On Thursday, October 21, 2021 at 2:09:25 PM UTC-4, WM wrote:
> Dan Christensen schrieb am Mittwoch, 20. Oktober 2021 um 23:29:18 UTC+2:
> > On Wednesday, October 20, 2021 at 4:42:55 PM UTC-4, WM wrote:
>
> > > Try mathematics. Infinitely many endsegments exhaust all natnumbers 1, 2, 3, ... as indexes.
> > Wrong
> Indfinitely many but not all natnumbers?
> > The infinitely many end-segments are a mathematical reality.

> If so, then most are dark because they are not infinite.

Liar. These "dark numbers" of yours are just your latest concoction to distract from your utter failure to demonstrate any inconsistencies in the ZFC axioms of set theory. What has it been 40 years of your life, wasted? If you can't dazzle them with brilliance, baffled them with bullshit, right, Mucke? But you are a failure at even at bullshit! Must be frustrating as hell for you.

> That is a basic reality: There are not two consecutive infinite sets in |N.

Sorry, Mucke, still no inconsistencies there either. Face, math just isn't your thing.

> > Numbers don't get "exhausted."

> ℕ \ ℕ = { }.

> ℕ \ {indexes of infinitely many endsegments} = { }, that is the empty set. All natnumbers are exhausted.
> > You can use them the same numbers over and over again.

> But not as index of endsegment E(n) and as contents of endsegments E(m) with m > n.

Wrong again, Mucke. Every natural number n is used as an index of the end-segment E(n) = {n, n+1, n+2, ... } and it occurs in each the end-segments E(1), E(2), E(3), ... E(n), each with infinitely many elements in them.

Once again, you have failed to prove your outrageous claim that there are only finitely many end-segments in N.

Once again, after 40+ years (?) of futile effort, you have failed to demonstrate any inconsistencies in the ZFC axioms of set theory. When will you learn???

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

torsdag 21 oktober 2021 kl. 20:25:59 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 21. Oktober 2021 um 07:05:25 UTC+2:
> > onsdag 20 oktober 2021 kl. 23:10:51 UTC+2 skrev Transfinity:
> > > zelos...@gmail.com schrieb am Dienstag, 19. Oktober 2021 um 11:47:24 UTC+2:
> >
> > > > So Zorns lemma is not applicable to natural numbers.
> > > Wrong. Note: This successorship would not stop at natural indices but would run through all ordinal numbers.
> > >
> > Sorry but it doesn't apply with natural numbers because the chain 1,2,3,4,5,6,... has no maximal element so the if condition has failed.
> we are going to define elements a0 < a1 < a2 < a3 < ... in P. This sequence is really long: the indices are not just the natural numbers, but all ordinals. In fact, the sequence is too long for the set P; there are too many ordinals (a proper class), more than there are elements in any set, and the set P will be exhausted before long and then we will run into the desired contradiction. (Wiki)
>
> Regards, WM

ordinals have no maximal element either so even then it fucking fails.

torsdag 21 oktober 2021 kl. 20:33:53 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 21. Oktober 2021 um 07:06:32 UTC+2:
> > onsdag 20 oktober 2021 kl. 23:13:51 UTC+2 skrev Transfinity:
> > >
> > > If there were infinitely many infinite endsegments, then this would happen. Of course it can't. Therefore there are not infinitely many infinite endsegments.
> > no it wouldn't, you state it but proves nothing of it.
> If the set of endsegments is infinite, then its indices are all natural numbers. Nothing remains for infinite contents. That is trivial:
> ℕ \ ℕ = { }.
> ℕ \ {indexes of infinitely many endsegments} = { }, that is the empty set.
> > There are infinitely many endsegments because they are in bijection with N
> Yes, but not infinitely many infinite endsegments.
>
> Regards, WM

All you're doing here is stating N\N={} which does not in any way prove your claim.

And yes, all endsegments are infinite because the endsegment E(n) is in bijection with N

On Thursday, October 21, 2021 at 8:33:53 PM UTC+2, WM wrote:
> zelos...@gmail.com schrieb am Donnerstag, 21. Oktober 2021 um 07:06:32 UTC+2:
> >
> > There are infinitely many endsegments because they are in bijection with N
> >
> Yes, but not infinitely many infinite endsegments.

Nonsense.

There are infinitely many infinite endsegment:

An e IN: card(E(n)) = aleph_0
"Each and every endsegment is infinite."

and

card({E(n) : n e IN}) = aleph_0.
"There are infinitely many (infinite) endsegments."

On Thursday, October 21, 2021 at 8:20:35 PM UTC+2, WM wrote:

> There is nothing between {[0,] 1, 2, 3, ...} and ω

Indeed, especially since (due to von Neumann) ω = {0, 1, 2, 3, ...}.

> but much between every [...] natnumber and ω.

Indeed! After all

An e IN: card({k e IN : n < k < ω}) = ℵo .

On Wednesday, October 20, 2021 at 11:04:03 PM UTC+2, Transfinity wrote:

> For ***all*** FISONs: |ℕ \ {1, 2, 3, ..., n}| = ℵo .

Can't you use PROPER mathematical langage (any more) you imbecile?

Ever heard of "universal quantification" and of bound variables?

See: https://en.wikipedia.org/wiki/Universal_quantification
and. https://en.wikipedia.org/wiki/Free_variables_and_bound_variables

I guess what you MEANT to state is

either

(a) For all n in IN: Iℕ \ {1, 2, 3, ..., n}| = ℵo

or

(b) For all F in Set_of_FISONs: Iℕ \ F| = ℵo

or

(c) For all FISONs F: Iℕ \ F| = ℵo, which is just short for (b).

Since
For all n in IN: Phi({1, 2, 3, ..., n}) <-> For all F in Set_of_FISONs: Phi(F) ,
it doesn't matter, which one.

Note (here): IN := {1, 2, 3, ...}, Set_of_FISONs := {{1}, {1, 2}, {1, 2, 3}, ...}.

Greg Cunt wrote on 10/22/2021 :
> On Wednesday, October 20, 2021 at 11:04:03 PM UTC+2, Transfinity wrote:
>
>> For ***all*** FISONs: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
>
> Can't you use PROPER mathematical langage (any more) you imbecile?
>
> Ever heard of "universal quantification" and of bound variables?
>
> See: https://en.wikipedia.org/wiki/Universal_quantification
> and. https://en.wikipedia.org/wiki/Free_variables_and_bound_variables
>
> I guess what you MEANT to state is
>
> either
>
> (a) For all n in IN: Iℕ \ {1, 2, 3, ..., n}| = ℵo
>
> or
>
> (b) For all F in Set_of_FISONs: Iℕ \ F| = ℵo
>
> or
>
> (c) For all FISONs F: Iℕ \ F| = ℵo, which is just short for (b).
>
> Since
> For all n in IN: Phi({1, 2, 3, ..., n}) <-> For all F in Set_of_FISONs:
> Phi(F) , it doesn't matter, which one.
>
> Note (here): IN := {1, 2, 3, ...}, Set_of_FISONs := {{1}, {1, 2}, {1, 2, 3},
> ...}.

You seem to be using an uppercase "I" for your leading pipe character
"|" for "absolute value" or in this case "cardinality". The glyph used
for the natural numbers is already 'blackboard bold' or 'doublestruck'
not needing the "I".

On Friday, October 22, 2021 at 11:27:04 AM UTC+2, FromTheRafters wrote:

> You seem to be using an uppercase "I" for your leading pipe character
> "|" for "absolute value" or in this case "cardinality". The glyph used
> for the natural numbers is already 'blackboard bold' or 'doublestruck'
> not needing the "I".

Jep. A typo.

On Friday, October 22, 2021 at 10:48:15 AM UTC+2, Greg Cunt wrote [typo fixed]:

> For ***all*** FISONs: |ℕ \ {1, 2, 3, ..., n}| = ℵo. [WM]

Can't you use PROPER mathematical langage, you imbecile?

Ever heard of _universal quantification_ and of _bound variables_?

See: https://en.wikipedia.org/wiki/Universal_quantification
and. https://en.wikipedia.org/wiki/Free_variables_and_bound_variables

I guess what you MEANT to state is either

(a) For all n in IN: |ℕ \ {1, 2, 3, ..., n}| = ℵo

or

(b) For all F in Set_of_FISONs: |ℕ \ F| = ℵo

or

(c) For all FISONs F: |ℕ \ F| = ℵo, which is just short for (b).

Since
For all n in IN: Phi({1, 2, 3, ..., n}) <-> For all F in Set_of_FISONs: Phi(F) ,
it doesn't matter, which one.

Note (here): ℕ := {1, 2, 3, ...}, Set_of_FISONs := {{1}, {1, 2}, {1, 2, 3}, ...}.

On Friday, October 22, 2021 at 10:48:15 AM UTC+2, Greg Cunt wrote [typo fixed]:

> For ***all*** FISONs: |ℕ \ {1, 2, 3, ..., n}| = ℵo. [WM]

Can't you use PROPER mathematical langage, you imbecile?

Ever heard of _universal quantification_ and of _bound variables_?

See: https://en.wikipedia.org/wiki/Universal_quantification
and. https://en.wikipedia.org/wiki/Free_variables_and_bound_variables

I guess what you MEANT to state is either

(a) For all n in ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo

or

(b) For all F in Set_of_FISONs: |ℕ \ F| = ℵo

or

(c) For all FISONs F: |ℕ \ F| = ℵo, which is just short for (b).

Since
For all n in ℕ: Phi({1, 2, 3, ..., n}) <-> For all F in Set_of_FISONs: Phi(F) ,
it doesn't matter, which one.

Note (here): ℕ := {1, 2, 3, ...}, Set_of_FISONs := {{1}, {1, 2}, {1, 2, 3}, ...}.

Greg Cunt brought next idea :
> On Friday, October 22, 2021 at 11:27:04 AM UTC+2, FromTheRafters wrote:
>
>> You seem to be using an uppercase "I" for your leading pipe character
>> "|" for "absolute value" or in this case "cardinality". The glyph used
>> for the natural numbers is already 'blackboard bold' or 'doublestruck'
>> not needing the "I".
>
> Jep. A typo.

Ah, just as | expected.

Dan Christensen schrieb am Donnerstag, 21. Oktober 2021 um 21:59:47 UTC+2:

> > That is a basic reality: There are not two consecutive infinite sets in |N.
> Irrelevant gibberish.

Think over it. Come back when you have recognized it, but not before.

Regards, WM

Gus Gassmann schrieb am Donnerstag, 21. Oktober 2021 um 21:10:38 UTC+2:
> On Thursday, 21 October 2021 at 15:20:35 UTC-3, WM wrote:
> > Fact is, whether you like it or not: There is nothing between {1, 2, 3, ...} and ω but much between every definable natnumber and ω.
> Since you have no clue what quantifier switching actually looks like

That does not change the facts: There is nothing between {1, 2, 3, ...} and ω but much between every definable natnumber and ω.

Regards, WM

Jim Burns schrieb am Freitag, 22. Oktober 2021 um 00:57:24 UTC+2:
> On 10/21/2021 2:17 PM, WM wrote:

> For each ordinal 𝛂 < 𝛚, the successorship runs
> _through_ 𝛂 on the way _to_ 𝛂+1.
> What the successorship runs through is
> the essence of what we mean by finite.

Zorn's lemma is explained by: every element would have another next element.. This successorship would not stop at natural indices but would run through all ordinal numbers. That shiws the direct way from 1 to 𝛚 and beyond,
>
> 𝛚 is the first infinite ordinal.
> The successorship does not run _to_ 𝛚
> If the successorship ran _to_ 𝛚, 𝛚 would be
> included in what we mean by finite.

Of course Zorn's lemma is as silly as the axiom of choice for uncountable sets. But since there are no uncountable sets ...
>
> ----
> The successorship runs to all the things which the
> successorship runs to. It does not run to anything
> it does not run to. This seems like a useless observation.

Of course we cannot count to omega. The numbers before omega are dark. But without that feature, we could do what the explanation of Zorn's lemma says..

Regards, WM

zelos...@gmail.com schrieb am Freitag, 22. Oktober 2021 um 06:39:18 UTC+2:

> And yes, all endsegments are infinite because the endsegment E(n) is in bijection with N

If all indexes are an infinite subset of |N and the contents is infinite too. Where is the caesura?

Regards, WM

Crank Wolfgang Mueckenheim, WM wrote:
> zelos...@gmail.com schrieb am Freitag, 22. Oktober 2021 um 06:39:18 UTC+2:
>
>
>> And yes, all endsegments are infinite because the endsegment E(n) is in bijection with N
>
> If all indexes are an infinite subset of |N and the contents is infinite too. Where is the caesura?

Following your idiotic "argument", Crank Wolfgang Mueckenheim, from
Hochschule Augsburg, NxN does not exist because the "first" N in the
cartesian product "exhaust" N, so the second one is the empty set.

How can you be so asinine, Crank Wolfgang Mueckenheim, from Hochschule
Augsburg?

On Friday, October 22, 2021 at 11:30:08 AM UTC-4, WM wrote:
> Dan Christensen schrieb am Donnerstag, 21. Oktober 2021 um 21:59:47 UTC+2:
>
> > > That is a basic reality: There are not two consecutive infinite sets in |N.
> > Irrelevant gibberish.
> Think over it. Come back when you have recognized it, but not before.
>

Once again, you have failed to prove your outrageous claim that there are only finitely many end-segments in N.

Once again, after 40+ years (?) of futile effort, you have failed to demonstrate any inconsistencies in the ZFC axioms of set theory. When will you learn, Mucke???

Dan

On Friday, 22 October 2021 at 12:30:08 UTC-3, WM wrote:
> Dan Christensen schrieb am Donnerstag, 21. Oktober 2021 um 21:59:47 UTC+2:
>
> > > That is a basic reality: There are not two consecutive infinite sets in |N.
> > Irrelevant gibberish.
> Think over it. Come back when you have recognized it, but not before.

Since you can't even formulate this simple idea properly: I have thought it over, and I still recognize your shit as irrelevant gibberish.

On Friday, 22 October 2021 at 12:31:54 UTC-3, WM wrote:
> Gus Gassmann schrieb am Donnerstag, 21. Oktober 2021 um 21:10:38 UTC+2:
> > On Thursday, 21 October 2021 at 15:20:35 UTC-3, WM wrote:
> > > Fact is, whether you like it or not: There is nothing between {1, 2, 3, ...} and ω but much between every definable natnumber and ω.
> > Since you have no clue what quantifier switching actually looks like
> That does not change the facts: There is nothing between {1, 2, 3, ...} and ω but much between every definable natnumber and ω.

Sure. For *EVERY* natural number n the set {m in N: n <= m < ω} has cardinality aleph_0. This set is called an *ENDSEGMENT*.

But it is interesting that you are once again trying to hide our glaring mistake that led to our exchange. Let me refresh your memory:

Gus Gassmann schrieb am Donnerstag, 21. Oktober 2021 um 01:58:14 UTC+2:
[...]
> > ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo (*)
> > But ℕ \ {1, 2, 3, ...} = { }. (**)
> > The last line shows that not all natnumbers have an infinite distance from ω.
> The last line shows again that you have no idea about mathematics. You switched quantifiers again.

You *STILL* have no qualms about treating (*) and (**) as the same statement. You don't have a *FUCKING CLUE* about mathematics.

On Friday, October 22, 2021 at 6:30:30 PM UTC+2, Gus Gassmann wrote:
> On Friday, 22 October 2021 at 12:30:08 UTC-3, WM wrote:
> > Dan Christensen schrieb am Donnerstag, 21. Oktober 2021 um 21:59:47 UTC+2:
> > > >
> > > > That is a basic reality: There are not two consecutive infinite sets in |N.
> > > >
> > > Irrelevant gibberish.
> > >
> > Think over it. Come back when you have recognized it, but not before.
> >
> Since you can't even formulate this simple idea properly: I have thought it over, and I still recognize your shit as irrelevant gibberish.

Well, the reason for this might be the fact that it *is* irrelevant gibberish, I'd say.

Greg Cunt schrieb am Freitag, 22. Oktober 2021 um 10:48:15 UTC+2:
> On Wednesday, October 20, 2021 at 11:04:03 PM UTC+2, Transfinity wrote:
>
> > For ***all*** FISONs: |ℕ \ {1, 2, 3, ..., n}| = ℵo .

> I guess what you MEANT to state is
>
> either
>
> (a) For all n in IN: Iℕ \ {1, 2, 3, ..., n}| = ℵo

No, that could misdirect the reader. Not all natnumbers can be last elements of FISONs.
>
> or
>
> (b) For all F in Set_of_FISONs: Iℕ \ F| = ℵo

Yes, but that is the same as for all FISONs.
>
> (c) For all FISONs F: Iℕ \ F| = ℵo, which is just short for (b).
>
> Since
> For all n in IN: Phi({1, 2, 3, ..., n}) <-> For all F in Set_of_FISONs: Phi(F) ,
> it doesn't matter, which one.

That is wrong because not every n can appear as last element of a FISON.
>
> Note (here): IN := {1, 2, 3, ...}, Set_of_FISONs := {{1}, {1, 2}, {1, 2, 3}, ...}.

But the set of FISONs is rather a collection.

Regards, WM

Python schrieb am Freitag, 22. Oktober 2021 um 17:49:04 UTC+2:
> Wolfgang Mueckenheim, WM wrote:
> > zelos...@gmail.com schrieb am Freitag, 22. Oktober 2021 um 06:39:18 UTC+2:
> >
> >
> >> And yes, all endsegments are infinite because the endsegment E(n) is in bijection with N
> >
> > If all indexes are an infinite subset of |N and the contents is infinite too. Where is the caesura?
> Following your"argument", Wolfgang Mueckenheim, from
> Hochschule Augsburg, NxN does not exist because the "first" N in the
> cartesian product "exhaust" N, so the second one is the empty set.

No, you have obviously not understood. You can write as many 1's or 2' or n's as you like. But the numbers in an endsegment cannot serve as indices for the predecessors of this endsegment.
>
Regards, WM

Gus Gassmann schrieb am Freitag, 22. Oktober 2021 um 18:30:30 UTC+2:
> I have thought it over, and I still

not understood. The numbers contained in an endsegment cannot serve as indexes for its predecessors. Example E(17) = {17, 18, 19, ..}. The numbers 17, 18, 19, ... cannot serbe as indices for endegments E(1) to E(16).

If every endsegment is infinite, then infinitely many numbers cannot serve as indices for endsegments. Then only the complement of this infinite set is available as indices. The complement of an infinite contents is finite.

Regards, WM

Gus Gassmann schrieb am Freitag, 22. Oktober 2021 um 18:38:34 UTC+2:
> On Friday, 22 October 2021 at 12:31:54 UTC-3, WM wrote:

> > > ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo (*)
> > > But ℕ \ {1, 2, 3, ...} = { }. (**)
> > > The last line shows that not all natnumbers have an infinite distance from ω.
> You *STILL* have no qualms about treating (*) and (**) as the same statement.

No, these are very different statements. (*) concerns all definable natnumbers, (**) concerns all natnumbers.

Regards, WM

Dan Christensen schrieb am Freitag, 22. Oktober 2021 um 18:21:44 UTC+2:
> On Friday, October 22, 2021 at 11:30:08 AM UTC-4, WM wrote:
> > Dan Christensen schrieb am Donnerstag, 21. Oktober 2021 um 21:59:47 UTC+2:
> >
> > > > That is a basic reality: There are not two consecutive infinite sets in |N.
> > > Irrelevant gibberish.
> > Think over it. Come back when you have recognized it, but not before.
> >
> Once again, you have failed to prove your outrageous claim that there are only finitely many end-segments in N.

Once again: Come back when you have recognized it, but not before.

Regards, WM