample.

--
William Hughes

William schrieb am Sonntag, 8. August 2021 um 03:30:08 UTC+2:
> > B is a counterexample.
>
> >No. It is only an example of switching between actual infinity and potential infinity
>
> Piffle. Every B(i) is an unchangeing entity.

So are the endsegments and the set of natural numbers.

> The sequence B is an unchanging entity. The words "Potential infinitu" are meaningless when applied to something that does not change. You are staring at a simple counterexample.

Actual infinity actually exhausts all natural numbers n in the bijection with all endsegments E(n). Since every endsegment contains only natural numbers equal to or larger than its index, it is impossible that every endsegment contains all natural numbers or an equivalent infinite set.

Your example would only be possible if not all natural numbers were exhausted as indices but every index was followed by almost all natural numbers.

Regards, WM

WM wrote:

> Actual infinity actually exhausts all natural numbers n in the bijection
> with all endsegments E(n). Since every endsegment contains only natural
> numbers equal to or larger than its index, it is impossible that every
> endsegment contains all natural numbers or an equivalent infinite set.

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On Monday, August 9, 2021 at 7:27:28 AM UTC-4, WM wrote:

> Your example would only be possible if not all [elements of |N_F] were exhausted as indices but every index was followed by almost all [elements of |N_F].

He said staring at an example in which every element of |N_F is used as an index, i, and every set B(i) has cardinality aleph_0.

(Note, every element of |N_F is finite and followed by almost all elements of |N_F.)

--
William Hughes

On Monday, 9 August 2021 at 08:27:28 UTC-3, WM wrote:
> William schrieb am Sonntag, 8. August 2021 um 03:30:08 UTC+2:
> > > B is a counterexample.
> >
> > >No. It is only an example of switching between actual infinity and potential infinity
> >
> > Piffle. Every B(i) is an unchangeing entity.
> So are the endsegments and the set of natural numbers.
> > The sequence B is an unchanging entity. The words "Potential infinitu" are meaningless when applied to something that does not change. You are staring at a simple counterexample.
> Actual infinity actually exhausts all natural numbers n in the bijection with all endsegments E(n). Since every endsegment contains only natural numbers equal to or larger than its index, it is impossible that every endsegment contains all natural numbers or an equivalent infinite set.

You are denser than a neutron star! *EVERY* natural number n can be used to start an end segment E(n), and every end segment E(n) can be bijected with |N trivially: 1 <-> n, 2 <-> n+1, 3 <-> n+2, ...
On the left every natural number is used (none is left unused), and on the right the end segment is similarly exhausted. (You may recall, oh exalted one, that E(n) has cardinality aleph_0.

> Your example would only be possible if not all natural numbers were exhausted as indices but every index was followed by almost all natural numbers.

Your comprehension of anything infinite is subterranean.

Gus Gassmann explained on 8/9/2021 :
> On Monday, 9 August 2021 at 08:27:28 UTC-3, WM wrote:
>> William schrieb am Sonntag, 8. August 2021 um 03:30:08 UTC+2:
>>>> B is a counterexample.
>>>
>>>> No. It is only an example of switching between actual infinity and
>>>> potential infinity
>>>
>>> Piffle. Every B(i) is an unchangeing entity. So are the endsegments and the
>>> set of natural numbers. The sequence B is an unchanging entity. The words
>>> "Potential infinitu" are meaningless when applied to something that does
>>> not change. You are staring at a simple counterexample.
>> Actual infinity actually exhausts all natural numbers n in the bijection
>> with all endsegments E(n). Since every endsegment contains only natural
>> numbers equal to or larger than its index, it is impossible that every
>> endsegment contains all natural numbers or an equivalent infinite set.
>
> You are denser than a neutron star! *EVERY* natural number n can be used to
> start an end segment E(n), and every end segment E(n) can be bijected with |N
> trivially: 1 <-> n, 2 <-> n+1, 3 <-> n+2, ... On the left every natural
> number is used (none is left unused), and on the right the end segment is
> similarly exhausted. (You may recall, oh exalted one, that E(n) has
> cardinality aleph_0.
>
>> Your example would only be possible if not all natural numbers were
>> exhausted as indices but every index was followed by almost all natural
>> numbers.
>
> Your comprehension of anything infinite is subterranean.

Yes, he has a deep misunderstanding of the subject.

William schrieb am Montag, 9. August 2021 um 17:55:55 UTC+2:
> On Monday, August 9, 2021 at 7:27:28 AM UTC-4, WM wrote:
>
> > Your example would only be possible if not all [elements of |N_F] were exhausted as indices but every index was followed by almost all [elements of |N_F].
>
> He said staring at an example in which every element of |N_F is used as an index, i,

No, only those indexes i which are followed by infinitely many indexes are used. But these indexes are by far not all, because if you use all indexes, then none remains:
|N_F \ |N_F = { }.

Therefore you use only the definable indexes.

> and every set B(i) has cardinality aleph_0.

That is another proof that not all have been used as indexes.

> (Note, every element of |N_F is finite and followed by almost all elements of |N_F.)

Every definable index, yes. But all indexes are not followed by indexes:
|N_F \ |N_F = { }.

Regards, WM

Gus Gassmann schrieb am Montag, 9. August 2021 um 18:52:11 UTC+2:
> (You may recall that E(n) has cardinality aleph_0.

That is only so for definable indexes.
If really all indexes were used, then nothing would be left:
|N \ |N = { }.

> > Your example would only be possible if not all natural numbers were exhausted as indices but every index was followed by almost all natural numbers.
> Your comprehension of anything infinite is subterranean.

Your idea of ||N \ |N| = aleph_0 is simply selfcontradictory.

Regards, WM

FromTheRafters schrieb am Montag, 9. August 2021 um 19:21:06 UTC+2:
>
> Yes, he has a deep misunderstanding of the subject.

There are two opinions: Either all endsegments which are definable as a last one have infinite intersections with all their predecessors

|∩{E(k) : k ∈ ℕ_def}| = ℵo (*)

can also be used in an infinite set such that

E(1) ∩ E(2) ∩ E(3) ∩ ... = { }.

Or the infinite set must contain more endsegments than are in (*).

The first opinion is wrong because (1) an infinite set of endsegments contains more endsegments than every finite set. And (2) the intersection is independent of the order.

There it is refuted that only all endsegments which in finite sets leave infinite intersections can constitute an infinite set yielding an empty intersection. Are you really unable to understand these two simple arguments?

Regards, WM

On Tuesday, 10 August 2021 at 06:38:42 UTC-3, WM wrote:
> Gus Gassmann schrieb am Montag, 9. August 2021 um 18:52:11 UTC+2:
> > (You may recall that E(n) has cardinality aleph_0.
>
> That is only so for definable indexes.
> If really all indexes were used, then nothing would be left:
> |N \ |N = { }.
> > > Your example would only be possible if not all natural numbers were exhausted as indices but every index was followed by almost all natural numbers.
> > Your comprehension of anything infinite is subterranean.
> Your idea of ||N \ |N| = aleph_0 is simply selfcontradictory.

The only one who thinks that anyone forms |N \ |N in any of this is you. I add to the many things that you do not comprehend (any longer) the forming of intersections.

On Tuesday, 10 August 2021 at 06:49:39 UTC-3, WM wrote:
> FromTheRafters schrieb am Montag, 9. August 2021 um 19:21:06 UTC+2:
> >
> > Yes, he has a deep misunderstanding of the subject.
> There are two opinions: Either all endsegments which are definable as a last one have infinite intersections with all their predecessors
>
> |∩{E(k) : k ∈ ℕ_def}| = ℵo (*)
>
> can also be used in an infinite set such that
>
> E(1) ∩ E(2) ∩ E(3) ∩ ... = { }.
Gibberish.
> Or the infinite set must contain more endsegments than are in (*).

*WHAT* infinite set? Can't you even write ordinary language clearly any longer?

> The first opinion is wrong because (1) an infinite set of endsegments contains more endsegments than every finite set. And (2) the intersection is independent of the order.
>
> There it is refuted that only all endsegments which in finite sets leave infinite intersections can constitute an infinite set yielding an empty intersection. Are you really unable to understand these two simple arguments?

The third, majority opinion is that WM has lost his marbles.

Gus Gassmann schrieb am Dienstag, 10. August 2021 um 12:53:59 UTC+2:
> On Tuesday, 10 August 2021 at 06:38:42 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Montag, 9. August 2021 um 18:52:11 UTC+2:
> > > (You may recall that E(n) has cardinality aleph_0.
> >
> > That is only so for definable indexes.
> > If really all indexes were used, then nothing would be left:
> > |N \ |N = { }.
> > > > Your example would only be possible if not all natural numbers were exhausted as indices but every index was followed by almost all natural numbers.
> > Your idea of ||N \ |N| = aleph_0 is simply selfcontradictory.
> The only one who thinks that anyone forms |N \ |N in any of this is you.

It can be used. It exhausts all natural numbers. That is enough.

Regards, WM

Gus Gassmann schrieb am Dienstag, 10. August 2021 um 12:59:05 UTC+2:
> On Tuesday, 10 August 2021 at 06:49:39 UTC-3, WM wrote:

> > There are two opinions: Either all endsegments which are definable as a last one have infinite intersections with all their predecessors
> >
> > |∩{E(k) : k ∈ ℕ_def}| = ℵo (*)
> >
> > can also be used in an infinite set such that
> >
> > E(1) ∩ E(2) ∩ E(3) ∩ ... = { }.
> Gibberish.

You cannot even understand this simple thought?

> > Or the infinite set must contain more endsegments than are in (*).
> *WHAT* infinite set?

The set intersected above: E(1), E(2), E(3), ...

> > The first opinion is wrong because (1) an infinite set of endsegments contains more endsegments than every finite set. And (2) the intersection is independent of the order.
> >
> > There it is refuted that only all endsegments which in finite sets leave infinite intersections can constitute an infinite set yielding an empty intersection. Are you really unable to understand these two simple arguments?
> The third, majority opinion is that WM has lost his marbles.

Among fools of matheology this may be a majority opinion.

Regards, WM

On Tuesday, 10 August 2021 at 08:33:49 UTC-3, WM wrote:
> Gus Gassmann schrieb am Dienstag, 10. August 2021 um 12:53:59 UTC+2:
> > On Tuesday, 10 August 2021 at 06:38:42 UTC-3, WM wrote:
> > > Gus Gassmann schrieb am Montag, 9. August 2021 um 18:52:11 UTC+2:
> > > > (You may recall that E(n) has cardinality aleph_0.
> > >
> > > That is only so for definable indexes.
> > > If really all indexes were used, then nothing would be left:
> > > |N \ |N = { }.
> > > > > Your example would only be possible if not all natural numbers were exhausted as indices but every index was followed by almost all natural numbers.
> > > Your idea of ||N \ |N| = aleph_0 is simply selfcontradictory.
> > The only one who thinks that anyone forms |N \ |N in any of this is you.
> It can be used. It exhausts all natural numbers. That is enough.
Do you even know how to think any longer? (Can you do that?) Instead of spewing out such a word salad, why don't you take the time and formulate properly? What is "it"? Where can "it" be used? Why is any of "it" relevant?

On Tuesday, 10 August 2021 at 08:36:58 UTC-3, WM wrote:
> Gus Gassmann schrieb am Dienstag, 10. August 2021 um 12:59:05 UTC+2:
> > On Tuesday, 10 August 2021 at 06:49:39 UTC-3, WM wrote:
>
> > > There are two opinions: Either all endsegments which are definable as a last one have infinite intersections with all their predecessors
> > >
> > > |∩{E(k) : k ∈ ℕ_def}| = ℵo (*)
> > >
> > > can also be used in an infinite set such that
> > >
> > > E(1) ∩ E(2) ∩ E(3) ∩ ... = { }.
> > Gibberish.
> You cannot even understand this simple thought?

What thought? That finite intersections of end segments have cardinality aleph_0, while infinite intersections of end segments are empty? This seems to have nothing to do with your gratuitous and frankly insulting "can also be used". I called your phrasing "gibberish", which in retrospect was very generous.
[...]
Took out the rest of the garbage.

On Tuesday, August 10, 2021 at 5:35:20 AM UTC-4, WM wrote:
> William schrieb am Montag, 9. August 2021 um 17:55:55 UTC+2:

> > (Note, every element of |N_F is finite and followed by almost all elements of |N_F.)
> Every definable index, yes

Indeed, every index you can write down, and every index you cannot write down too.

> But all indexes are not followed by indexes:

Another use of the ambiguous "all".
"All indexes" are not followed by indexes. Every element of |N_F is not followed by indexes . False
"All indexes" are not followed by indexes. The set |N_F is not followed by indexes. True.

--
William Hughes

William schrieb am Dienstag, 10. August 2021 um 15:14:35 UTC+2:
> On Tuesday, August 10, 2021 at 5:35:20 AM UTC-4, WM wrote:
> > William schrieb am Montag, 9. August 2021 um 17:55:55 UTC+2:
>
> > > (Note, every element of |N_F is finite and followed by almost all elements of |N_F.)
> > Every definable index, yes
> Indeed, every index you can write down, and every index you cannot write down too.
> > But all indexes are not followed by indexes:
> Another use of the ambiguous "all".

It isn't ambiguous but is the basis of set theory: finished infinity (vollendete Unendlichkeit).

> "All indexes" are not followed by indexes. Every element of |N_F is not followed by indexes . False
> "All indexes" are not followed by indexes. The set |N_F is not followed by indexes. True.

Not only "the set". If all indexes have been issued, then none remains. When the enumeration has been finished (vollendet) then none remains. If it can't be finished, then set theory is nonsense.

Regards, WM

Gus Gassmann schrieb am Dienstag, 10. August 2021 um 13:56:10 UTC+2:
> On Tuesday, 10 August 2021 at 08:36:58 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Dienstag, 10. August 2021 um 12:59:05 UTC+2:
> > > On Tuesday, 10 August 2021 at 06:49:39 UTC-3, WM wrote:
> >
> > > > There are two opinions: Either all endsegments which are definable as a last one have infinite intersections with all their predecessors
> > > >
> > > > |∩{E(k) : k ∈ ℕ_def}| = ℵo (*)
> > > >
> > > > can also be used in an infinite set such that
> > > >
> > > > E(1) ∩ E(2) ∩ E(3) ∩ ... = { }.
> > > Gibberish.
> > You cannot even understand this simple thought?
> What thought? That finite intersections of end segments have cardinality aleph_0, while infinite intersections of end segments are empty?

No. The antinomy is that all endsegments which can finish a set shall be sufficient to form an infinite set (actually infinite!) and there have an empty intersection. Note that only such endsegments are available if ℕ_def = ℕ.

Note that an infinite set requires many more elements than every finite set.. Same with FISONs:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

This however is a contradiction:
∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo
because it says that not all natural numbers can be subtracted from ℕ. Always ℵo will remain. But it is possible to subtract all natural numbers such that none remains:
ℕ \ ℕ = { }.

Regards, WM

On Wednesday, 11 August 2021 at 11:04:20 UTC-3, WM wrote:
> Gus Gassmann schrieb am Dienstag, 10. August 2021 um 13:56:10 UTC+2:
> > On Tuesday, 10 August 2021 at 08:36:58 UTC-3, WM wrote:
> > > Gus Gassmann schrieb am Dienstag, 10. August 2021 um 12:59:05 UTC+2:
> > > > On Tuesday, 10 August 2021 at 06:49:39 UTC-3, WM wrote:
> > >
> > > > > There are two opinions: Either all endsegments which are definable as a last one have infinite intersections with all their predecessors
> > > > >
> > > > > |∩{E(k) : k ∈ ℕ_def}| = ℵo (*)
> > > > >
> > > > > can also be used in an infinite set such that
> > > > >
> > > > > E(1) ∩ E(2) ∩ E(3) ∩ ... = { }.
> > > > Gibberish.
> > > You cannot even understand this simple thought?
> > What thought? That finite intersections of end segments have cardinality aleph_0, while infinite intersections of end segments are empty?
> No. The antinomy is that all endsegments which can finish a set shall be sufficient to form an infinite set (actually infinite!) and there have an empty intersection. Note that only such endsegments are available if ℕ_def = ℕ.
Boring... It has been proven to you many times that ℕ_def = ℕ. That you do not understand or accept the proof is immaterial.

On Wednesday, August 11, 2021 at 9:57:42 AM UTC-4, WM wrote:
> William schrieb am Dienstag, 10. August 2021 um 15:14:35 UTC+2:
> > On Tuesday, August 10, 2021 at 5:35:20 AM UTC-4, WM wrote:
> > > William schrieb am Montag, 9. August 2021 um 17:55:55 UTC+2:
> >
> > > > (Note, every element of |N_F is finite and followed by almost all elements of |N_F.)
> > > Every definable index, yes
> > Indeed, every index you can write down, and every index you cannot write down too.
> > > But all indexes are not followed by indexes:
> > Another use of the ambiguous "all".
> It isn't ambiguous

He said staring at two different meanings of the phrase " All indexes are not followed by indexes"

Do not use the natural language term "All". Better yet, do not use natural language.

> Not only "the set". If all indexes have been issued, then none remains.

"None remains" of the copy of |N_F you are using. As you note, you can use as many copies of |N_F as you like.

--
William Hughes

Gus Gassmann schrieb am Mittwoch, 11. August 2021 um 16:32:32 UTC+2:
> On Wednesday, 11 August 2021 at 11:04:20 UTC-3, WM wrote:
> > The antinomy is that all endsegments which can finish a set shall be sufficient to form an infinite set (actually infinite!) and there have an empty intersection. Note that only such endsegments are available if ℕ_def = ℕ.
> Boring... It has been proven to you many times that ℕ_def = ℕ. That you do not understand or accept the proof is immaterial.

I understand that the theory this "proof" is based upon is inconsistent, because it is this same theory which claims that intersections are indepedent of the order of endsegments.

Regards, WM

William schrieb am Mittwoch, 11. August 2021 um 16:50:30 UTC+2:
> On Wednesday, August 11, 2021 at 9:57:42 AM UTC-4, WM wrote:
> > William schrieb am Dienstag, 10. August 2021 um 15:14:35 UTC+2:
> > > On Tuesday, August 10, 2021 at 5:35:20 AM UTC-4, WM wrote:
> > > > William schrieb am Montag, 9. August 2021 um 17:55:55 UTC+2:
> > >
> > > > > (Note, every element of |N_F is finite and followed by almost all elements of |N_F.)
> > > > Every definable index, yes
> > > Indeed, every index you can write down, and every index you cannot write down too.
> > > > But all indexes are not followed by indexes:
> > > Another use of the ambiguous "all".
> > It isn't ambiguous
> He said staring at two different meanings of the phrase " All indexes are not followed by indexes"

There is only one meaning here. All indexes are used up in the bijection of indexes and endsegments. No additional natural numbers are available as contents of endsegments.
>
> Do not use the natural language term "All". Better yet, do not use natural language.

All is a term which denotes all with no exception in every sober environment.

> > Not only "the set". If all indexes have been issued, then none remains.
> "None remains" of the copy of |N_F you are using. As you note, you can use as many copies of |N_F as you like.

Not when only the part remaining from the first copy is available.

Regards, WM

On 8/11/2021 2:46 PM, WM wrote:
> William schrieb am Mittwoch, 11. August 2021 um 16:50:30 UTC+2:
>> On Wednesday, August 11, 2021 at 9:57:42 AM UTC-4, WM wrote:
>>> William schrieb am Dienstag, 10. August 2021 um 15:14:35 UTC+2:
>>>> On Tuesday, August 10, 2021 at 5:35:20 AM UTC-4, WM wrote:
>>>>> William schrieb am Montag, 9. August 2021 um 17:55:55 UTC+2:
>>>>
>>>>>> (Note, every element of |N_F is finite and followed by almost all elements of |N_F.)
>>>>> Every definable index, yes
>>>> Indeed, every index you can write down, and every index you cannot write down too.
>>>>> But all indexes are not followed by indexes:
>>>> Another use of the ambiguous "all".
>>> It isn't ambiguous
>> He said staring at two different meanings of the phrase " All indexes are not followed by indexes"
>
> There is only one meaning here. All indexes are used up in the bijection of indexes and endsegments. No additional natural numbers are available as contents of endsegments.

you ran out of natural numbers ? you can "re-use" them you know.

In case you run out, here are some extras 1's 1 1 1 1 1 1 1 1 1 1 1
2's 2 2 2 2 2 2 2 2 2 2 2 2
3's 3 3 3 3 3 3 3 3 3

>>
>> Do not use the natural language term "All". Better yet, do not use natural language.
>
> All is a term which denotes all with no exception in every sober environment.

wrong. "All" is a word, not a term. try again.

>
>>> Not only "the set". If all indexes have been issued, then none remains.
>> "None remains" of the copy of |N_F you are using. As you note, you can use as many copies of |N_F as you like.
>
> Not when only the part remaining from the first copy is available.

see extras above

>
> Regards, WM
>

On 8/11/2021 2:43 PM, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 11. August 2021 um 16:32:32 UTC+2:
>> On Wednesday, 11 August 2021 at 11:04:20 UTC-3, WM wrote:
>>> The antinomy is that all endsegments which can finish a set shall be sufficient to form an infinite set (actually infinite!) and there have an empty intersection. Note that only such endsegments are available if ℕ_def = ℕ.
>> Boring... It has been proven to you many times that ℕ_def = ℕ. That you do not understand or accept the proof is immaterial.
>
> I understand that the theory this "proof" is based upon is inconsistent, because it is this same theory which claims that intersections are indepedent of the order of endsegments.
>
> Regards, WM
>

intersections are indepedent of the order of endsegments.

On Wednesday, August 11, 2021 at 3:46:42 PM UTC-4, WM wrote:
> William schrieb am Mittwoch, 11. August 2021 um 16:50:30 UTC+2:
> > On Wednesday, August 11, 2021 at 9:57:42 AM UTC-4, WM wrote:
> > > William schrieb am Dienstag, 10. August 2021 um 15:14:35 UTC+2:
> > > > On Tuesday, August 10, 2021 at 5:35:20 AM UTC-4, WM wrote:
> > > > But all indexes are not followed by indexes:
> > > > Another use of the ambiguous "all".
> > > It isn't ambiguous
> > He said staring at two different meanings of the phrase " All indexes are not followed by indexes"
> There is only one meaning here.

He said staring at one true statement and one false statement.

--
William Hughes