On 10/20/2021 4:13 PM, Transfinity wrote:
> zelos...@gmail.com schrieb am Dienstag, 19. Oktober 2021 um 11:48:44 UTC+2:
>> tisdag 19 oktober 2021 kl. 11:36:43 UTC+2 skrev WM:
>
>>> If there is an infinite initial segment in |N which is followed by an infinite endsegment,
>> There are no such things so your point is moot.
>
> If there were infinitely many infinite endsegments, then this would happen. Of course it can't. Therefore there are not nfinitely many infinite endsegments.

thupppp!!! WRONG-O (again)

1 All Endsegments are infinite, E(n) where n is a natural number.

2. There are an infinity of Endsegments

>
>>> Logic says that an infinite sequence followed by an infinite sequence with no or at most finite overlap yields two infinite sequences. What do you think it says?
>> There is nothing in mathematics that says that is so in natural numbers.
>
> If there were infinitely many infinite endsegments, then this would happen. Of course it can't.

thupppp!!! WRONG-O (again)

1 All Endsegments are infinite, E(n) where n is a natural number.

2. There are an infinity of Endsegments

>
> Regards, WM
>

On Wednesday, October 20, 2021 at 4:42:55 PM UTC-4, WM wrote:
> Dan Christensen schrieb am Dienstag, 19. Oktober 2021 um 18:14:57 UTC+2:
> > On Tuesday, October 19, 2021 at 1:53:34 AM UTC-4, WM wrote:
>
> > > Sorry for inconvenience. But do you really believe that an infinite sequence of indices 1, 2, 3, ... of endsegments can be followed by an infinite sequence of members of endsegments?
> > >
> > It is a fact that there are infinitely many end-segments {n, n+1, n+2, .... } in N.

> It is fact that this is impossible.

Liar.

> > And each one of them is infinite.

> Try mathematics. Infinitely many endsegments exhaust all natnumbers 1, 2, 3, ... as indexes.

Wrong again, Mucke. The infinitely many end-segments are a mathematical reality. Numbers don't get "exhausted." You can use them the same numbers over and over again. Clearly your goofy little system is broken beyond repair if you have convinced yourself otherwise.

Dan

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

On 10/20/2021 3:50 PM, WM wrote:
> Greg Cunt schrieb am Dienstag, 19. Oktober 2021 um 15:41:24 UTC+2:
>> On Monday, October 18, 2021 at 11:23:05 AM UTC+2, WM wrote:
>>
>>>> ALL natural numbers have aleph_0 successors
>
>>> Here is the refutation: |ℕ \ {1, 2, 3, ...}| = 0 .
>
>> since "{1, 2, 3, ...}" denotes the very same set as "ℕ", you just stated that
>>
>> |ℕ \ ℕ| = 0 ,
>>
>> which is A TRIVIAL theorem,
>
> Of course. It is trivial that there is nothing between ℕ, i.e., all natural numbers, and ω.

Wrong. ω is not in ℕ

>>
>> Is this the refutation of the claim
>
> Of course. It shows that nothing is between all natural numbers and ω. Not all natural numbers have ℵ₀ successors.

which numbers ? you are the one making outrageous false claims.

On 10/20/2021 3:52 PM, WM wrote:
> Greg Cunt schrieb am Dienstag, 19. Oktober 2021 um 15:36:28 UTC+2:
>
>> you explicitly claimed: "n isn't a natural number" and "n+1 is not a natural number too."
>> Hence the claim
>>
>> "There are natnumbers in {n, n+1}"
>>
>> implies a contradiction.
>
> Thank you for your attention. But I told you already that this phrase "the natural number n has property X" is merely an abbreviation for "every natural number which can be inserted in place of n has property X".

and you are wrong, as you have added more constraints upon n

On 10/20/2021 5:10 PM, Transfinity wrote:
> 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.

Only if omega is finite.

Since omega is NOT finite, running to omega would
require a crossing-pair j,j+1 such that j is finite
and j+1 is infinite.

( Read this as:
( {0,...,j} for which each split of {0,...,j}
( has a crossing-pair i,i+1
( but NOT for which each split of {0,...,j}∪{j+1}
( has a crossing-pair i,i+1. Such a j,j+1 can't be.

On Wednesday, 20 October 2021 at 18:04:03 UTC-3, Transfinity wrote:
[...]

> No, there are far less than aleph_0. Simply adhere to mathematics:
> For ***all*** FISONs: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
> If you don't trust this simple and clear theorem, then I cannot help you.
> Similarly
> ∀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. Par for the course, I guess.

On 10/20/2021 6:58 PM, Gus Gassmann wrote:
> On Wednesday, 20 October 2021 at 18:04:03 UTC-3, Transfinity wrote:
> [...]
>
>> No, there are far less than aleph_0. Simply adhere to mathematics:
>> For ***all*** FISONs: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
>> If you don't trust this simple and clear theorem, then I cannot help you.
>> Similarly
>> ∀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. Par for the course, I guess.
>

WM is out to lunch with ω

On Wednesday, 20 October 2021 at 18:04:03 UTC-3, Transfinity wrote:
> But ℕ \ {1, 2, 3, ...} = { }.
> The last line shows that not all natnumbers have an infinite distance from ω.

Of course it does not. You have no clue about distances, since you clearly have no idea about even far simpler mathematical concepts. Of course one (obviously not you!) *can* define a distance on |N in such a way that the sequence {1, 2, 3, 4, ...} is a Cauchy sequence. That doesn't show anything along the lines you want, any more than lim{n->oo} 1/n = 0 even *suggests* that 1/n must be zero for n large enough. The mere hint of a thought that this might happen shows that you are insane.

On Wednesday, 20 October 2021 at 20:02:00 UTC-3, Serg io wrote:
> On 10/20/2021 3:50 PM, WM wrote:
[...]
> > Of course. It shows that nothing is between all natural numbers and ω. Not all natural numbers have ℵ₀ successors.

I constantly marvel at how easy and naturally it comes to you to exchange quantifiers. This clearly is a mental defect. It may not be Alzheimer's, because you had this quantifier dyslexia for a long time now, but your Alzheimer's does not help things on that front, either.

onsdag 20 oktober 2021 kl. 22:42:55 UTC+2 skrev WM:
> Dan Christensen schrieb am Dienstag, 19. Oktober 2021 um 18:14:57 UTC+2:
> > On Tuesday, October 19, 2021 at 1:53:34 AM UTC-4, WM wrote:
>
> > > Sorry for inconvenience. But do you really believe that an infinite sequence of indices 1, 2, 3, ... of endsegments can be followed by an infinite sequence of members of endsegments?
> > >
> > It is a fact that there are infinitely many end-segments {n, n+1, n+2, ... } in N.
> It is fact that this is impossible.
> > And each one of them is infinite.
> Try mathematics. Infinitely many endsegments exhaust all natnumbers 1, 2, 3, ... as indexes. Or could there be a smaller infinity? No. What remains a contents?
>
> Regards, WM

Try mathematics? YOU'RE THE ONE FAILING AT BASIC MATHEMATICS! you are so fucking retarded!

onsdag 20 oktober 2021 kl. 22:50:13 UTC+2 skrev WM:
> Greg Cunt schrieb am Dienstag, 19. Oktober 2021 um 15:41:24 UTC+2:
> > On Monday, October 18, 2021 at 11:23:05 AM UTC+2, WM wrote:
> >
> > >> ALL natural numbers have aleph_0 successors
> > > Here is the refutation: |ℕ \ {1, 2, 3, ...}| = 0 .
> > since "{1, 2, 3, ...}" denotes the very same set as "ℕ", you just stated that
> >
> > |ℕ \ ℕ| = 0 ,
> >
> > which is A TRIVIAL theorem,
> Of course. It is trivial that there is nothing between ℕ, i.e., all natural numbers, and ω.
> >
> > Is this the refutation of the claim
> Of course. It shows that nothing is between all natural numbers and ω. Not all natural numbers have ℵ₀ successors. But all definable numbers have ℵ₀ successors between themselves and omega:
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> Regards, WM

How mentally retarded are you? N\N={} proves NOTHING about natural numbers not having successors which they in fact do! Holy shit are you getting dumber by the day?

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:
> > tisdag 19 oktober 2021 kl. 11:32:12 UTC+2 skrev WM:
> > >
> > > Sorry that is as wrong and cannot serve as an argument. There are infinitely many real numbers beyond every definable one. But there is a point where the cursor encounters zero. Why should it, if always infinitely many points would lie between the cursor and zero? Can you imagine a reason?
> > >The axiom of choice is equivalent to Zorn's lemma: I
> > It is but this won't help you.
> > >If a partially ordered set S has the property that every chain has an upper bound in S, then the set S contains at least one maximal element. Without this feature, every element would have another next element. This successorship would not stop at natural indices but would run through all ordinal numbers.
> > And it does not. It doesn't apply to natural numbers as a whole because not every chain has it.
> But many chains have it. Note: This successorship would not stop at natural indices but would run through all ordinal numbers. That means there is a direct transition from natnumbers to omega and larger.
> > 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.
>
> Regards, WM

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.

onsdag 20 oktober 2021 kl. 23:13:51 UTC+2 skrev Transfinity:
> zelos...@gmail.com schrieb am Dienstag, 19. Oktober 2021 um 11:48:44 UTC+2:
> > tisdag 19 oktober 2021 kl. 11:36:43 UTC+2 skrev WM:
>
> > >If there is an infinite initial segment in |N which is followed by an infinite endsegment,
> > There are no such things so your point is moot.
> If there were infinitely many infinite endsegments, then this would happen. Of course it can't. Therefore there are not nfinitely many infinite endsegments.
> > >Logic says that an infinite sequence followed by an infinite sequence with no or at most finite overlap yields two infinite sequences. What do you think it says?
> > There is nothing in mathematics that says that is so in natural numbers.
> If there were infinitely many infinite endsegments, then this would happen. Of course it can't.
>
> Regards, WM

> If there were infinitely many infinite endsegments, then this would happen. Of course it can't. Therefore there are not nfinitely many infinite endsegments.

no it wouldn't, you state it but proves nothing of it. You have not proven your claim so fuck off.

There are infinitely many endsegments because they are in bijection with N

On Wednesday, October 20, 2021 at 10:52:22 PM UTC+2, WM wrote:

> Thank you for <bla>

Look, you silly crank, you explicitly claimed: "n isn't a natural number" and "n+1 is not a natural number too."

Hence the claim

"There are natnumbers in {n, n+1}"

implies a contradiction.

Hint: To express "for every ..." the _universal quantifier_ is used in mathematics.

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

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

> 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.

Regards, WM

Jim Burns schrieb am Donnerstag, 21. Oktober 2021 um 01:07:28 UTC+2:
> On 10/20/2021 5:10 PM, Transfinity wrote:
> > 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.
> Only if omega is finite.

It is not. Nevertheless cp. Wikipedia Zorn's lemma:

If you are building a mathematical object in stages and find that (i) you have not finished even after infinitely many stages, and (ii) there seems to be nothing to stop you continuing to build, then Zorn’s lemma may well be able to help you.— William Timothy Gowers, "How to use Zorn’s lemma"

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. 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.
>
> Since omega is NOT finite, running to omega would
> require a crossing-pair j,j+1 such that j is finite
> and j+1 is infinite.

You see your crossing-pair argument is invalid.
>
Regards, WM

Gus Gassmann schrieb am Donnerstag, 21. Oktober 2021 um 01:58:14 UTC+2:
> On Wednesday, 20 October 2021 at 18:04:03 UTC-3, Transfinity wrote:
> [...]
> > No, there are far less than aleph_0. Simply adhere to mathematics:
> > For ***all*** FISONs: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
> > If you don't trust this simple and clear theorem, then I cannot help you.
> > Similarly
> > ∀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 cannot put parrot always the same sentence which you obviously have not understood at all. 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.

Regards, 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

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

On 10/21/2021 1:09 PM, 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. That is a basic reality: There are not two consecutive infinite sets in |N.

agree, you are totally in the dark. That is a basic reality.

Your math is dark, no one can see it.

>
>> Numbers don't get "exhausted."
>
> ℕ \ ℕ = { }.
> ℕ \ {indexes of infinitely many endsegments} = { }, that is the empty set. All natnumbers are exhausted.

you don't even understand what = means.

"exhausted" is not used in Math.

"contents" is not used in math.

all natnumbers do not come from one bucket.

>
>> 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.
>
> Regards, WM
>

you are counting rocks, emptying buckets, exhausting sheeps type person

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

On Thursday, 21 October 2021 at 15:20:35 UTC-3, WM wrote:
> Gus Gassmann schrieb am Donnerstag, 21. Oktober 2021 um 01:58:14 UTC+2:
> > On Wednesday, 20 October 2021 at 18:04:03 UTC-3, Transfinity wrote:
> > [...]
> > > No, there are far less than aleph_0. Simply adhere to mathematics:
> > > For ***all*** FISONs: |ℕ \ {1, 2, 3, ..., n}| = ℵo ..
> > > If you don't trust this simple and clear theorem, then I cannot help you.
> > > Similarly
> > > ∀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 cannot put parrot always the same sentence which you obviously have not understood at all. 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.

Please do not simply parrot your old recycled garbage over and over. Since you have no clue what quantifier switching actually looks like, you should quit here. EOD.

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:

> > The infinitely many end-segments are a mathematical reality.

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

Liar. For every natural number n, there exists a unique end-segment E(n) = {n, n+1, n+2, ... }.

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

Irrelevant gibberish. ​

> > Numbers don't get "exhausted."

> ℕ \ ℕ = { }.
> ℕ \ {indexes of infinitely many endsegments} = { }, that is the empty set. All natnumbers are exhausted.

Liar.

> > 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.
>

Liar. Every natural number n is both an index of the end-segment E(n) = {n, n+1, n+2, ... } and element of each of E(1), E(2), ... , E(n). Contrary to your desperate lies, there is no contradiction here.

We are left with the fact that, contrary to your lies, there are infinitely many end-segments {n, n+1, n+2, ... } in N. That you believe otherwise shows just how broken your goofy little system is, Mucke. Deal with it.

Dan

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

On 10/21/2021 1:20 PM, WM wrote:
> Gus Gassmann schrieb am Donnerstag, 21. Oktober 2021 um 01:58:14 UTC+2:
>> On Wednesday, 20 October 2021 at 18:04:03 UTC-3, Transfinity wrote:
>> [...]
>>> No, there are far less than aleph_0. Simply adhere to mathematics:
>>> For ***all*** FISONs: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
>>> If you don't trust this simple and clear theorem, then I cannot help you.
>>> Similarly
>>> ∀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 cannot put parrot always the same sentence which you obviously have not understood at all. 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.
>
> Regards, WM
>

SQUACK! So much discussion of on a chopped set of natural numbers, it is nauseating to see such stupidity, and WM cannot get it right.

He is like Chumley in Tennessee Tuxedo playing with 3 blocks, endsegment, FISON, an mystery block, over and over and over again.

https://www.youtube.com/watch?v=F3aDMChYPss

On 10/21/2021 2:17 PM, WM wrote:
> Jim Burns schrieb
> am Donnerstag, 21. Oktober 2021 um 01:07:28 UTC+2:
>> On 10/20/2021 5:10 PM, Transfinity wrote:

>>> This successorship would not stop at natural indices
>>> but would run through all ordinal numbers.
>>
>> Only if omega is finite.

>> Since omega is NOT finite, running to omega would
>> require a crossing-pair j,j+1 such that j is finite
>> and j+1 is infinite.
>
> You see your crossing-pair argument is invalid.

If the successorship runs _through_ ordinal 𝛂
on the way to ordinal 𝛃, then 𝛂 ∈ {0,...,𝛃}\{𝛃}
for {0,...,𝛃} which is totally-ordered, in which,
for each split, there is a crossing-pair 𝛄,𝛄+1,
and which begins at 0 and ends at 𝛃.

( If you mean something else, please say what it is.

𝛚 is the first infinite ordinal.

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.

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.

----
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.

It is slightly more useful to point out that the
successorship runs to the things we count with.
That is a good enough reason for us to distinguish between
things the successorship runs to and things it
does not run to.

That bit I keep repeating about splits and crossing-pairs
and beginning and ending is a description of being
something the successorship runs to. A good description
is useful for reasoning from truth-preserving-ly.