On Monday, April 11, 2022 at 3:38:09 PM UTC-3, 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.

True and completely irrelevant, all endsements have cardinality aleph_0 >100. Mininum is based on set inclusion. not on cardinality. Recall B can be a proper subset of A even though both sets have the same cardinality, aleph_0. Since for all n, E(n+1) is a proper subset of E(n) there is no minimum.

--
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?

Nope, the limit is 1. But we are not considering limits but infinite intersections. True. the infinite intersection can be shown to be equal to a limit (not a limit based on metric so "close" does not apply) but we do not have to use limits. We use a definition of infinite intersection that is not based on a limit so the cardinality of a limit is irrelevant. In particular, your non-standard "limit" where the limit of a sequence of sets with cardinality aleph_0 has cardinality aleph_0 is irrelevant.

--
William Hughes

måndag 11 april 2022 kl. 20:18:48 UTC+2 skrev WM:
> 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

It is an infinite cardinality, your "actual" is meaningless

måndag 11 april 2022 kl. 20:23:26 UTC+2 skrev 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.

non-squitor

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

You being a retard does not mean it is non'sense

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

False {1,2},{2,3},{3,1} all have something, any 2 have a non-empty intersection, but the intersection of all is empty.

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

But there is no subset other than the empty set that is a subset to all endsegments

>
> Regards, WM

On Tuesday, April 12, 2022 at 6:49:31 AM UTC+2, zelos...@gmail.com wrote:
> måndag 11 april 2022 kl. 20:23:26 UTC+2 skrev WM:
> >
> > Inclusion monotony means we have a decreasing sequence of sets. If all sets are containg somethimg, then [the intersection of the set of] all [these] set[s is] non-empty[.]

Of course not.

Counterexample: We may consider the sequence (I_n) with I_n = (0, 1/n], its terms and the intersection of the set of all terms of this sequence.

> False {1,2},{2,3},{3,1} all have something, any 2 have a non-empty intersection, but the intersection of all is empty.

On the other hand, no "inclusion monotony". :-P

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

Nonsense.

> But there is no subset other than the empty set that is a subset to all endsegments.

I'm sure you will convince him THIS TIME!

horand....@gmail.com schrieb am Montag, 11. April 2022 um 20:55:09 UTC+2:
> 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",

It is the same because every endsegment can be inserted. Which one would be disallowed?

Regards, WM

William schrieb am Montag, 11. April 2022 um 21:40:49 UTC+2:
> 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.

Impossible for FISONs. Note that in
o oo
ooo
....
height ℵ₀ implies width ℵ₀.

Regards, WM

William schrieb am Montag, 11. April 2022 um 22:10:02 UTC+2:
> On Monday, April 11, 2022 at 3:38:09 PM UTC-3, 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.
> True and completely irrelevant, all endsements have cardinality aleph_0 >100. Mininum is based on set inclusion. not on cardinality.

> If set inclusion is true, then cardinality is true and vice versa.

> Recall B can be a proper subset of A even though both sets have the same cardinality, aleph_0. Since for all n, E(n+1) is a proper subset of E(n) there is no minimum.
>
There are at least 100 elements per infinite endsegment. They are in the intersection.

Regards, WM

William schrieb am Montag, 11. April 2022 um 22:29:55 UTC+2:
> your non-standard "limit" where the limit of a sequence of sets with cardinality aleph_0 has cardinality aleph_0 is irrelevant.
>
Why? Does card aleph_0 not imply that there are as many elements?

Regards, WM

On Tuesday, April 12, 2022 at 2:20:46 PM UTC+2, WM wrote:
>
> There are at least 100 elements per endsegment. They are in the intersection.

Quantifier shift.

For each and every endsegment: there are at least 100 elements in it. (true)

WM concludes (by a quantifier shift):

There are at least 100 elements, which are in each and every endsegement. (false)

"Hence": They (!) are in the intersection (of all endsegments).

_______________________

You are such a fool, man.

tisdag 12 april 2022 kl. 13:01:42 UTC+2 skrev Fritz Feldhase:
> On Tuesday, April 12, 2022 at 6:49:31 AM UTC+2, zelos...@gmail.com wrote:
> > måndag 11 april 2022 kl. 20:23:26 UTC+2 skrev WM:
> > >
> > > Inclusion monotony means we have a decreasing sequence of sets. If all sets are containg somethimg, then [the intersection of the set of] all [these] set[s is] non-empty[.]
>
> Of course not.
>
> Counterexample: We may consider the sequence (I_n) with I_n = (0, 1/n], its terms and the intersection of the set of all terms of this sequence.
> > False {1,2},{2,3},{3,1} all have something, any 2 have a non-empty intersection, but the intersection of all is empty.
> On the other hand, no "inclusion monotony". :-P
> > > Note that the intersection is the minimum. Infinite sets have an unknown minimum but it is larger than every finite set.
> Nonsense.
>
> > But there is no subset other than the empty set that is a subset to all endsegments.
>
> I'm sure you will convince him THIS TIME!

Of course not, he is an idiot

tisdag 12 april 2022 kl. 14:17:27 UTC+2 skrev WM:
> William schrieb am Montag, 11. April 2022 um 21:40:49 UTC+2:
> > 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.
> Impossible for FISONs. Note that in
> o
> oo
> ooo
> ...
> height ℵ₀ implies width ℵ₀.
>
> Regards, WM

Here you go trying to extrapolate from the finite, the infinite.

IT DOES NOT WORK!

On 4/12/2022 7:17 AM, WM wrote:
> William schrieb am Montag, 11. April 2022 um 21:40:49 UTC+2:
>> 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.
>
> Impossible for FISONs. Note that in
> o
> oo
> ooo
> ...
> height ℵ₀ implies width ℵ₀.
>
> Regards, WM
>

Wrong, each and every horizontal line is a finite set.

On Tuesday, 12 April 2022 at 09:11:26 UTC-3, WM wrote:
> horand....@gmail.com schrieb am Montag, 11. April 2022 um 20:55:09 UTC+2:
> > 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",
> It is the same because every endsegment can be inserted. Which one would be disallowed?

You are spouting nonsense, as usual. Not all end segments are predecessors of E(n). How dense are you this morning? Quantifier dyslexia strike again? Hit the sauce early, perhaps?

On Tuesday, April 12, 2022 at 9:17:27 AM UTC-3, WM wrote:
> William schrieb am Montag, 11. April 2022 um 21:40:49 UTC+2:
> > 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.
> Impossible for FISONs.

Piffle.

Note that in
> o
> oo
> ooo
> ...
> height ℵ₀ implies width ℵ₀.

Piffle. Width is supremum of row width, width is ℵ₀.

--
William Hughes

On Tuesday, April 12, 2022 at 3:28:01 PM UTC+2, sergio wrote:
> On 4/12/2022 7:17 AM, WM wrote:
> > William schrieb am Montag, 11. April 2022 um 21:40:49 UTC+2:
> >> 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.
> >
> > Impossible for FISONs. Note that in
> > o
> > oo
> > ooo
> > ...
> > height ℵ₀ implies width ℵ₀.
> >
> > Regards, WM
> >
> Wrong, each and every horizontal line is a finite set.

Actually, he is (or may be) right. After defining width and height for such "figures" (comprising all finite cases as well as "the infinite case") in a propper way. Of course he's too dumb to comprehend that...

On Tuesday, April 12, 2022 at 9:22:49 AM UTC-3, WM wrote:
> William schrieb am Montag, 11. April 2022 um 22:29:55 UTC+2:
> > your non-standard "limit" where the limit of a sequence of sets with cardinality aleph_0 has cardinality aleph_0 is irrelevant.
> >
> Why?

Because no limit is relevant.

--
William Hughes

On Tuesday, April 12, 2022 at 9:20:46 AM UTC-3, WM wrote:
>
> There are at least 100 elements per infinite endsegment.

but not the same 100 elements in every endsegment

> They are in the intersection

Nope, the 100 elements changes. There is no element that is in every endsegment.

--
William Hughes

On 4/12/2022 7:22 AM, WM wrote:
> William schrieb am Montag, 11. April 2022 um 22:29:55 UTC+2:
>> your non-standard "limit" where the limit of a sequence of sets with cardinality aleph_0 has cardinality aleph_0 is irrelevant.
>>
> Why? Does card aleph_0 not imply that there are as many elements?
>
> Regards, WM

no. aleph_0 = aleph_0 + k

On 4/12/2022 7:11 AM, WM wrote:
> horand....@gmail.com schrieb am Montag, 11. April 2022 um 20:55:09 UTC+2:
>> 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",
>
> It is the same because every endsegment can be inserted. Which one would be disallowed?
>
> Regards, WM

Wrong, Try Again.

On 4/12/2022 8:50 AM, Fritz Feldhase wrote:
> On Tuesday, April 12, 2022 at 3:28:01 PM UTC+2, sergio wrote:
>> On 4/12/2022 7:17 AM, WM wrote:
>>> William schrieb am Montag, 11. April 2022 um 21:40:49 UTC+2:
>>>> 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.
>>>
>>> Impossible for FISONs. Note that in
>>> o
>>> oo
>>> ooo
>>> ...
>>> height ℵ₀ implies width ℵ₀.
>>>
>>> Regards, WM
>>>
>> Wrong, each and every horizontal line is a finite set.
>
> Actually, he is (or may be) right. After defining width and height for such "figures" (comprising all finite cases as well as "the infinite case") in a propper way. Of course he's too dumb to comprehend that...
>

I think each o means a coconut.

If WM is gonna do that, he needs to paint the counting number on the side of each coconut, so he doesn't double or triple count them.

On 4/9/2022 7:23 AM, 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 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

"Now extend this to a number larger than all n" Fail.

"distinct" Fail.

On 4/12/2022 7:20 AM, WM wrote:
> William schrieb am Montag, 11. April 2022 um 22:10:02 UTC+2:
>> On Monday, April 11, 2022 at 3:38:09 PM UTC-3, 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.
>> True and completely irrelevant, all endsements have cardinality aleph_0 >100. Mininum is based on set inclusion. not on cardinality.
>
>> If set inclusion is true, then cardinality is true and vice versa.
>
>> Recall B can be a proper subset of A even though both sets have the same cardinality, aleph_0. Since for all n, E(n+1) is a proper subset of E(n) there is no minimum.
>>
> There are at least 100 elements per infinite endsegment. They are in the intersection.

None of those 100 elements are in all endsegments.

>
> Regards, WM

On 4/10/2022 4:10 PM, WM wrote:
> 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.

you are using VARIABLES, n and m. You said you do not believe in them!

Understand: All non-empty endsegments have at least one natnumber in common.

and the intersection of all endsegments is not an endsegment, and is empty.

>
> Regards, WM

On Tuesday, April 12, 2022 at 6:59:54 PM UTC+2, sergio wrote:
> On 4/10/2022 4:10 PM, WM wrote:
> > 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.
> you are using VARIABLES, n and m. You said you do not believe in them!
>
>
> Understand: All non-empty endsegments have at least one natnumber in common.

Errr? Any two endsegments?

> and the intersection of all endsegments is not an endsegment, and is empty.