FromTheRafters schrieb am Freitag, 8. April 2022 um 08:33:25 UTC+2:
> It happens that WM formulated :

> > There are not ℵo FISONs without an infinite FISON.
> There you go with your infinite FISONs again. Did you forget the
> definition of FISON?

No. But I know that ℵo-infinitely many FISON cannot exist without at least one ℵo-infinite FISON.

Regards, WM

zelos...@gmail.com schrieb am Freitag, 8. April 2022 um 07:08:13 UTC+2:
> torsdag 7 april 2022 kl. 19:12:02 UTC+2 skrev WM:

> > > All FINITE unions, I say union of ALL INFINITELY MANY FISONs is N
> >
> > There are not ℵo FISONs without an infinite FISON.
> False, that is a pure non-sequitor.

It is higher mathematics, too high for you.

Start wit finite sets of FISONs: Every set of n FISONs has at least one FISON with n symbols. Now extend this to a number larger than all n. How can we distinguish more FISONs than distinct by all n? Impossible.

Regards, WM

FromTheRafters schrieb am Donnerstag, 7. April 2022 um 21:41:04 UTC+2:
> WM wrote :

> > Better you can see dark numbers with the complementary endsegments E(n) = {n,
> > n+1, n+2, ...}. The sequence of endsegments is inclusion-monotonic.
> Each successive endsegmment in your sequence excludes the element "n-1"
> which the previous endsegment had. How does this make the sequence
> *inclusion* monotonic?

F(n+1) is superset of F(n+2). Every FISON is the subset of its predecessors. Every FISON is infinite. That means infinitely many natnumbers are in every intersected FISON. Hence the intersection cannot be empty.

Regards, WM

Fritz Feldhase schrieb am Donnerstag, 7. April 2022 um 21:39:04 UTC+2:
> On Thursday, April 7, 2022 at 7:22:54 PM UTC+2, WM wrote:
> >
> > But since the sequence contains all unions of FISONs
> Nope. It does not contain the union of *all* FISONs.

Name any missing FISON (but don't use the matheologial miracle "all").

> > If the union of FISONs is ℕ, then the intersection of endsegments is empty.
> Well, the union of all FISONs is ℕ and the intersection of all endsegments is empty.

Note that we use mathematics here: Every FISON is the subset of its predecessors. Every FISON is infinite. That means infinitely many natnumbers are in every intersected FISON. Hence the intersection cannot be empty.

Spare you matheologial delusions. They are only for cranks and from cranks.

Regards, WM

Fritz Feldhase schrieb am Donnerstag, 7. April 2022 um 21:33:16 UTC+2:
> On Thursday, April 7, 2022 at 6:41:33 PM UTC+2, WM wrote:
> >
> > All unions of FISONs are unions of finitely many FISONs
> he union of *all* FISONs is an union of infinitely many FISONs, because there are infinitely many FISONs.
>
Start wit finite sets of FISONs: Every set of n FISONs has at least one FISON with n symbols. Now extend this to a number larger than all n. How can we distinguish more FISONs than distinct by all n? Impossible.

Regards, WM

William schrieb am Donnerstag, 7. April 2022 um 20:19:46 UTC+2:
> On Thursday, April 7, 2022 at 1:37:51 PM UTC-3, WM wrote:
> > William schrieb am Mittwoch, 6. April 2022 um 21:11:56 UTC+2:
> > > On Wednesday, April 6, 2022 at 3:37:18 PM UTC-3, WM wrote:
> > >
> > > > Show at least one FISON that is in the union of all FISONs but not in any finite union of the FISONs in the sequence.
> > > Piffle. As I have pointed out many times it is not a singleton that is not in a finite union, but an infininte subset.
> > If there is not even any FISON not in a finite union, then you are wrong with an infinite subset.
> Piffle. It is trivial to note the every FISON is a finite union. It is trivial to see that an infinite set of FISONs is not in a finite union.

Of course. That is because there is none.

Start wit finite sets of FISONs: Every set of n FISONs has at least one FISON with n symbols. Now extend this to a number larger than all n. How can we distinguish more FISONs than distinct by all n? Impossible.

Regards, WM

William schrieb am Donnerstag, 7. April 2022 um 20:24:41 UTC+2:
> On Thursday, April 7, 2022 at 2:23:08 PM UTC-3, WM wrote:
>
> > If the union of FISONs is ℕ, then the intersection of endsegments is empty.
> Correct, Indeed both statements are trivially true
> (The sequence of endsegments has no last element).

That is irrelevant. All definable endsegments are infinite, that means they contain an infinite set of natnumbers together. Therefore their intersection is infinite.

Your claim is as wrong as the claim the the sequence a_n = 1 + 1/n has limit 0.

Regards, WM

On 4/9/2022 7:13 AM, WM wrote:
> FromTheRafters schrieb am Donnerstag, 7. April 2022 um 21:41:04 UTC+2:
>> WM wrote :
>
>>> Better you can see dark numbers with the complementary endsegments E(n) = {n,
>>> n+1, n+2, ...}. The sequence of endsegments is inclusion-monotonic.
>> Each successive endsegmment in your sequence excludes the element "n-1"
>> which the previous endsegment had. How does this make the sequence
>> *inclusion* monotonic?
>
> F(n+1) is superset of F(n+2). Every FISON is the subset of its predecessors. Every FISON is infinite. That means infinitely many natnumbers are in every intersected FISON. Hence the intersection cannot be empty.
>
> Regards, WM

*This is not the newsgroup to lie in*.

1. "F(n+1) is superset of F(n+2)." Wrong, subset.

2. "Every FISON is the subset of its predecessors". Wrong

3. "Every FISON is infinite". Wrong

4. "That means infinitely many natnumbers are in every intersected FISON." Wrong.

5. "Hence the intersection cannot be empty." Wrong

0 of 5 correct, 0% which is F, for Fail.

FromTheRafters schrieb am Freitag, 8. April 2022 um 08:33:25 UTC+2:
> It happens that WM formulated :
> > zelos...@gmail.com schrieb am Donnerstag, 7. April 2022 um 06:46:30 UTC+2:
> >> onsdag 6 april 2022 kl. 15:10:44 UTC+2 skrev WM:
> >
> >>> How should ℵo result if all unios of existing FISONs are in the sequence of
> >>> FISONs and are therefore finite?
> >>
> >> All FINITE unions, I say union of ALL INFINITELY MANY FISONs is N
> >
> > There are not ℵo FISONs without an infinite FISON.
> There you go with your infinite FISONs again. Did you forget the
> definition of FISON?

Start with finite sets of FISONs: Every set of n FISONs has at least one FISON with n symbols. Now extend this to a number larger than all n. How can we distinguish more FISONs than distinct by all n? Impossible.

Regards, WM

FromTheRafters schrieb am Donnerstag, 7. April 2022 um 21:41:04 UTC+2:
> WM wrote :
> > William schrieb am Mittwoch, 6. April 2022 um 21:11:56 UTC+2:
> >> On Wednesday, April 6, 2022 at 3:37:18 PM UTC-3, WM wrote:
> >>
> >>> Show at least one FISON that is in the union of all FISONs but not in any
> >>> finite union of the FISONs in the sequence.
> >> Piffle. As I have pointed out many times it is not a singleton that is not
> >> in a finite union, but an infininte subset.
> >
> > If there is not even any FISON not in a finite union, then you are wrong with
> > an infinite subset.
> >
> > Better you can see dark numbers with the complementary endsegments E(n) = {n,
> > n+1, n+2, ...}. The sequence of endsegments is inclusion-monotonic.
> Each successive endsegmment in your sequence excludes the element "n-1"
> which the previous endsegment had. How does this make the sequence
> *inclusion* monotonic?

E(n+1) is superset of E(n+2). Every endsegment is the subset of its predecessors. Every endsegment is infinite. That means infinitely many natnumbers are in every intersected endsegment. Hence the intersection cannot be empty.

Regards, WM

On 4/9/2022 9:04 AM, WM wrote:
> William schrieb am Donnerstag, 7. April 2022 um 20:24:41 UTC+2:
>> On Thursday, April 7, 2022 at 2:23:08 PM UTC-3, WM wrote:
>>
>>> If the union of FISONs is ℕ, then the intersection of endsegments is empty.
>> Correct, Indeed both statements are trivially true
>> (The sequence of endsegments has no last element).
>
> That is irrelevant.

No. It is the point.

> All endsegments are infinite,

trivial.

> that means they contain an infinite set of natnumbers together.

No. what do you mean by "together" ?

> Therefore their intersection is infinite.

Define "together" mathematically, first.

>
> Your claim is as wrong as the claim the the sequence a_n = 1 + 1/n has limit 0.

Wrong
Lim as n => oo of a_n = 1 + 1/n, has limit 1

>
> Regards, WM

On 4/9/2022 7:06 AM, WM wrote:
> FromTheRafters schrieb am Freitag, 8. April 2022 um 08:33:25 UTC+2:
>> It happens that WM formulated :
>
>>> There are not ℵo FISONs without an infinite FISON.
>> There you go with your infinite FISONs again. Did you forget the
>> definition of FISON?
>
> No. But I know that ℵo-infinitely many FISON cannot exist without at least one ℵo-infinite FISON.

You are in error.

>
> Regards, WM

On 4/9/2022 9:27 AM, WM wrote:
> FromTheRafters schrieb am Freitag, 8. April 2022 um 08:33:25 UTC+2:
>> It happens that WM formulated :
>>> zelos...@gmail.com schrieb am Donnerstag, 7. April 2022 um 06:46:30 UTC+2:
>>>> onsdag 6 april 2022 kl. 15:10:44 UTC+2 skrev WM:
>>>
>>>>> How should ℵo result if all unios of existing FISONs are in the sequence of
>>>>> FISONs and are therefore finite?
>>>>
>>>> All FINITE unions, I say union of ALL INFINITELY MANY FISONs is N
>>>
>>> There are not ℵo FISONs without an infinite FISON.
>> There you go with your infinite FISONs again. Did you forget the
>> definition of FISON?
>
> Start with finite sets of FISONs: Every set of n FISONs has at least one FISON with n symbols.

>
>Now extend this to a number larger than all n.

OK, chose p > n

> How can we distinguish more FISONs than distinct by all n?

F(p)...

Do you know your use of "distinguished" and "distinct" makes your sentences meaningless?

> Impossible.
>
> Regards, WM

On Saturday, April 9, 2022 at 9:23:13 AM UTC-3, WM wrote:
> William schrieb am Donnerstag, 7. April 2022 um 20:19:46 UTC+2:
> > On Thursday, April 7, 2022 at 1:37:51 PM UTC-3, WM wrote:
> > > William schrieb am Mittwoch, 6. April 2022 um 21:11:56 UTC+2:
> > > > On Wednesday, April 6, 2022 at 3:37:18 PM UTC-3, WM wrote:
> > > >
> > > > > Show at least one FISON that is in the union of all FISONs but not in any finite union of the FISONs in the sequence.
> > > > Piffle. As I have pointed out many times it is not a singleton that is not in a finite union, but an infininte subset.
> > > If there is not even any FISON not in a finite union, then you are wrong with an infinite subset.
> > Piffle. It is trivial to note the every FISON is a finite union. It is trivial to see that an infinite set of FISONs is not in a finite union.
> Of course. That is because there is none.
> Start wit finite sets of FISONs: Every set of n FISONs has at least one FISON with n symbol
Piffle. This is only true for n finite.

--
William Hughes

On Saturday, April 9, 2022 at 11:04:26 AM UTC-3, WM wrote:
> William schrieb am Donnerstag, 7. April 2022 um 20:24:41 UTC+2:
> > On Thursday, April 7, 2022 at 2:23:08 PM UTC-3, WM wrote:
> >
> > > If the union of FISONs is ℕ, then the intersection of endsegments is empty.
> > Correct, Indeed both statements are trivially true
> > (The sequence of endsegments has no last element).
> That is irrelevant.
Piffle. The intersetion is empty iff the sequence has no last element.
--
William Hughes

On Saturday, April 9, 2022 at 2:06:07 PM UTC+2, WM wrote:
>
> But I know that

You know shit, you silly crank.

On Saturday, April 9, 2022 at 2:20:25 PM UTC+2, WM wrote:

> don't use the matheologial miracle "all".

Oh, in your psychotic kingdom universal quantification is a "matheologial miracle". Fascinating, but irrelavant in this context.

> matheologial delusions. They are only [from a] crank[.]

Yeah, we know that.

WM submitted this idea :
> FromTheRafters schrieb am Donnerstag, 7. April 2022 um 21:41:04 UTC+2:
>> WM wrote :
>
>>> Better you can see dark numbers with the complementary endsegments E(n) =
>>> {n, n+1, n+2, ...}. The sequence of endsegments is inclusion-monotonic.
>> Each successive endsegmment in your sequence excludes the element "n-1"
>> which the previous endsegment had. How does this make the sequence
>> *inclusion* monotonic?
>
> F(n+1) is superset of F(n+2).

What?

> Every FISON is the subset of its predecessors.

What??

> Every FISON is infinite.

What??? Infinite FISONs again????

> That means infinitely many natnumbers are in every
> intersected FISON. Hence the intersection cannot be empty.

What?????

That's not what I asked anyway. We were talking about endsegments and
you claimed that *they* are 'inclusion monotonic'. Are you counting a
sequence of endsegments backward from infinity?

sergio used his keyboard to write :
> On 4/9/2022 7:13 AM, WM wrote:
>> FromTheRafters schrieb am Donnerstag, 7. April 2022 um 21:41:04 UTC+2:
>>> WM wrote :
>>
>>>> Better you can see dark numbers with the complementary endsegments E(n) =
>>>> {n,
>>>> n+1, n+2, ...}. The sequence of endsegments is inclusion-monotonic.
>>> Each successive endsegmment in your sequence excludes the element "n-1"
>>> which the previous endsegment had. How does this make the sequence
>>> *inclusion* monotonic?
>>
>> F(n+1) is superset of F(n+2). Every FISON is the subset of its
>> predecessors. Every FISON is infinite. That means infinitely many
>> natnumbers are in every intersected FISON. Hence the intersection cannot be
>> empty.
>>
>> Regards, WM
>
>
> *This is not the newsgroup to lie in*.

Maybe he is developing a Lie Algebra.

> 1. "F(n+1) is superset of F(n+2)." Wrong, subset.
>
>
> 2. "Every FISON is the subset of its predecessors". Wrong
>
>
> 3. "Every FISON is infinite". Wrong
>
>
> 4. "That means infinitely many natnumbers are in every intersected FISON."
> Wrong.
>
>
> 5. "Hence the intersection cannot be empty." Wrong
>
>
>
>
> 0 of 5 correct, 0% which is F, for Fail.

Should get an extra 'wrong' for not talking about the topic of
inclusion monotonous endsegments which he is replying to.

WM formulated the question :
> FromTheRafters schrieb am Freitag, 8. April 2022 um 08:33:25 UTC+2:
>> It happens that WM formulated :
>>> zelos...@gmail.com schrieb am Donnerstag, 7. April 2022 um 06:46:30 UTC+2:
>>>> onsdag 6 april 2022 kl. 15:10:44 UTC+2 skrev WM:
>>>>> How should ℵo result if all unios of existing FISONs are in the sequence
>>>>> of FISONs and are therefore finite?
>>>>
>>>> All FINITE unions, I say union of ALL INFINITELY MANY FISONs is N
>>>
>>> There are not ℵo FISONs without an infinite FISON.
>> There you go with your infinite FISONs again. Did you forget the
>> definition of FISON?
>
> Start with finite sets of FISONs: Every set of n FISONs has at least one
> FISON with n symbols. Now extend this to a number larger than all n.

There is no natural number larger than all n.

On Saturday, 9 April 2022 at 09:06:07 UTC-3, WM wrote:
> FromTheRafters schrieb am Freitag, 8. April 2022 um 08:33:25 UTC+2:
> > It happens that WM formulated :
> > > There are not ℵo FISONs without an infinite FISON.
> > There you go with your infinite FISONs again. Did you forget the
> > definition of FISON?
> No. But I know that ℵo-infinitely many FISON cannot exist without at least one ℵo-infinite FISON.

It's not what we don't know that dooms us, but rather what we think we know that just isn't so.

On Saturday, 9 April 2022 at 09:20:25 UTC-3, WM wrote:
> Fritz Feldhase schrieb am Donnerstag, 7. April 2022 um 21:39:04 UTC+2:
> > On Thursday, April 7, 2022 at 7:22:54 PM UTC+2, WM wrote:
> Spare you matheologial delusions. They are only for cranks and from cranks.

Right back atcha, moron.

lördag 9 april 2022 kl. 14:09:14 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 8. April 2022 um 07:08:13 UTC+2:
> > torsdag 7 april 2022 kl. 19:12:02 UTC+2 skrev WM:
>
> > > > All FINITE unions, I say union of ALL INFINITELY MANY FISONs is N
> > >
> > > There are not ℵo FISONs without an infinite FISON.
> > False, that is a pure non-sequitor.
> It is higher mathematics, too high for you.

comes from the idiot that do not understand basic mathematics.

>
> Start wit finite sets of FISONs: Every set of n FISONs has at least one FISON with n symbols. Now extend this to a number larger than all n. How can we distinguish more FISONs than distinct by all n? Impossible.

And here we have you trying to extrapolate from the finite to the infinite.

This is an invalid approach.

>
> Regards, WM

lördag 9 april 2022 kl. 16:04:26 UTC+2 skrev WM:
> William schrieb am Donnerstag, 7. April 2022 um 20:24:41 UTC+2:
> > On Thursday, April 7, 2022 at 2:23:08 PM UTC-3, WM wrote:
> >
> > > If the union of FISONs is ℕ, then the intersection of endsegments is empty.
> > Correct, Indeed both statements are trivially true
> > (The sequence of endsegments has no last element).
> That is irrelevant. All definable endsegments are infinite, that means they contain an infinite set of natnumbers together. Therefore their intersection is infinite.
>
> Your claim is as wrong as the claim the the sequence a_n = 1 + 1/n has limit 0.
>
> Regards, WM
false, as usual. The intersection is infinite only if the largest commob subset in all is infinite.

There is no subset that is common to all endsegments except for the empty set.

William schrieb am Samstag, 9. April 2022 um 17:10:19 UTC+2:
> On Saturday, April 9, 2022 at 9:23:13 AM UTC-3, WM wrote:
> > > It is trivial to see that an infinite set of FISONs is not in a finite union.
> > Of course. That is because there is none.
> > Start with finite sets of FISONs: Every set of n FISONs has at least one FISON with n symbols.
> This is only true for n finite.

Of course it is true for finite n, even for *all* finite n. Therefore no number of n symbols is sufficient to distuínguish an ℵo-infinite set of FISONs. Note that ℵo is larger than all finite n.

Regards, WM