FromTheRafters schrieb am Sonntag, 10. April 2022 um 21:34:51 UTC+2:
> 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.

Infinite endsegments have an infinite intersection, namely the infinmum of the set which in this case is infinite.

Regards, WM

WM presented the following explanation :
> 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,

Each endsegment has a least element and induction.

> ℵo of which

What does that mean? Are you trying to use a trasnsfinite cardinal as
if it were a natural number?

on 4/10/2022, WM supposed :
> 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.

No, but 2^aleph_null is 'larger'.

On Sunday, April 10, 2022 at 11:07:39 PM UTC+2, WM wrote:
>
> Every n has ℵo successors, ℵo of which can not be chosen as an arbitrary natural number.

Try to look up "arbitrary" in a dictionary, you silly crank.

On Sunday, April 10, 2022 at 11:10:10 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 10. April 2022 um 21:54:20 UTC+2:
> >
> > 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.

Find a natnumber that is in E(n+1) where n is this number. Fail.

No natnumber is in all endsegments.

In other words, there is not even a single element "common to all (infinite) endsegments".

On Sunday, April 10, 2022 at 11:07:39 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 10. April 2022 um 22:06:49 UTC+2:
> >
> > Ah, _a proof_ proves nothing in your psychotic kingdom.
> >
> An insufficient proof

Look, you silly idiot. Assume that the set of natural numbers that are elements in all endsegments is *not* empty. Let WM be the smallest element in this set. Then WM is not an element in the endsegment {WM+1, WM+2, WM+3, ...} Hence WM is not in _all_ endsegments. Contradiction! This means that our assumption is wrong. Hence the the set of natural numbers that are elements in all endsegments *is* empty. qed

Too dumb to comprehend a simple/elementary proof?

On 4/10/2022 4:11 PM, WM wrote:
> FromTheRafters schrieb am Sonntag, 10. April 2022 um 21:34:51 UTC+2:
>> 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.
>
> Infinite endsegments have an infinite intersection, namely the infinmum of the set which in this case is infinite.
>
> Regards, WM

Nope, try again.

On 4/10/2022 4:28 PM, Fritz Feldhase wrote:
> On Sunday, April 10, 2022 at 11:10:10 PM UTC+2, WM wrote:
>> Fritz Feldhase schrieb am Sonntag, 10. April 2022 um 21:54:20 UTC+2:
>>>
>>> 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.
>
> Find a natnumber that is in E(n+1) where n is this number. Fail.
>
> No natnumber is in all endsegments.
>
> In other words, there is not even a single element "common to all (infinite) endsegments".

which means the intersection of all endsegments is empty.

On Sunday, April 10, 2022 at 2:44:40 PM UTC-3, 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.

Piffle. Indeed, for all n finite it true. However it is not true for n infinite.

--
William Hughes

On Sunday, April 10, 2022 at 2:47:12 PM UTC-3, WM wrote:
> 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.
Nope, very relevent to the question of when the intersection is empty.

--
William Hughes

On Sunday, April 10, 2022 at 7:58:05 PM UTC+2, WM wrote:

> There is an infinite subset common to all infinite endsegments.

No, there is no such set. After all this would imply, that the set of natural numbers that are elements in all infinite endsegments is *not* empty.

Assume that the set of natural numbers that are elements in all infinite endsegments is *not* empty. Let WM be the smallest element in this set. Then WM is not an element in the infinite endsegment {WM+1, WM+2, WM+3, ...} Hence WM is not in _all_ infinite endsegments. Contradiction! This means that our assumption is wrong. Hence the the set of natural numbers that are elements in all infinite endsegments *is* empty. qed

So no, there is no such set.

On Sunday, April 10, 2022 at 2:52:33 PM UTC-3, WM wrote:

> For *all* finite

Says nothing about infinite.

--
William Hughes

On Sunday, April 10, 2022 at 2:55:24 PM UTC-3, WM wrote:

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

Yes, if n is infinite.

--
William Hughes

On Sunday, April 10, 2022 at 2:58:05 PM UTC-3, WM wrote:
> zelos...@gmail.com schrieb am Sonntag, 10. April 2022 um 14:22:07 UTC+2:

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

iff the sequence has last element.

On Sunday, April 10, 2022 at 3:27:58 PM UTC-3, WM wrote:

>... all endsegments that can be identified

The sequence of endsegments that can be identified has no last element.

--
William Hughes

On Sunday, April 10, 2022 at 6:07:39 PM UTC-3, WM wrote:
>...The set of infinite endsegments has a minimum which is infinite.

Nope, there is no minimum infinite endsegment (An endsegment is just the complemnt of a FISON and there is no maximum FISON).

--
William Hughes

On 4/10/2022 5:31 PM, Fritz Feldhase wrote:
> On Sunday, April 10, 2022 at 11:07:39 PM UTC+2, WM wrote:
>> Fritz Feldhase schrieb am Sonntag, 10. April 2022 um 22:06:49 UTC+2:
>>>
>>> Ah, _a proof_ proves nothing in your psychotic kingdom.
>>>
>> An insufficient proof
>
> Look, you silly idiot. Assume that the set of natural numbers that are elements in all endsegments is *not* empty. Let WM be the smallest element in this set. Then WM is not an element in the endsegment {WM+1, WM+2, WM+3, ...} Hence WM is not in _all_ endsegments. Contradiction! This means that our assumption is wrong. Hence the the set of natural numbers that are elements in all endsegments *is* empty. qed
>
> Too dumb to comprehend a simple/elementary proof?

yes, WM is too dumb, he wants to keep trolling.

söndag 10 april 2022 kl. 20:27:58 UTC+2 skrev 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.

It is, you are just too much of an idiot.

>Try first to understand

I understand mathematics far better than you.

>it for all endsegments that can be identified:

"identified" has no meaning in mathematics.

>Every infinite endsegment receives its contents from the first one

Nothing "recieves" anything.

> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀

Trivially true, but those are FINITE intersection so not relevant here.

> For them we can prove an infinite intersection, can't we?

Correct, but you intersection only finitely many sets, not all endsegments.

> > n is not present in E(n+1), ergo n is not in the intersection, but as this is arbitrary,
> it proves nothing.

It proves that n cannot be in the intersection from definition.

>We know that every arbitrary natnumber has ℵ₀ successors which cannot be exhausted

"exhausted" has no mathematical meaning.

>because otherwise we had a natnumber with less successors.

Non-sequitor.

>
> Regards, WM
> >

Are you done with your logical fallacies?

söndag 10 april 2022 kl. 23:07:39 UTC+2 skrev 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.

It is very sufficient as it proves that ALL natural numbers, are excluded. Thus it is empty.

Your stuff is insufficient as they are pure non-sequitors.

>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

There is no "minimum" there. You assume it without proof

William was thinking very hard :
> On Sunday, April 10, 2022 at 2:44:40 PM UTC-3, 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.
>
> Piffle. Indeed, for all n finite it true. However it is not true for n
> infinite.

I thought n was an element of the naturals, none of which can ever be
infinite.

FromTheRafters schrieb am Sonntag, 10. April 2022 um 23:15:12 UTC+2:
> WM presented the following explanation :

> > An insufficient proof proves nothing. Every n has ℵo successors,
> Each endsegment has a least element and induction.
>
> > ℵo of which
>
> What does that mean? Are you trying to use a trasnsfinite cardinal as
> if it were a natural number?

I use ℵo to denote actual infinity.

Regards, WM

Fritz Feldhase schrieb am Sonntag, 10. April 2022 um 23:28:26 UTC+2:
> On Sunday, April 10, 2022 at 11:10:10 PM UTC+2, WM wrote:
> > Fritz Feldhase schrieb am Sonntag, 10. April 2022 um 21:54:20 UTC+2:
> > >
> > > 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.
> Find a natnumber that is in E(n+1) where n is this number. Fail.
>
> No natnumber is in all endsegments.

If so, then you had to find a natnumber whih is in E(n) but not in all its predecessors.
>
> In other words, there is not even a single element "common to all (infinite) endsegments".

That is nosense. Inclusion monotony means we have a decreasing sequence of sets. If all sets are containg somethimg, then all set have a non-empty intersection.

Note that the intersection is the minimum. Infinite sets have an unknown minimum but it is larger than every finite set.

Regards, WM

Fritz Feldhase schrieb am Montag, 11. April 2022 um 00:31:37 UTC+2:
> Assume that the set of natural numbers that are elements in all endsegments is *not* empty. Let WM be the smallest element in this set. Then WM is not an element in the endsegment {WM+1, WM+2, WM+3, ...} Hence WM is not in _all_ endsegments. Contradiction! This means that our assumption is wrong. Hence the the set of natural numbers that are elements in all endsegments *is* empty. qed

As long as all sets are infinite WM is not required.

Regards, WM

William schrieb am Montag, 11. April 2022 um 00:40:11 UTC+2:
> On Sunday, April 10, 2022 at 2:44:40 PM UTC-3, 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.
>
> Piffle.

No, a simple mathematical fact.

> Indeed, for all n finite it true. However it is not true for n infinite.

Assume you were right. Then a finite number of n symbols would be sufficient to distinguish more than n FISONs, namely ℵo. That is too obviously wrong that you could really believe it.

Regards, WM

On Monday, April 11, 2022 at 8:23:26 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 10. April 2022 um 23:28:26 UTC+2:
> >
> > Find a natnumber that is in E(n+1) where n is this number. Fail.
> >
> > No natnumber is in all endsegments.
> >
> If so, then you had to find a natnumber which is in E(n) but not in all its predecessors.

Huh?! Utter nonsense.

> > In other words, there is not even a single element "common to all (infinite) endsegments".
> >
> That is

A provable fact.

> Note that the intersection is the

empty set. Yes, we know that.