zelos...@gmail.com schrieb am Donnerstag, 2. Dezember 2021 um 11:34:31 UTC+1:
> torsdag 2 december 2021 kl. 10:37:58 UTC+1 skrev WM:

> >For 7 and 3 we need no "or". For x and 3 we need it.
> "7 is greater than 3 or 7 is equal to 3 or 7 is less than three"
>
> is still a valid sentence even in logic :) So it still applies.

There are many valid sentences in logic. Irrelevant for the present case.

> >No, you know not even the difference between a constant, like 7, and a variable, like x.
> A constant is somethign that doesn't changes. A variable changes.

Therefore it is not fixed.

Regards, WM

WM explained :
> William schrieb am Donnerstag, 2. Dezember 2021 um 17:36:02 UTC+1:
>> On Thursday, December 2, 2021 at 4:30:47 AM UTC-5, WM wrote:>
>>
>>> Every element of ℕ_P has ℵ₀ successors. That means we can collect all
>>> elements of ℕ_P which have ℵ₀ successors by the axiom of separation. Now
>>> the question is: Does this collection have ℵ₀ successors too?
>> The answer is no. The axiom of separation does not imply that if every
>> *element* of a set has property P then the *set* has property P.
>
> Here we have only a finite collection because only elements with ℵ₀
> successors are collected.

*EVERY* element of N has a successor. With induction *EVERY* element of
N has infinitely many (aleph_zero) successive successors.

Stop lying.

fredag 3 december 2021 kl. 10:44:48 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 2. Dezember 2021 um 11:34:31 UTC+1:
> > torsdag 2 december 2021 kl. 10:37:58 UTC+1 skrev WM:
>
> > >For 7 and 3 we need no "or". For x and 3 we need it.
> > "7 is greater than 3 or 7 is equal to 3 or 7 is less than three"
> >
> > is still a valid sentence even in logic :) So it still applies.
> There are many valid sentences in logic. Irrelevant for the present case.
> > >No, you know not even the difference between a constant, like 7, and a variable, like x.
> > A constant is somethign that doesn't changes. A variable changes.
> Therefore it is not fixed.
>
> Regards, WM

but x does not need to be a variable you moron.

fredag 3 december 2021 kl. 10:41:34 UTC+1 skrev WM:
> Fritz Feldhase schrieb am Donnerstag, 2. Dezember 2021 um 11:29:09 UTC+1:
> > On Thursday, December 2, 2021 at 10:33:23 AM UTC+1, WM wrote:
> >
> > Concerning /trichotomy/.
> > No. For each and every x,y e IN: either x < y, or x = y, or x > y.
> For 7 we can fix 7 > 3. For x we cannot fix x > 3. That's an important difference.
> Hint: x is called a variable, meaning not being fixed.
>
> Regards, WM

x can be a constant, x can be a variable.

It depends on context you moron!

FromTheRafters schrieb am Freitag, 3. Dezember 2021 um 12:20:32 UTC+1:
> WM explained :
> > William schrieb am Donnerstag, 2. Dezember 2021 um 17:36:02 UTC+1:
> >> On Thursday, December 2, 2021 at 4:30:47 AM UTC-5, WM wrote:>
> >>
> >>> Every element of ℕ_P has ℵ₀ successors. That means we can collect all
> >>> elements of ℕ_P which have ℵ₀ successors by the axiom of separation. Now
> >>> the question is: Does this collection have ℵ₀ successors too?
> >> The answer is no. The axiom of separation does not imply that if every
> >> *element* of a set has property P then the *set* has property P.
> >
> > Here we have only a finite collection because only elements with ℵ₀
> > successors are collected.
> *EVERY* element of N has a successor. With induction *EVERY* element of
> N has infinitely many (aleph_zero) successive successors.

Then there are less elements. We are interested only in these less elements having ℵ₀ successors. Collecting them cannot yield an ℵ₀-set because there are not two consecutive ℵ₀-sets in ℕ possible. Claiming the contrary does make it possible.. It only shows a lack of logical thinking.

Collecting them cannot yield a set without successors. This would require magic.

Regards, WM

fredag 3 december 2021 kl. 12:42:38 UTC+1 skrev WM:
> FromTheRafters schrieb am Freitag, 3. Dezember 2021 um 12:20:32 UTC+1:
> > WM explained :
> > > William schrieb am Donnerstag, 2. Dezember 2021 um 17:36:02 UTC+1:
> > >> On Thursday, December 2, 2021 at 4:30:47 AM UTC-5, WM wrote:>
> > >>
> > >>> Every element of ℕ_P has ℵ₀ successors. That means we can collect all
> > >>> elements of ℕ_P which have ℵ₀ successors by the axiom of separation. Now
> > >>> the question is: Does this collection have ℵ₀ successors too?
> > >> The answer is no. The axiom of separation does not imply that if every
> > >> *element* of a set has property P then the *set* has property P.
> > >
> > > Here we have only a finite collection because only elements with ℵ₀
> > > successors are collected.
> > *EVERY* element of N has a successor. With induction *EVERY* element of
> > N has infinitely many (aleph_zero) successive successors.
> Then there are less elements. We are interested only in these less elements having ℵ₀ successors. Collecting them cannot yield an ℵ₀-set because there are not two consecutive ℵ₀-sets in ℕ possible. Claiming the contrary does make it possible. It only shows a lack of logical thinking.
>
> Collecting them cannot yield a set without successors. This would require magic.
>
> Regards, WM

No one is saying there are 2 consecutive aleph-0 because that is pure non-sense.

The fact that each E(n) is of the cardinality aleph-0 does and that N is aleph-0 does not in anyway imply there are anything consecutive. You are making that up dishonestly

On Friday, December 3, 2021 at 10:41:34 AM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Donnerstag, 2. Dezember 2021 um 11:29:09 UTC+1:
> > On Thursday, December 2, 2021 at 10:33:23 AM UTC+1, WM wrote:
> >
> > Concerning /trichotomy/.
> >
> > No. For each and every x,y e IN: either x < y, or x = y, or x > y.
> >
> For 7 <bla>

Learn the meaning of the notion "trichotomy", you silly crank:

https://en.wikipedia.org/wiki/Trichotomy_(mathematics)

On Friday, December 3, 2021 at 12:32:02 PM UTC+1, zelos...@gmail.com wrote:
> fredag 3 december 2021 kl. 10:41:34 UTC+1 skrev WM:
> >
> > Hint: x is called a variable, meaning not being fixed.
> >
> x can be a constant, x can be a variable.
>
> It depends on context you moron!

He's clearly too dumb to get that.

Though it might be helpful usimg " in the present context:

| "x" can be a constant, "x" can be a variable. It depends on context!

:-P

zelos...@gmail.com schrieb am Freitag, 3. Dezember 2021 um 12:50:24 UTC+1:
> fredag 3 december 2021 kl. 12:42:38 UTC+1 skrev WM:
> > FromTheRafters schrieb am Freitag, 3. Dezember 2021 um 12:20:32 UTC+1:
> > > WM explained :
> > > > William schrieb am Donnerstag, 2. Dezember 2021 um 17:36:02 UTC+1:
> > > >> On Thursday, December 2, 2021 at 4:30:47 AM UTC-5, WM wrote:>
> > > >>
> > > >>> Every element of ℕ_P has ℵ₀ successors. That means we can collect all
> > > >>> elements of ℕ_P which have ℵ₀ successors by the axiom of separation. Now
> > > >>> the question is: Does this collection have ℵ₀ successors too?
> > > >> The answer is no. The axiom of separation does not imply that if every
> > > >> *element* of a set has property P then the *set* has property P.
> > > >
> > > > Here we have only a finite collection because only elements with ℵ₀
> > > > successors are collected.
> > > *EVERY* element of N has a successor. With induction *EVERY* element of
> > > N has infinitely many (aleph_zero) successive successors.
> > Then there are less elements. We are interested only in these less elements having ℵ₀ successors. Collecting them cannot yield an ℵ₀-set because there are not two consecutive ℵ₀-sets in ℕ possible. Claiming the contrary does make it possible. It only shows a lack of logical thinking.
> >
> > Collecting them cannot yield a set without successors. This would require magic.

> No one is saying there are 2 consecutive aleph-0 because that is pure non-sense.

The contents of endsegments that is not any index n either is ℵ₀-infinite or not. If it is ℵ₀-infinite then there are not ℵ₀ endsegments. If the contents, that is not any index, is not ℵ₀-infinite, then not all endsegments are ℵ₀-infinite.
>
> The fact that each E(n) is of the cardinality aleph-0 does and that N is aleph-0 does not in anyway imply there are anything consecutive.

The contents of endsegments that is not any index n either is ℵ₀-infinite or not. What is it?

Regards, WM

On Friday, December 3, 2021 at 10:37:52 AM UTC+1, WM wrote:

> Here we have only a finite collection

No.

> because only elements with ℵ₀ successors are collected.

Non sequitur.

Hint: Let IN' = {n e IN : card({m e IN : m > n}) = ℵ₀}. Then card(IN') = ℵ₀.

fredag 3 december 2021 kl. 13:14:54 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 3. Dezember 2021 um 12:50:24 UTC+1:
> > fredag 3 december 2021 kl. 12:42:38 UTC+1 skrev WM:
> > > FromTheRafters schrieb am Freitag, 3. Dezember 2021 um 12:20:32 UTC+1:
> > > > WM explained :
> > > > > William schrieb am Donnerstag, 2. Dezember 2021 um 17:36:02 UTC+1:
> > > > >> On Thursday, December 2, 2021 at 4:30:47 AM UTC-5, WM wrote:>
> > > > >>
> > > > >>> Every element of ℕ_P has ℵ₀ successors. That means we can collect all
> > > > >>> elements of ℕ_P which have ℵ₀ successors by the axiom of separation. Now
> > > > >>> the question is: Does this collection have ℵ₀ successors too?
> > > > >> The answer is no. The axiom of separation does not imply that if every
> > > > >> *element* of a set has property P then the *set* has property P.
> > > > >
> > > > > Here we have only a finite collection because only elements with ℵ₀
> > > > > successors are collected.
> > > > *EVERY* element of N has a successor. With induction *EVERY* element of
> > > > N has infinitely many (aleph_zero) successive successors.
> > > Then there are less elements. We are interested only in these less elements having ℵ₀ successors. Collecting them cannot yield an ℵ₀-set because there are not two consecutive ℵ₀-sets in ℕ possible. Claiming the contrary does make it possible. It only shows a lack of logical thinking.
> > >
> > > Collecting them cannot yield a set without successors. This would require magic.
> > No one is saying there are 2 consecutive aleph-0 because that is pure non-sense.
> The contents of endsegments that is not any index n either is ℵ₀-infinite or not. If it is ℵ₀-infinite then there are not ℵ₀ endsegments. If the contents, that is not any index, is not ℵ₀-infinite, then not all endsegments are ℵ₀-infinite.
> >
> > The fact that each E(n) is of the cardinality aleph-0 does and that N is aleph-0 does not in anyway imply there are anything consecutive.
> The contents of endsegments that is not any index n either is ℵ₀-infinite or not. What is it?
>
> Regards, WM

>The contents of endsegments that is not any index n either is ℵ₀-infinite or not.

all endsegments have the same cardinality as N

>If it is ℵ₀-infinite then there are not ℵ₀ endsegments. If the contents, that is not any index, is not ℵ₀-infinite, then not all endsegments are ℵ₀-infinite.

Here we have your shitty wording making this non-sense. Speak mathematics.

>The contents of endsegments that is not any index n either is ℵ₀-infinite or not. What is it?

|E(n)|=|N|=ℵ₀

Get it=?

On Friday, 3 December 2021 at 08:14:54 UTC-4, WM wrote:
> zelos...@gmail.com schrieb am Freitag, 3. Dezember 2021 um 12:50:24 UTC+1:
[...]
> > No one is saying there are 2 consecutive aleph-0 because that is pure non-sense.
> The contents of endsegments that is not any index n either is ℵ₀-infinite or not.

Note that you are using the plural here. There is not one end segment, but ℵ₀ of them: E(n) for every natural number n. It should have further come to your attention that E(n) =/= E(n+1) or in general any E(m) with m =/= n.

> If it is ℵ₀-infinite then there are not ℵ₀ endsegments.

For that reason "it" is nonsense, and the rest is gibberish that I removed.

[...]

Fritz Feldhase schrieb am Freitag, 3. Dezember 2021 um 13:19:35 UTC+1:
> On Friday, December 3, 2021 at 10:37:52 AM UTC+1, WM wrote:
>
> > Here we have only a finite collection
> No.

Where does the infinite set of successors start?

> > because only elements with ℵ₀ successors are collected.
> Non sequitur.

Simply answer the question: If there are all ℵ₀ natnumbers used, what is the first number of the infinite set of ℵ₀ successors?
>
> Hint: Let IN' = {n e IN : card({m e IN : m > n}) = ℵ₀}. Then card(IN') = ℵ₀.

Obviously matheologians have never thought about the nonsense they proclaim..

Regards, WM

On Friday, December 3, 2021 at 12:42:38 PM UTC+1, WM wrote:
> FromTheRafters schrieb am Freitag, 3. Dezember 2021 um 12:20:32 UTC+1:
> > WM explained :
> > >
> > > Here we have only a finite collection because only elements with ℵ₀
> > > successors are collected.
> > >
> > *EVERY* element of N has a successor. With induction *EVERY* element of
> > N has infinitely many (aleph_zero) successive successors.
> >
> Then there are less elements.

Holy shit!

"less elements"??? Huh?!

Anyway: card(N) = aleph_zero BY DEFINITION of aleph_zero, dumbo.

> We are interested only in [...] elements having ℵ₀ successors.

You just have been told (you silly crank):

"*EVERY* element of N has aleph_zero successors."

> Collecting them cannot yield an ℵ₀-set

Nonsense.

Hint: Let IN' = {n e IN : card({m e IN : m > n}) = ℵ₀}. Then card(IN') = ℵ₀. (Since IN' = IN.)

> because there are not two consecutive ℵ₀-sets in ℕ possible.

Non sequitor. The fact that "there are not two consecutive ℵ₀-sets in ℕ" is in no way related to the fact that card({n e IN : card({m e IN : m > n}) = ℵ₀}) = ℵ₀. (The latter is implied by the fact that {n e IN : card({m e IN : m > n}) = ℵ₀} = IN.)

> Claiming the contrary [...] shows a lack of logical thinking.

Indeed! :-)

> Collecting them cannot yield a set without successors.

Please define the notion "succesor of set X". Then prove your claim. (it might be correct, after all.)

On Friday, December 3, 2021 at 1:24:51 PM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Freitag, 3. Dezember 2021 um 13:19:35 UTC+1:
> > On Friday, December 3, 2021 at 10:37:52 AM UTC+1, WM wrote:
> > >
> > > Here we have only a finite collection
> > >
> > No.
> >
> Where does the infinite set of successors start?

Relevance? None. Btw. Please don't ask silly and/or nonsensical questions.

> > > because only elements with ℵ₀ successors are collected.
> > >
> > Non sequitur.
> >
> Simply answer the question: <bla>

Please don't ask silly and/or nonsensical questions.

> > Hint: Let IN' = {n e IN : card({m e IN : m > n}) = ℵ₀}. Then card(IN') = ℵ₀.

Since IN' = IN.

> Obviously bla>

Just too dense to comprehend the trivial proof? (See my other post.)

Hint: {n e IN : card({m e IN : m > n}) = ℵ₀} = IN since for all n e IN: card({m e IN : m > n}) = ℵ₀.

WM was thinking very hard :
> FromTheRafters schrieb am Freitag, 3. Dezember 2021 um 12:20:32 UTC+1:
>> WM explained :
>>> William schrieb am Donnerstag, 2. Dezember 2021 um 17:36:02 UTC+1:
>>>> On Thursday, December 2, 2021 at 4:30:47 AM UTC-5, WM wrote:>
>>>>
>>>>> Every element of ℕ_P has ℵ₀ successors. That means we can collect all
>>>>> elements of ℕ_P which have ℵ₀ successors by the axiom of separation. Now
>>>>> the question is: Does this collection have ℵ₀ successors too?
>>>> The answer is no. The axiom of separation does not imply that if every
>>>> *element* of a set has property P then the *set* has property P.
>>>
>>> Here we have only a finite collection because only elements with ℵ₀
>>> successors are collected.
>> *EVERY* element of N has a successor. With induction *EVERY* element of
>> N has infinitely many (aleph_zero) successive successors.
>
> Then there are less elements.

What does that even mean?

On Friday, December 3, 2021 at 2:14:35 PM UTC+1, FromTheRafters wrote:
> WM was thinking very hard :
> > >
> > > *EVERY* element of N has a successor. With induction *EVERY* element of
> > > N has infinitely many (aleph_zero) successive successors.
> > >
> > Then there are less elements.
> >
> What does that even mean?

Seems he's losing it.

WM has brought this to us :
> Fritz Feldhase schrieb am Freitag, 3. Dezember 2021 um 13:19:35 UTC+1:
>> On Friday, December 3, 2021 at 10:37:52 AM UTC+1, WM wrote:
>>
>>> Here we have only a finite collection
>> No.
>
> Where does the infinite set of successors start?

At bthe beginning.

>>> because only elements with ℵ₀ successors are collected.
>> Non sequitur.
>
> Simply answer the question: If there are all ℵ₀ natnumbers used,

By 'used' do you mean exhausted -- like they are no longer available?

> what is the first number of the infinite set of ℵ₀ successors?

It is n+1 for each n.

>> Hint: Let IN' = {n e IN : card({m e IN : m > n}) = ℵ₀}. Then card(IN') = ℵ₀.
>
> Obviously matheologians have never thought about the nonsense they proclaim.

Obviously, but don't be too hard on yourself, it can't be easy at your
advanced age.

On Friday, December 3, 2021 at 4:37:52 AM UTC-5, WM wrote:
> William schrieb am Donnerstag, 2. Dezember 2021 um 17:36:02 UTC+1:
> > On Thursday, December 2, 2021 at 4:30:47 AM UTC-5, WM wrote:>
> >
> > > Every element of ℕ_P has ℵ₀ successors. That means we can collect all elements of ℕ_P which have ℵ₀ successors by the axiom of separation.
> > > Now the question is: Does this collection have ℵ₀ successors too?
> > The answer is no. The axiom of separation does not imply that if every *element* of a set has property P then the *set* has property P.
> Here we have only a finite collection because only elements with ℵ₀ successors are collected.

Nope. N_P has no largest element so every element of N_P has ℵ₀ successors ( a different set each time). The collection is infinite.

--
William Hughes

On Friday, December 3, 2021 at 6:42:38 AM UTC-5, WM wrote:
> Collecting them cannot yield an ℵ₀-set because there are not two consecutive ℵ₀-sets in ℕ possible

Nope. The *elements* of the set have ℵ₀- successors. The *set* does not have ℵ₀- successors.

--
William Hughes

On 12/3/2021 5:42 AM, WM wrote:
> FromTheRafters schrieb am Freitag, 3. Dezember 2021 um 12:20:32 UTC+1:
>> WM explained :
>>> William schrieb am Donnerstag, 2. Dezember 2021 um 17:36:02 UTC+1:
>>>> On Thursday, December 2, 2021 at 4:30:47 AM UTC-5, WM wrote:>
>>>>
>>>>> Every element of ℕ_P has ℵ₀ successors. That means we can collect all
>>>>> elements of ℕ_P which have ℵ₀ successors by the axiom of separation. Now
>>>>> the question is: Does this collection have ℵ₀ successors too?
>>>> The answer is no. The axiom of separation does not imply that if every
>>>> *element* of a set has property P then the *set* has property P.
>>>
>>> Here we have only a finite collection because only elements with ℵ₀
>>> successors are collected.
>> *EVERY* element of N has a successor. With induction *EVERY* element of
>> N has infinitely many (aleph_zero) successive successors.
>
> Then there are less elements.

wrong.

> We are interested only in these less elements having ℵ₀ successors.

you have wandered off the path and into the woods again, there are no "less elements".

> Collecting them cannot yield an ℵ₀-set because there are not two consecutive ℵ₀-sets in ℕ possible. Claiming the contrary does make it possible.

wrong again, on 3 counts.

>It only shows a lack of logical thinking.

your thinking is not logical, nor mathematics

>
> Collecting them cannot yield a set without successors. This would require magic.

meaningless

>
> Regards, WM
>

On 12/3/2021 6:24 AM, WM wrote:
> Fritz Feldhase schrieb am Freitag, 3. Dezember 2021 um 13:19:35 UTC+1:
>> On Friday, December 3, 2021 at 10:37:52 AM UTC+1, WM wrote:
>>
>>> Here we have only a finite collection
>> No.
>
> Where does the infinite set of successors start?

at the beginning

>
>>> because only elements with ℵ₀ successors are collected.
>> Non sequitur.
>
> Simply answer the question: If there are all ℵ₀ natnumbers used, what is the first number of the infinite set of ℵ₀ successors?

what number did you STOP AT ? add 1, and there you go.

>>
>> Hint: Let IN' = {n e IN : card({m e IN : m > n}) = ℵ₀}. Then card(IN') = ℵ₀.
>
> Obviously matheologians have never thought about the nonsense they proclaim.

then you are a matheologian.

>
> Regards, WM
>

William schrieb am Freitag, 3. Dezember 2021 um 14:36:09 UTC+1:
> On Friday, December 3, 2021 at 6:42:38 AM UTC-5, WM wrote:
> > Collecting them cannot yield an ℵ₀-set because there are not two consecutive ℵ₀-sets in ℕ possible
> Nope. The *elements* of the set have ℵ₀- successors. The *set* does not have ℵ₀- successors.

The elements are the set!

Collecting natural numbers yields this set. Collecting only definable natural numbers yields finite sets with ℵo successors before ω. You will agree that this cannot be changed.

Collecting all natnumbers however yields a set with no successors before ω.

The (more or less sudden) change cannot be observed.
Anyhow, this shows that something here cannot be obeserved.
Do you agree that something here cannot be observed?

Regards, WM

On 12/3/2021 8:08 AM, WM wrote:
> William schrieb am Freitag, 3. Dezember 2021 um 14:36:09 UTC+1:
>> On Friday, December 3, 2021 at 6:42:38 AM UTC-5, WM wrote:
>>> Collecting them cannot yield an ℵ₀-set because there are not two consecutive ℵ₀-sets in ℕ possible
>> Nope. The *elements* of the set have ℵ₀- successors. The *set* does not have ℵ₀- successors.
>
> The elements are the set!

wrong. elements are inside a set, they are members of a set, they are not the set.

>
> Collecting natural numbers yields this set.

"Collecting" is not used in Math. Perhaps you are indirectly referring to ℕ ?

> Collecting only definable natural numbers yields finite sets with ℵo successors before ω.

"Collecting" and your "definable" is not used in Math. So your results are moot.

> You will agree that this cannot be changed.

no. your sentence makes no sense in Math

>

FromTheRafters schrieb am Freitag, 3. Dezember 2021 um 14:26:01 UTC+1:
> WM has brought this to us :

> > Simply answer the question: If there are all ℵ₀ natnumbers used,
> By 'used' do you mean exhausted -- like they are no longer available?

Every index of an endsegment is no longer available as contents of further endsegments.

> > what is the first number of the infinite set of ℵ₀ successors?
> It is n+1 for each n.

The set of contents, if existing, is a fixed quantity. All n will get exhausted by indices. But a fixed quantity is claimed to remain in all endsegments - if all endsegments are infinite.

Regards, WM