William schrieb am Samstag, 9. April 2022 um 17:13:10 UTC+2:
> 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.
> The intersetion is empty iff the sequence has no last element.

Irrelevant. The intersection is empty, if no element is in all sets. But here we assume that ℵo elements are in all sets because all are infinite and inclusion monotonic. That is the mathematical argument, stronger than matheologial belief.

Regards, WM

Fritz Feldhase schrieb am Samstag, 9. April 2022 um 17:24:53 UTC+2:
> On Saturday, April 9, 2022 at 2:20:25 PM UTC+2, WM wrote:
>
> > don't use the matheologial miracle "all".
>
> Oh,

Show a missing FISON. Fail.

Regards, WM

horand....@gmail.com schrieb am Samstag, 9. April 2022 um 21:40:58 UTC+2:
> 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.

For *all* finite n: n FISONs require at least n symbols to distimguish them.. 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

zelos...@gmail.com schrieb am Sonntag, 10. April 2022 um 14:20:54 UTC+2:
> lördag 9 april 2022 kl. 14:09:14 UTC+2 skrev WM:

> > 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.

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

This kind is valid. If not, then the infinite has no mathematical contents.

Can infinitely many FISONs be distinguished by any set of n symbols?

Regards, WM

zelos...@gmail.com schrieb am Sonntag, 10. April 2022 um 14:22:07 UTC+2:
> 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.
> >
> 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.

There is an infinite subset common to all infinite endsegments, by inclusion monotony.

Regards, WM

On Sunday, April 10, 2022 at 7:47:12 PM UTC+2, WM wrote:
> >
> > The intersection is empty, if no element is in all sets.

Right.

But let's be precise: The intersection is empty, if there is no element such that it is in all sets (which are intersected).

> But here we assume that ℵo elements are in all sets because

You can't just /assume/ that, idiot, you have to _show/prove_ it.

Hint: There is no element such that it is in all sets (which are intersected).

> all are infinite and inclusion monotonic.

They are, but these facts do not imply that there are elements that are in all (intersected) sets.

> That is [...]

Yeah, whatever.

On 4/10/2022 12:44 PM, WM wrote:
> 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

Wrong, there are no "ℵo-infinite set of FISONs"

söndag 10 april 2022 kl. 19:47:12 UTC+2 skrev WM:
> William schrieb am Samstag, 9. April 2022 um 17:13:10 UTC+2:
> > 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.
> > The intersetion is empty iff the sequence has no last element.
> Irrelevant. The intersection is empty, if no element is in all sets. But here we assume that ℵo elements are in all sets because all are infinite and inclusion monotonic. That is the mathematical argument, stronger than matheologial belief.
>
> Regards, WM
and here we have you demonstrating you do not know mathematics.

The intersection is empty **IF AND ONLY IF all elements are not present in at least one of the sets**

Which for endsegments is trivial

n is not present in E(n+1), ergo n is not in the intersection, but as this is arbitrary, NO element is in the intersection

On 4/10/2022 12:58 PM, WM wrote:
> zelos...@gmail.com schrieb am Sonntag, 10. April 2022 um 14:22:07 UTC+2:
>> 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.
>>>
>> 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.
>
> There is an infinite subset common to all infinite endsegments, by inclusion monotony.
>
> Regards, WM

Wrong.

Pick a number k, it is not in endsegment E(k+1)

{WM-You have failed yet again, but if you learn how to actually do math, you can reverse your downward trend.}

söndag 10 april 2022 kl. 19:55:24 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Sonntag, 10. April 2022 um 14:20:54 UTC+2:
> > lördag 9 april 2022 kl. 14:09:14 UTC+2 skrev WM:
> > > 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.
> > And here we have you trying to extrapolate from the finite to the infinite.
> This kind is valid.

It isn't valid because it is trivial to show such reasoning is faulty.

>If not, then the infinite has no mathematical contents.

Non-sequitor again, infinite has mathematical meaning and definitions depending on contexts.

>
> Can infinitely many FISONs be distinguished by any set of n symbols?

what does "distinguished" even mean here?

>
> Regards, WM

söndag 10 april 2022 kl. 19:58:05 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Sonntag, 10. April 2022 um 14:22:07 UTC+2:
> > 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.
> > >
> > 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.
> There is an infinite subset common to all infinite endsegments, by inclusion monotony.
>
> Regards, WM
false.

Easily proven

assume it exists. call it S, as in "shit". and is Non-empty.

By natural numbers, we have a least element in S, call it n.

n is not in E(n+1)

so S is not a subset of E(n+1), therefore non-empty S does not exist as it leads to a contradiction.

Fritz Feldhase schrieb am Sonntag, 10. April 2022 um 20:05:37 UTC+2:
> On Sunday, April 10, 2022 at 7:47:12 PM UTC+2, WM wrote:
> > >
> The intersection is empty, if there is no element such that it is in all sets (which are intersected).
> > But here we assume that ℵo elements are in all sets because
> You can't just /assume/ that, you have to _show/prove_ it.

We can assume it because we intersect, by definition, only infinite endsegments.
>
> Hint: There is no element such that it is in all sets (which are intersected).
> > all are infinite and inclusion monotonic.
> They are, but these facts do not imply that there are elements that are in all (intersected) sets.

They do. Try first to understand it for all endsegments that can be identified:
Every infinite endsegment receives its contents from the first one
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
For them we can prove it, can't we?

Regards, WM

zelos...@gmail.com schrieb am Sonntag, 10. April 2022 um 20:15:07 UTC+2:

> The intersection is empty **IF AND ONLY IF all elements are not present in at least one of the sets**
>
> Which for endsegments is trivial

But not for infinite endsegments. Try first to understand it for all endsegments that can be identified: Every infinite endsegment receives its contents from the first one
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
For them we can prove an infinite intersection, can't we?

> n is not present in E(n+1), ergo n is not in the intersection, but as this is arbitrary,

it proves nothing. We know that every arbitrary natnumber has ℵ₀ successors which cannot be exhausted because otherwise we had a natnumber with less successors.

Regards, WM
>

WM formulated the question :
> zelos...@gmail.com schrieb am Sonntag, 10. April 2022 um 20:15:07 UTC+2:
>
>> The intersection is empty **IF AND ONLY IF all elements are not present in
>> at least one of the sets**
>>
>> Which for endsegments is trivial
>
> But not for infinite endsegments.

As if there was another kind of endsegment.

The farce is strong in this one.

WM formulated the question :
> horand....@gmail.com schrieb am Samstag, 9. April 2022 um 21:40:58 UTC+2:
>> 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.
>
> For *all* finite n:

As if there were any other kind of n.

On Sunday, April 10, 2022 at 7:58:05 PM UTC+2, WM wrote:
> zelos...@gmail.com schrieb am Sonntag, 10. April 2022 um 14:22:07 UTC+2:
> >
> > There is no subset that is common to all endsegments except for the empty set.
> >
> There is an infinite subset common to all infinite endsegments.

Nope. There is not even a single element "common to all infinite endsegments".

On Sunday, April 10, 2022 at 8:25:10 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 10. April 2022 um 20:05:37 UTC+2:
> > On Sunday, April 10, 2022 at 7:47:12 PM UTC+2, WM wrote:
> >
> > The intersection is empty, if there is no element such that it is in all sets (which are intersected). [FF]
> >
> > > But here we assume that ℵo elements are in all sets because
> >
> > You can't just /assume/ that, you have to _show/prove_ it.
> >
> We can assume it because <bla>

No, we can't just "assume" it in math. If a claim (in math) is not an axiom, you have to prove it.

> we intersect [...] only infinite endsegments.

Right, since there are no other endsegments.

Hint: There is no element such that it is in all sets (which are intersected).

On Sunday, April 10, 2022 at 8:27:58 PM UTC+2, WM wrote:
> zelos...@gmail.com schrieb am Sonntag, 10. April 2022 um 20:15:07 UTC+2:
> >
> > n is not present in E(n+1), ergo n is not in the intersection, but as this is arbitrary, NO element is in the intersection.

> it proves nothing.

Ah, _a proof_ proves nothing in your psychotic kingdom.

Well, though this be madness, yet there is method in't.

On 4/10/2022 1:27 PM, WM wrote:
> zelos...@gmail.com schrieb am Sonntag, 10. April 2022 um 20:15:07 UTC+2:
>
>> The intersection is empty **IF AND ONLY IF all elements are not present in at least one of the sets**
>>
>> Which for endsegments is trivial
>
> But not for infinite endsegments. Try first to understand it for all endsegments that can be identified: Every infinite endsegment receives its contents from the first one

gad, back to second grade...

Endsegments are FIXED SETS. No Endsegment "receives its contents" from any other set. USE MATH !!

<snip bla>

> Regards, WM
>>

On Sunday, 10 April 2022 at 14:52:33 UTC-3, WM wrote:
[...]
> For *all* finite n: n FISONs require at least n symbols to distimguish them. 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.

Cute, but stupid. You are also switching quantifiers. There are infinitely many finite natural numbers, no matter how you care to represent them. End of story.

On 4/10/2022 12:55 PM, WM wrote:
> zelos...@gmail.com schrieb am Sonntag, 10. April 2022 um 14:20:54 UTC+2:
>> lördag 9 april 2022 kl. 14:09:14 UTC+2 skrev WM:
>
>>> 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.
>
>> And here we have you trying to extrapolate from the finite to the infinite.
>
> This kind is valid. If not, then the infinite has no mathematical contents.
>
> Can infinitely many FISONs be distinguished by any set of n symbols?
>
> Regards, WM

do the math yourself, it is very simple, why do you keep asking us for help on such simple stuff ?

[Hint: take n to be in the set of natural numbers, let FISON(n) be the FISON ending in n, How many FISON(n) are there ?]

On 4/10/2022 12:48 PM, WM wrote:
> Fritz Feldhase schrieb am Samstag, 9. April 2022 um 17:24:53 UTC+2:
>> On Saturday, April 9, 2022 at 2:20:25 PM UTC+2, WM wrote:
>>
>>> don't use the matheologial miracle "all".
>>
>> Oh,
>
> Show a missing FISON. Fail.
>
> Regards, WM

the missing one is FISON(oo), right ?

YES ! I WIN, I WIN !!!

horand....@gmail.com schrieb am Sonntag, 10. April 2022 um 22:43:27 UTC+2:
> On Sunday, 10 April 2022 at 14:52:33 UTC-3, WM wrote:
> [...]
> > For *all* finite n: n FISONs require at least n symbols to distimguish them. 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.
> There are infinitely many finite natural numbers, no matter how you care to represent them.

But ℵo is larger.

Regards, WM

Fritz Feldhase schrieb am Sonntag, 10. April 2022 um 22:06:49 UTC+2:
> On Sunday, April 10, 2022 at 8:27:58 PM UTC+2, WM wrote:
> > zelos...@gmail.com schrieb am Sonntag, 10. April 2022 um 20:15:07 UTC+2:
> > >
> > > n is not present in E(n+1), ergo n is not in the intersection, but as this is arbitrary, NO element is in the intersection.
>
> > it proves nothing.
>
> Ah, _a proof_ proves nothing in your

An insufficient proof proves nothing. Every n has ℵo successors, ℵo of which can not be chosen as an arbitrary natural number. The set of infinite endsegments has a minimum which is infinite.

Regards, WM

Fritz Feldhase schrieb am Sonntag, 10. April 2022 um 21:54:20 UTC+2:
> On Sunday, April 10, 2022 at 7:58:05 PM UTC+2, WM wrote:
> > zelos...@gmail.com schrieb am Sonntag, 10. April 2022 um 14:22:07 UTC+2:
> > >
> > > There is no subset that is common to all endsegments except for the empty set.
> > >
> > There is an infinite subset common to all infinite endsegments.
>
> Nope. There is not even a single element "common to all infinite endsegments".

Find a natnumber that is in E(n+m) but not in E(n). Fail. Understand: All non-empty endsegments have at least one natnumber in common.

Regards, WM