On 10/22/2021 11:37 AM, WM wrote:
> Jim Burns schrieb
> am Freitag, 22. Oktober 2021 um 00:57:24 UTC+2:

>> 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.

<wiki>
| Motivation
| | To prove the existence of a mathematical object that
| can be viewed as a maximal element in some partially
| ordered set in some way, one can try proving the
| existence of such an object by assuming there is no
| maximal element and using transfinite induction and
| the assumptions of the situation to get a contradiction.
| Zorn's lemma tidies up the conditions a situation needs
| to satisfy in order for such an argument to work and
| enables mathematicians to not have to repeat the
| transfinite induction argument by hand each time,
| but just check the conditions of Zorn's lemma.

> Zorn's lemma is explained by:
> every element would have another next element.

Let P have an order '=<' which is reflexive, antisymmetric,
and transitive.

Zorn's lemma
| | Either there is an element in P with no followers in P,
| or there is a chain in P that with no end in the chain.

Define
N such that
if k ends a FISON, then k is in N, and
if k is in N, then k ends a FISON.

In the case of N, the second option applies:
there is a chain in N that with no end in the chain.
(That chain is all of N.)

> This successorship would not stop at natural indices

This successorship does not stop. Anywhere.
It is a chain with no end in the chain.

> 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,

This successorship does not run through anything beyond
all natural indices.

Things beyond all natural indices can have successors.
First infinite 𝛚 can have 𝛚+1. First inaccessible 𝛋
can have 𝛋+1.

However, "𝛂 has a successor" is not the same as "this
successorship runs through 𝛂". You do not _explicitly_
define the concept, but one conclusion you draw from
your claim that this successorship runs through 𝛚
is that 𝛚 has an immediate predecessor 𝛚-1.

That conclusion is only valid if this successorship is
limited to running through only such 𝛂 as have an
immediate predecessor 𝛂-1 or 𝛂 = 0. And that is only valid
if what this successorship runs through is limited to
the natural indices (AKA finite ordinals).

https://en.wikipedia.org/wiki/Natural_number#Modern_definitions
| | [Peano arithmetic] is based on an axiomatization of the
| properties of ordinal numbers: each natural number has
| a successor and every non-zero natural number has a
| unique predecessor.

https://en.wikipedia.org/wiki/Axiom_of_infinity#Extracting_the_natural_numbers_from_the_infinite_sect
| | The natural numbers can be defined in a way which
| does not assume any axioms except the axiom of
| extensionality and the axiom of induction --
| a natural number is either zero or a successor and
| each of its elements is either zero or a successor
| of another of its elements.

This successorship does not run through 𝛚.
If this successorship ran through 𝛚, then 𝛚,𝛚+1,𝛚+2,...
would be natural indices, and _what we mean_ by 𝛚
is that 𝛚 follows all natural indices.

>> 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.

The numbers before omega can be counted to.
All of them.
It is their defining characteristic.

We cannot count to omega because,
if we could count to omega, it would be just another
natural number, a thing we can count to, and not
after all the things we can count to.

> But without that feature,
> we could do what the explanation of Zorn's lemma says.

On Friday, October 22, 2021 at 4:28:00 PM UTC-4, WM wrote:
> 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.
>

I will be back to kick your lying ass, Mucke! Get used to it. This is not your Crank Institute. You will not get a free ride for your lies and nonsense here.

Dan

On Friday, October 22, 2021 at 11:32:02 PM UTC+2, Dan Christensen wrote:

@WM:

> This is not your Crank Institute. You will not get a free ride for your lies and nonsense here.

Agree.

On Friday, October 22, 2021 at 10:12:57 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Freitag, 22. Oktober 2021 um 10:48:15 UTC+2:

> > (b) For all F in Set_of_FISONs: |ℕ \ F| = ℵo
> >
[...] that is the same as for all FISONs.

No, you silly idtiot, this is equivalent to

| For all FISONs F: |ℕ \ F| = ℵo

(note the bound variable "F" after the phrase "For all FISONs").

> > For all n in IN: Phi({1, 2, 3, ..., n}) <-> For all F in Set_of_FISONs: Phi(F) ,
> >
> That is wrong

No, this is not wrong, you silly idiot.

> because not every n [in IN] can appear as last element of a FISON.

Each and every n in IN "appears" as last element of a FISON, you psychotic asshole fill of shit.

Hint: An e IN: {1, ..., n} e Set_of_FISONs

crank Wolfgang Mueckenheim, aka WM wrote:
> 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.

There is nothing to understood but for a psychiatrist, fool.

> 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.

Of course the can, idiot crank Wolfgang Mueckenheim, from
Hochschule Augsburg. What prevent me to index {n,n+1,...}
by another number then n or n-1 in another sequence...

Moreover your gibberish doesn't even anymore try to make sense,
crank Wolfgang Mueckenhein, from Hochschule Augsburg;

crank Wolfgang Mueckenheim, aka WM wrote:
> 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).

but they can for E(17), E(18), etc. Are you really that dumb?

> 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.

Oh dear... 'then infinitely many numbers cannot serve as indices for
*some* endsegments' but 'all can serve as indicies for *others*
endsegments".

You're going mad, aren't you crank Wolfgang Mueckenheim, from Hochschule
Augsburg?

Crank Wolfgang Mueckenheim, aka WM wrote:
> 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.

Once again: shut the f*ck up, crank. You'll be hit in the face sooner
or later, at Hochschule Augsburg.

You have NO authority here. Don't confuse Usenet or real Math with the
classrooms where you abuse students at Hochschule Augsburg for years.

On 10/22/2021 3:12 PM, WM wrote:
> 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.

"Not all natnumbers can be last elements of FISONs"

Wrong. Name them. You cant.

>>
>> 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.

Wrong. Name them.

>>
>> Note (here): IN := {1, 2, 3, ...}, Set_of_FISONs := {{1}, {1, 2}, {1, 2, 3}, ...}.
>
> But the set of FISONs is rather a collection.

no it is a set. You call it a set.

>
> Regards, WM
>

On 10/21/2021 1:33 PM, WM wrote:
> 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
>

you have confused yourself totally.

Every one of your posts, has errors in it. That is what internet cranks do.

On 10/22/2021 3:16 PM, WM wrote:
> 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.

you are mumbling mumbo jumbo. If you used equations, you could be clear. Give it a try!

>>
> Regards, WM
>

On 10/21/2021 2:07 PM, WM wrote:
> Serg io schrieb am Donnerstag, 21. Oktober 2021 um 20:46:15 UTC+2:
>
>> "exhausted" is not used in Math.
>
> Who told you? Tell him that he was wrong. "the set P will be exhausted before long" (Wiki)
>
> Regards, WM
>

you quote wiki! "exhausted" is not used in Math.

Sets are fixed. You do not understand that.

One does operations to fixed sets, and they become a different set.

On 10/22/2021 3:27 PM, WM wrote:
> 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
>

It is a boldface Lie, by one that hates Math.

On 10/22/2021 3:21 PM, WM wrote:
> 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).

Wrong. Lets try it again.

Example E(17) = {17, 18, 19, ..}. The numbers 17, 18, 19, ... are elements in the Endegments E(1) to E(16).

or did you mean: E(1) = {1,2,3,...} and the numbers 1,2,3 are not elements in the endsegments E(4) and above.

Anyhow you are using the words "indexes", "contained", and "serve" very poorly. Please use the correct nomenclature!

I give you an F for Fail, because you are not using correct nomenclature.

>
> 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.

that statement is meaningless mush, AND you use incorrect nomenclature, such as "cannot serve", "indices", "available", "contents"

no math here at all.

>
> Regards, WM
>

Jim Burns schrieb am Freitag, 22. Oktober 2021 um 23:10:23 UTC+2:
> On 10/22/2021 11:37 AM, WM wrote:

> > This successorship would not stop at natural indices
> > but would run through all ordinal numbers.
> > That shows the direct way from 1 to 𝛚 and beyond,
> This successorship does not run through anything beyond
> all natural indices.

Then try to correct Wikipedia.

Using the function b, 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.

> This successorship does not run through 𝛚.

Then try to correct Wikipedia.

> > Of course we cannot count to omega.
> > The numbers before omega are dark.
> The numbers before omega can be counted to.
> All of them.

Wrong. Provably wrong. Every number you count has a distance of aleph_0 from omega. But you claim that all numbers are removed in case of endsegments. No distance aleph_0 remains. Hence there are more numbers than can be counted.

> We cannot count to omega because,
> if we could count to omega, it would be just another
> natural number,

How can you empty the intersection of endsegments?

Regards, WM

Dan Christensen schrieb am Freitag, 22. Oktober 2021 um 23:32:02 UTC+2:

> > Once again: Come back when you have recognized it, but not before.
> >
> I will be back to kick your lying ass,

That is the field of your expertise?

Regards, WM

Python schrieb am Samstag, 23. Oktober 2021 um 01:37:58 UTC+2:
> WM wrote:

the numbers in an endsegment cannot serve as indices for the predecessors of this endsegment.
> Of course the can,

No, but you are obviously unable to understand even the simple example:
The numbers in E(4) = {4, 5, 6, ...} cannot serve as indices for E(1), E(2), an E(3).

Regards, W;

On Saturday, October 23, 2021 at 1:54:09 PM UTC-4, WM wrote:
> Dan Christensen schrieb am Freitag, 22. Oktober 2021 um 23:32:02 UTC+2:
>
> > > Once again: Come back when you have recognized it, but not before.
> > >
> > I will be back to kick your lying ass, Mucke!

> That is the field of your expertise?
>

One of them, maybe.

> > Get used to it. This is not your Crank Institute. You will not get a free ride for your lies and nonsense here.

Dan

On Saturday, October 23, 2021 at 7:57:46 PM UTC+2, WM wrote:

> The numbers in E(4) = {4, 5, 6, ...} cannot serve as indices for E(1), E(2), an E(3).

They could, but actually, they don't (here: (E(1), E(2), E(3), E(4), E(5), E(6), ...)).

Proof: Reorder the sequence (E(1), E(2), E(3), E(4), E(5), E(6), E(7), E(8), E(9), ...) the following way:
(E(4), E(5), E(6), E(1), E(2), E(3), E(7), E(8), E(9), ...).

Now the indexes for the sets E(1), E(2), E(3) (in that order) are 4, 5, 6 (in that order). So some numbers in E(4) = {4, 5, 6, ...} (namely 4, 5, 6) "can" serve as indexes for E(1), E(2), an E(3), since they actually do (in this case).

Python schrieb am Samstag, 23. Oktober 2021 um 01:40:41 UTC+2:
> WM wrote:
> > 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 serve as indices for endegments E(1) to E(16).
> but they can for E(17), E(18), etc.

Of course. But if all endsegments are infinite then infinitely many numbers cannot serve as indices. Only the complement, namely finitely many numbers, can.

> > 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.
> Oh dear... 'then infinitely many numbers cannot serve as indices for
> *some* endsegments' but 'all can serve as indicies for *others*
> endsegments".

No, if there are no endsegments with less than infinitely many numbers, then infinitely many numbers cannot serve as indices.

Regards, WM

Greg Cunt schrieb am Samstag, 23. Oktober 2021 um 20:08:52 UTC+2:
> On Saturday, October 23, 2021 at 7:57:46 PM UTC+2, WM wrote:
>
> > The numbers in E(4) = {4, 5, 6, ...} cannot serve as indices for E(1), E(2), an E(3).
> They could, but actually, they don't (here: (E(1), E(2), E(3), E(4), E(5), E(6), ...)).
>
> Proof: Reorder the sequence

The sequence is and remains ordered in the natural way. Only then my proof is valid.

Regards, WM

On Saturday, October 23, 2021 at 8:12:00 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Samstag, 23. Oktober 2021 um 20:08:52 UTC+2:
> > On Saturday, October 23, 2021 at 7:57:46 PM UTC+2, WM wrote:
> > >
> > > The numbers in E(4) = {4, 5, 6, ...} cannot serve as indices for E(1), E(2), an E(3).
> > >
> > They could, but actually, they don't (here: (E(1), E(2), E(3), E(4), E(5), E(6), ...)).
> >
> > Proof: Reorder the sequence [...]
> >
> The sequence is and remains ordered in the natural way.

It cannot be reordered in mückenmath? Is this a dogma of mückenmath (par ordre du mufti)?

Fascinating! In the context of mathematics each and every sequence _can be_ reordered!

On Saturday, October 23, 2021 at 8:09:55 PM UTC+2, WM wrote:

> infinitely many numbers cannot serve as indices.

They can, because they do, you psychotic asshole full of shit.

Hint: IN is the index set of so called "infinite sequences", and hence of (E(1), E(2), E(3), ...) too.

This means, each and every natural number "serves as an index" here.

https://en.wikipedia.org/wiki/Sequence

Crank Wolfgang Mueckenheim, aka WM wrote:
> ... Only then my proof is valid.

You have never written a single proof in your whole life, crank
Wolfgang Mueckenheim from Hochschule Augsburg. And you'll never
do.

Greg Cunt schrieb am Samstag, 23. Oktober 2021 um 20:20:51 UTC+2:
> On Saturday, October 23, 2021 at 8:09:55 PM UTC+2, WM wrote:
>
> > infinitely many numbers cannot serve as indices.
> They can, because they do,

No, they don't. Only stupid matheologians cannot understand that if there are no endsegments with less than infinitely many numbers, then infinitely many numbers cannot serve as indices.
>
> This means, each and every natural number "serves as an index" here.

Then each and every number must leave the endsegments. Then they get empty. But that cannot be observed. They are dark.

Regards, WM

Greg Cunt schrieb am Samstag, 23. Oktober 2021 um 20:17:00 UTC+2:
> On Saturday, October 23, 2021 at 8:12:00 PM UTC+2, WM wrote:
> > Greg Cunt schrieb am Samstag, 23. Oktober 2021 um 20:08:52 UTC+2:
> > > On Saturday, October 23, 2021 at 7:57:46 PM UTC+2, WM wrote:
> > > >
> > > > The numbers in E(4) = {4, 5, 6, ...} cannot serve as indices for E(1), E(2), an E(3).
> > > >
> > > They could, but actually, they don't (here: (E(1), E(2), E(3), E(4), E(5), E(6), ...)).
> > >
> > > Proof: Reorder the sequence [...]
> > >
> > The sequence is and remains ordered in the natural way.

> Fascinating! In the context of mathematics each and every sequence _can be_ reordered!

It can be reordered, but I consider it in its natural order. There the sequence of infinite endsegments is finite. (It would be finite in every order, but why consider that?)

Regards, WM