On Tuesday, 12 April 2022 at 14:49:28 UTC-3, Fritz Feldhase wrote:
> 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?

Any finite number. That's practically all of them, at least according to our resident perfesser.

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

Fritz Feldhase schrieb am Dienstag, 12. April 2022 um 14:42:31 UTC+2:
> 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.

Correct in this case of inclusion monotony.
>
> 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). for a

Try to find a counter example for an inclusion-monotonic sequence of sets (without assuming the contrary).

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 12. April 2022 um 14:58:50 UTC+2:
> 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.

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

Of course. There is no other was to do *mathematics* of the infinite. Even in an infinite set of red apples all apples are red.
>
> IT DOES NOT WORK!

Then all your claims are invalid too.

Regards, WM

On 4/13/2022 8:38 AM, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 12. April 2022 um 14:42:31 UTC+2:
>> 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.
>
> Correct in this case of inclusion monotony.

No. "There are at least 100 elements per endsegment" has nothing to do with "They are in the intersection."

You have been caught in what most call "an obvious lie".

>>
>> 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). for a
>
> Try to find a counter example for an inclusion-monotonic sequence of sets (without assuming the contrary).

Try to present a full, complete and correct example of inclusion monotony applied to your 100.

Can you do that ?

>
> Regards, WM

horand....@gmail.com schrieb am Dienstag, 12. April 2022 um 15:28:01 UTC+2:
> 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?
> Not all end segments are predecessors of E(n).

It is not necessecary and not recommendedto choose a fixed n. All endsegments are predecessors of endsegments. My statement holds for all endsegments.

> Quantifier dyslexia strike again?

Your only reflex? Which endsegment is not a predecessor?

Regards, WM

William schrieb am Dienstag, 12. April 2022 um 15:50:18 UTC+2:
> 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.

ℵ₀ is a number (here of symbols) which is larger than every FISON.
Note that a "supremum" would not be sufficient to prove enumeration of all fractions. ℵ₀ must be realized by elements in every case.

Regards, WM

William schrieb am Dienstag, 12. April 2022 um 15:57:57 UTC+2:
> 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.

Why should they? Inclusion monotony means that the same 100 elements are in every endsegment.

> There is no element that is in every endsegment.

Wrong for inclusion monotoy. Proof is simple: There is no element a in A which is not in B when an element b exist in B which is not in A.

Regards, WM

Crank Wolfgang Mückenheim, aka WM wrote:
> William schrieb am Dienstag, 12. April 2022 um 15:57:57 UTC+2:
>> 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.
>
> Why should they? Inclusion monotony means that the same 100 elements are in every endsegment.
>
>> There is no element that is in every endsegment.
>
> Wrong for inclusion monotoy. Proof is simple: There is no element a in A which is not in B when an element b exist in B which is not in A.

This is not even a statement about "inclusion monotony", it is only
stating strict inclusion. Hardly a counter argument to the fact that
there is no element that is in every endsegment (which is trivial).

Fritz Feldhase schrieb am Dienstag, 12. April 2022 um 19:49:28 UTC+2:
> On Tuesday, April 12, 2022 at 6:59:54 PM UTC+2, sergio wrote:

> > Understand: All non-empty endsegments have at least one natnumber in common.
> Errr? Any two endsegments?

All non-empty endsegments. There is no element a in A which is not in B when an element b exist in B which is not in A.

Regards, WM

Python schrieb am Mittwoch, 13. April 2022 um 16:01:28 UTC+2:
> Crank Wolfgang Mückenheim, aka WM wrote:
> > There is no element a in A which is not in B when an element b exist in B which is not in A.
> This is not even a statement about "inclusion monotony", it is only
> stating strict inclusion.

That is what endsegments obey and what we abbreviate by inclusion monotony here.

> Hardly a counter argument to the fact that
> there is no element that is in every endsegment (which is trivial).

The statement is: There is one and the same element in every infinite endsegment.

Example:
A = {2, 3, 4,...}
B = {1, 2, 3, 4, ...}
Every element of A is in B since there exists b = 1 in B which is not in A.

Regards, WM

On 4/13/2022 8:54 AM, WM wrote:
> William schrieb am Dienstag, 12. April 2022 um 15:57:57 UTC+2:
>> 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.
>
> Why should they? Inclusion monotony means that the same 100 elements are in every endsegment.

Wrong, you confuse endsegments with FISONs. Fail.

>
>> There is no element that is in every endsegment.
>
> Wrong for inclusion monotoy. Proof is simple: There is no element a in A which is not in B when an element b exist in B which is not in A.

your "proof" is a spoof.

>
> Regards, WM
>
>

WM laid this down on his screen :
> Fritz Feldhase schrieb am Dienstag, 12. April 2022 um 14:42:31 UTC+2:
>> 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.
>
> Correct in this case of inclusion monotony.
>>
>> 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). for a
>
> Try to find a counter example for an inclusion-monotonic sequence of sets
> (without assuming the contrary).

How is your inclusion monotonic sequence structured? It looks to me
like you are starting at infinity and generating them in reverse order
to claim inclusion monotonicity when actually they spring forth as
complementary sets from the FISON generation. The FISONs are inclusion
monotonic not the endsegments.

WM brought next idea :
> Python schrieb am Mittwoch, 13. April 2022 um 16:01:28 UTC+2:
>> Crank Wolfgang Mückenheim, aka WM wrote:
>>> There is no element a in A which is not in B when an element b exist in B
>>> which is not in A.
>> This is not even a statement about "inclusion monotony", it is only
>> stating strict inclusion.
>
> That is what endsegments obey and what we abbreviate by inclusion monotony
> here.

Nothing to do with sequences, which we were supposed to be taliking
about. Does monotonic mean you've only had one drink?

On Wednesday, April 13, 2022 at 10:55:02 AM UTC-3, WM wrote:
> William schrieb am Dienstag, 12. April 2022 um 15:57:57 UTC+2:
> > 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.
> Why should they?
Any set of 100 elements has a largest element, call it M.
No element of a set with largest element M is contained in E(M+1)
So now you do,
Whoops. That set did not work lets try a set with maximum M+1,
Whoops. That set did not work lets try a set with maximum M+2,
Whoops. That set did not work lets try a set with maximumM+3,
Whoops, ...

You never get a set that works.

>Inclusion monotony means that the same 100 elements are in every endsegment.
Piffle
> > There is no element that is in every endsegment.
>

Wrong for inclusion monotoy. Proof is simple: There is no element a in A which is not in B when an element b exist in B which is not in A.

Indeed, But what you need is "There is an a(A) in A and in B when an element b exist in B which is not in A." When you change A you may have to change a(A). There is no element a which works for every A.

--
William Hughes

On Wednesday, 13 April 2022 at 10:38:06 UTC-3, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 12. April 2022 um 14:42:31 UTC+2:
> > 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.
> Correct in this case of inclusion monotony.

Bullshit. Quantifier shift always needs a proof. And when performed by WM, one can be sure that there is no proof possible.

On Wednesday, 13 April 2022 at 10:41:36 UTC-3, WM wrote:
> [...] Even in an infinite set of red apples all apples are red.

Yes, but the *SET* is not red. The SET of natural numbers is not a natural number, and the set of prime numbers is not prime, for just two simple examples.

On Wednesday, April 13, 2022 at 11:13:02 AM UTC-3, WM wrote:
> Python schrieb am Mittwoch, 13. April 2022 um 16:01:28 UTC+2:
> > Crank Wolfgang Mückenheim, aka WM wrote:
> > > There is no element a in A which is not in B when an element b exist in B which is not in A.
> > This is not even a statement about "inclusion monotony", it is only
> > stating strict inclusion.
> That is what endsegments obey and what we abbreviate by inclusion monotony here.
> > Hardly a counter argument to the fact that
> > there is no element that is in every endsegment (which is trivial).
> The statement is: There is one and the same element in every infinite endsegment.
>
> Example:
> A = {2, 3, 4,...}
> B = {1, 2, 3, 4, ...}
> Every element of A is in B since there exists b = 1 in B which is not in A.

Lets create the intersection I. 2 is in A and B maybe 2 is in I.

B = {1, 2, 3, 4, ...}
A = { 2, 3, 4, ...}
C = {3, 4, 5 ...}

Whoops, 2 is not in B maybe 3 is in I

B = {1, 2, 3, 4, ...}
A = { 2, 3, 4, ...}
C = {3, 4, 5 ...}
D = {4, 5, 6 ...}

Whoops, ....

--
William Hughes

On Wednesday, 13 April 2022 at 10:46:26 UTC-3, WM wrote:
[...]
> > Not all end segments are predecessors of E(n).
> It is not necessecary and not recommendedto choose a fixed n. All endsegments are predecessors of endsegments. My statement holds for all endsegments.

That, my friend, is the source of your quantifier shift. You truly have no idea how quantifiers are used.

> > Quantifier dyslexia strike again?
>
> Your only reflex? Which endsegment is not a predecessor?

Your only reflex: Deny, deny, deny. Every natural number has a finite number of predecessors and an infinite number of successors. This is the asymmetry that your "argument" cannot address. You are shifting quantifiers left and right and expect people to take you seriously. Now, please, shut up and go home.

On Wednesday, 13 April 2022 at 11:12:49 UTC-3, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 12. April 2022 um 19:49:28 UTC+2:
> > On Tuesday, April 12, 2022 at 6:59:54 PM UTC+2, sergio wrote:
>
> > > Understand: All non-empty endsegments have at least one natnumber in common.
> > Errr? Any two endsegments?
> All non-empty endsegments. There is no element a in A which is not in B when an element b exist in B which is not in A.

Quantifier dyslexia again. I'll call you on that every time I see it. (You make it really easy!)

On Wednesday, 13 April 2022 at 11:13:02 UTC-3, WM wrote:
[...]
> Example:
> A = {2, 3, 4,...}
> B = {1, 2, 3, 4, ...}
> Every element of A is in B since there exists b = 1 in B which is not in A.

That's the most idiotic attempt at proving that A is a subset of B that I have ever seen. Presumably this is also caused by your utter ignorance of quantifiers.

On 4/13/2022 9:12 AM, WM wrote:
> Python schrieb am Mittwoch, 13. April 2022 um 16:01:28 UTC+2:
>> Crank Wolfgang Mückenheim, aka WM wrote:
>>> There is no element a in A which is not in B when an element b exist in B which is not in A.
>> This is not even a statement about "inclusion monotony", it is only
>> stating strict inclusion.
>
> That is what endsegments obey and what we abbreviate by inclusion monotony here.

No. Endsegment is simply a definition of an infinite set of increasing natural numbers starting at k.

try to understand the basics.

>
>> Hardly a counter argument to the fact that
>> there is no element that is in every endsegment (which is trivial).
>
> The statement is: There is one and the same element in every infinite endsegment.

Liar. you have been shown many times a proof that disproves your statement.

>
> Example:
> A = {2, 3, 4,...}
> B = {1, 2, 3, 4, ...}
> Every element of A is in B since there exists b = 1 in B which is not in A.\

your logical thinking unit is off line.

>
> Regards, WM
>

On 4/13/2022 8:49 AM, WM wrote:
> William schrieb am Dienstag, 12. April 2022 um 15:50:18 UTC+2:
>> 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.
>
> ℵ₀ is a number (here of symbols) which is larger than every FISON.
> Note that a "supremum" would not be sufficient to prove enumeration of all fractions. ℵ₀ must be realized by elements in every case.
>
> Regards, WM

not at all.

pick a fraction, any fraction, is it in Cantors Enumeration ? Yes.

FromTheRafters schrieb am Mittwoch, 13. April 2022 um 16:32:25 UTC+2:
> WM laid this down on his screen :
> > Fritz Feldhase schrieb am Dienstag, 12. April 2022 um 14:42:31 UTC+2:
> >> 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.
> >
> > Correct in this case of inclusion monotony.
> >>
> >> 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). for a
> >
> > Try to find a counter example for an inclusion-monotonic sequence of sets
> > (without assuming the contrary).
> How is your inclusion monotonic sequence structured? It looks to me
> like you are starting at infinity and generating them in reverse order
> to claim inclusion monotonicity when actually they spring forth as
> complementary sets from the FISON generation. The FISONs are inclusion
> monotonic not the endsegments.

Both can be named inclusion-monotonic, one ascending, the other descending.

Regards, WM

William schrieb am Mittwoch, 13. April 2022 um 16:39:31 UTC+2:
> On Wednesday, April 13, 2022 at 10:55:02 AM UTC-3, WM wrote:

> > > Nope, the 100 elements changes.
> > Why should they?
> Any set of 100 elements has a largest element, call it M.

No. Any set of *definable* endsegments has a largest element.

> No element of a set with largest element M is contained in E(M+1)
> So now you do,
> Whoops. That set did not work lets try a set with maximum M+1,
> Whoops. That set did not work lets try a set with maximum M+2,
> Whoops. That set did not work lets try a set with maximumM+3,
> Whoops, ...

No whoops. We know inclusion-monotony. Therefore the set must exist. It cannot be named. Therefore dark numbers must exist.
>
> You never get a set that works.

But I know that either there is no actual infinity and hence no endsegments, or there are endsegments proving by your argument that dark numbers must exist - or this simple mathematics fails in your world: There is no element a in A which is not in B when an element b exists in B which is not in A.

> Indeed, But what you need is "There is an a(A) in A and in B when an element b exist in B which is not in A." When you change A you may have to change a(A).

No problem: Then the first A will also contain this changed a. We know this from inclusion-monotony. Therefore it is not necessary to specify the a. Would you need to specify every natural number that has a unique prime decomposition? Or would you trust in mathematics?

> There is no element a which works for every A.

For every definable*) endsegment E(n), there are infinitely many.

*) A definable endsegment is, by definition, an endsegment which has infinitely many natnumbers in common with all definable endsegments: |∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .

Regards, WM

horand....@gmail.com schrieb am Mittwoch, 13. April 2022 um 16:45:40 UTC+2:
> On Wednesday, 13 April 2022 at 10:38:06 UTC-3, WM wrote:
> > Fritz Feldhase schrieb am Dienstag, 12. April 2022 um 14:42:31 UTC+2:
> > > 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.
> > Correct in this case of inclusion monotony.
> Bullshit. Quantifier shift always needs a proof.

Here is no shift but a correct statement. If you wish to attack it, find a counter example.

Regards, WM