William schrieb am Montag, 11. April 2022 um 00:46:40 UTC+2:
> On Sunday, April 10, 2022 at 2:47:12 PM UTC-3, WM wrote:

> > > The intersetion is empty iff the sequence has no last element.
> > Irrelevant.
> Nope, very relevent to the question of when the intersection is empty.

The limit of the sequence (1 + 1/n) is zero when the sequence has no last element?

Regards, WM

On Monday, April 11, 2022 at 8:25:25 PM UTC+2, WM wrote:
> 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.

Huh?!

Wie meinen, Mückenheim?

William schrieb am Montag, 11. April 2022 um 00:48:59 UTC+2:
> On Sunday, April 10, 2022 at 2:52:33 PM UTC-3, WM wrote:
>
> > For *all* finite
>
> Says nothing about infinite.

The infinite is larger than all finite n. Agreed? If all finite n are required to distingusih all finite FISONs, an infinite FISON would need somewhat more.

Regards, WM

William schrieb am Montag, 11. April 2022 um 01:00:24 UTC+2:
> 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.

But every element of the infinite sequence contains more than 100 natnumbers.

Regards, WM

William schrieb am Montag, 11. April 2022 um 01:06:28 UTC+2:
> 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

But among all infinite endsegments there is none with less than 100 natnumbers.

Regards, WM

WM was thinking very hard :
> William schrieb am Montag, 11. April 2022 um 00:48:59 UTC+2:
>> On Sunday, April 10, 2022 at 2:52:33 PM UTC-3, WM wrote:
>>
>>> For *all* finite
>>
>> Says nothing about infinite.
>
> The infinite is larger than all finite n. Agreed? If all finite n are
> required to distingusih all finite FISONs, an infinite FISON would need
> somewhat more.

Infinite FISONs again?

On 4/11/2022 1:28 PM, WM wrote:
> 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.

zug zug.

>
>> 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.

Wrong again, just use n+1, then use n+2, then use n+3, ...

Study your counting numbers.

>
> Regards, WM

On 4/11/2022 1:25 PM, WM wrote:
> 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

you didn't understand that either.

On 4/11/2022 1:35 PM, WM wrote:
> William schrieb am Montag, 11. April 2022 um 01:00:24 UTC+2:
>> 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.
>
> But every element of the infinite sequence contains more than 100 natnumbers.
>
> Regards, WM

and the intersection of all endsegments is empty.

zelos...@gmail.com schrieb am Montag, 11. April 2022 um 13:47:16 UTC+2:

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

Minimum of infimum. No infinite endsegment contains less than infinitely many natnumbers in common with all infinite endsegments. What is it that makes all endsements infinite? Natumbers.

Regards, WM

On 4/11/2022 1:31 PM, WM wrote:
> William schrieb am Montag, 11. April 2022 um 00:46:40 UTC+2:
>> On Sunday, April 10, 2022 at 2:47:12 PM UTC-3, WM wrote:
>
>>>> The intersetion is empty iff the sequence has no last element.
>>> Irrelevant.
>> Nope, very relevent to the question of when the intersection is empty.
>
> The limit of the sequence (1 + 1/n) is zero when the sequence has no last element?
>
> Regards, WM

two wrongs in one post

The limit of the sequence (1 + 1/n) as n => 00 is 1, not zero. There is no last element

Fritz Feldhase schrieb am Montag, 11. April 2022 um 20:30:19 UTC+2:
> 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.

Only in the failed theory of a gang of fools an inclusion monotonic sequenced of infinite terms can have an empty intersection.

Regards, WM

On 4/11/2022 1:23 PM, WM wrote:
> 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.

not at all.

>>
>> 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.

as shown above, and many times before, months before, with easy proofs, there are no elements in the intersection of all endsegments.

[you need to locate that one braincell that is stuck, and reset it to null.]

>
> Note that the intersection is the minimum.

you need to show that. I bet you cannot.

> Infinite sets have an unknown minimum but it is larger than every finite set.

what are you talking about ? number of elements, the value of elements,
so far that sentence is nonsense.

>
> Regards, WM

On 4/11/2022 1:33 PM, WM wrote:
> William schrieb am Montag, 11. April 2022 um 00:48:59 UTC+2:
>> On Sunday, April 10, 2022 at 2:52:33 PM UTC-3, WM wrote:
>>
>>> For *all* finite
>>
>> Says nothing about infinite.
>
> The infinite is larger than all finite n. Agreed? If all finite n are required to distingusih all finite FISONs, an infinite FISON would need somewhat more.
>
> Regards, WM

you mean FISON(oo) ?

On Monday, 11 April 2022 at 15:23:26 UTC-3, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 10. April 2022 um 23:28:26 UTC+2:
[...]
> > 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.

"Not in all end segments" is *NOT* the same as "not in all predecessors", except, it seems, in Wolkenmuckenheim. You are such a DUMB *FUCK*!!

On 4/11/2022 1:48 PM, WM wrote:
> Fritz Feldhase schrieb am Montag, 11. April 2022 um 20:30:19 UTC+2:
>> 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.
>
> Only in the failed theory of a gang of fools an inclusion monotonic sequenced of infinite terms can have an empty intersection.
>
> Regards, WM

Only a fool that cannot follow a simple proof that *no element is in the intersection of all endsegments* would say so.

On 4/11/2022 1:38 PM, WM wrote:
> William schrieb am Montag, 11. April 2022 um 01:06:28 UTC+2:
>> 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
>
> But among all infinite endsegments there is none with less than 100 natnumbers.
>
> Regards, WM

Reminder: all endsegments have infinite number of elements.

On 4/11/2022 1:46 PM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 11. April 2022 um 13:47:16 UTC+2:
>
>> There is no "minimum" there. You assume it without proof
>
> Minimum of infimum. No infinite endsegment contains less than infinitely many natnumbers in common with all infinite endsegments. What is it that makes all endsements infinite? Natumbers.
>
> Regards, WM
>
>

Still Wrong. is k in the intersection ?

On Monday, 11 April 2022 at 15:33:40 UTC-3, WM wrote:
[...]
> If all finite n are required to distingusih all finite FISONs, an infinite FISON would need somewhat more.

Yeah, well, if there *WERE* an infinite FISON, then yes. I hope the esteemed perfesser can still figure out for himself why that won't be necessary.

On Monday, April 11, 2022 at 8:48:19 PM UTC+2, WM wrote:
>
> Only in the failed theory of a gang of fools an inclusion monotonic sequence of infinite terms can have an empty intersection.

So real analysis is wrong too in your psychotic kingdom?

lim_(n->oo) (0, 1/n] =/= { } there?

Hint: In real analysis lim_(n->oo) [0, 1/n] = {0}, hence lim_(n->oo) (0, 1/n] = { }.

On Monday, 11 April 2022 at 15:46:13 UTC-3, WM wrote:
[...]
> No infinite endsegment contains less than infinitely many natnumbers in common with all infinite endsegments.
Bullshit. Nonsensical blabbering, probably with an implied quantifier switch tossed in for good measure.

On Monday, 11 April 2022 at 15:46:52 UTC-3, sergio wrote:
[...]
> two wrongs in one post

That's not bad, actually. His normal rate is at least three per post...

On Monday, April 11, 2022 at 8:48:19 PM UTC+2, WM wrote:
>
> Only in the failed theory of a gang of fools an inclusion monotonic sequence of infinite terms can have an empty intersection.

So real analysis is wrong too in your psychotic kingdom?

INTERSECTION_(n e IN) (0, 1/n] =/= { } there?

Hint: In real analysis INTERSECTION_(n e IN) [0, 1/n] = {0}, hence INTERSECTION_(n e IN) (0, 1/n] = { } (since for all n e IN: 0 !e (0, 1/n]).

On Monday, April 11, 2022 at 3:28:51 PM UTC-3, WM wrote:
>.. Then a finite number of n

Nope, n is not finite, n is the infinite cardinality ℵo. A set of symbols with cardinality ℵo can distinguish the set of FISONs which has cardinality ℵo (like any Peano set). The set is infinite, the *elements* of the set are finite.

--
William Hughes

On Monday, April 11, 2022 at 3:31:13 PM UTC-3, WM wrote:
> William schrieb am Montag, 11. April 2022 um 00:46:40 UTC+2:
> > On Sunday, April 10, 2022 at 2:47:12 PM UTC-3, WM wrote:
>
> > > > The intersetion is empty iff the sequence has no last element.
> > > Irrelevant.
> > Nope, very relevent to the question of when the intersection is empty.
> The limit of the sequence (1 + 1/n) is zero when the sequence has no last element?

Trivially true. Has very limited relevance. [Do not confuse a metric limit with a set limit. If there is no metric the concept of "close" does
not apply, Set limits are not covered in Calc 101.]

--
William Hughes