William schrieb am Freitag, 8. Juli 2022 um 23:36:20 UTC+2:
> On Friday, July 8, 2022 at 6:05:23 PM UTC-3, WM wrote:
> > William schrieb am Freitag, 8. Juli 2022 um 17:08:33 UTC+2:
> > > >... their intersection contains more than 100 elements.
> > > Nope does not follow and is not true.
> > Try to think better. If all contain more than 100 elements, then all contain the same 100 elements.
> Nope not true. All contain more than 100 elements but there is no set of 100 elements common to all endsegments.

For definable elements it is certainly true. Endsegments are terms of a decreasing sequence. Compare with a decreasing sequence of sets of only definable elements. There all set with more than 100 elements have an intersection with more than 100 elements. Try to find a counterexample with definable elemehts. Fail.

If there are dark elements this logic may be violated. But only if the elements are really dark.
>
> 1, For any finite set R of endsegments the is a set S(R) of 100 natural numbers such that S(R) is contained in each element of R

For every set of definable elements.

> 2. There is a set S of 100 natural numbers such that for any finite set R of endsegments, S is contained in each element of R.
>
> 1. is true, 2. is false.

2. is obviously true. What should the endsegments consist of, if not of infinitely many numbers, 100 of which are in all of them?

What you address here is the potential infinity of natural numbers. You cannot define 100 natural numbers in all endsegments because for every defined set of numbers you can find an endsegment not containing them. But that is a matter of potential infinity only because you can for "given" (i.e. defined) endsegments always prove that 100 numbers are contained in all of them..

Fact is: All definable endsegments have an infinite intersection. You cannot define that intersection because every defined number is deleted in some endsegment. Nevertheless its infinite contents is in all its predecessors.

And if only infinite endsegments are considered, then all endsegments are predecessors of an infinite endsegment and therefore sharing infinitely many numbers.

The intersectiopn of infinite endsegments is infinite.

Regards, WM

FromTheRafters schrieb am Freitag, 8. Juli 2022 um 23:41:06 UTC+2:
> WM pretended :
> > William schrieb am Freitag, 8. Juli 2022 um 17:08:33 UTC+2:
> >> On Friday, July 8, 2022 at 11:59:02 AM UTC-3, WM wrote:
> >>
> >>
> >>>>
> >>>> ... All endsgements are large than 100 elements but their intersection is
> >>>> smaller than the minimum 100 elements.
> >>>>
> >>>> So what? This is not a contradiction (even noting inclusion monotony) as
> >>>> there is no last endsegment.
> >>> There is no minimum endsegment, but all are infinite endsegments. They
> >>> [each] contain more than 100 elements.
> >> Correct (but not the same 100 elements).
> >
> > Wrong. The endsegments get emptier step by step.
>
> Some particular symbols may be missing or shifted, but the endsegment
> is still inductive and its size is Aleph_zero.

And remains so for every definable endsegment. Every definable endsegment is infinite and has an infinite intersection with all its predecessors.

And if we consider only definale (= infinite) endsegments, then necessarily all are predecessors of an infinite endsegment and therefore all definable endsegments have an infinite intersection.

Regards, WM

On Saturday, July 9, 2022 at 9:41:57 AM UTC-3, WM wrote:

> potential infinity of natural numbers.

Absolute nonsense. |N does not change and |N has cardinality aleph_0

--
William Hughes

On Saturday, July 9, 2022 at 2:41:57 PM UTC+2, WM wrote:
> William schrieb am Freitag, 8. Juli 2022 um 23:36:20 UTC+2:
> >
> > 2. There is a set S of 100 natural numbers such that for any finite set R of endsegments, S is contained in each element of R.
> >
> > 2. is false.
> >
> 2. is obviously true.

Really?

Let S = {n_1, n_2, ..., n_100} with n_i e IN for all i e {1, ..., n} and n_i =/= n_j for all i, j e {1, ..., 100}, i =/= j.

Then S has a largest element max(S). Let n0 = max(S).

Now consider the following finite set of endsegments { (n0+1, n0+2, n0+3, ...) }. Is S contained in (n0+1, n0+2, n0+3, ...)?

Obviously not. Hence 2 is false.

On Saturday, July 9, 2022 at 9:41:57 AM UTC-3, WM wrote:

>...
> Fact is: All definable endsegments have an infinite intersection.

By "definable" you mean a finite So 1 is true.

1.For any definable set R of endsegments there is a set S(R) of 100 natural numbers such that S(R) is contained in each element of R

2. There is a set S of 100 natural numbers such that for any definable set R of endsegments, S is contained in each element of R.

> ... You cannot define [S] .

So 2. is false. ( if S exists [and, like any set, does not change] then S is definable).
--
William Hughes

On 7/9/2022 7:41 AM, WM wrote:
> William schrieb am Freitag, 8. Juli 2022 um 23:36:20 UTC+2:
>> On Friday, July 8, 2022 at 6:05:23 PM UTC-3, WM wrote:
>>> William schrieb am Freitag, 8. Juli 2022 um 17:08:33 UTC+2:
>>>>> ... their intersection contains more than 100 elements.
>>>> Nope does not follow and is not true.
>>> Try to think better. If all contain more than 100 elements, then all contain the same 100 elements.
>> Nope not true. All contain more than 100 elements but there is no set of 100 elements common to all endsegments.
>
> For definable "CLOWN SHOES" elements it is certainly true. Endsegments are terms of a decreasing sequence. Compare with a decreasing sequence of sets of only definable "CLOWN SHOES" elements. There all set with more than 100 elements have an intersection with more than 100 elements. Try to find a counterexample with definable "CLOWN SHOES" elemehts. Fail.
>
> If there are dark "CLOWN SHOES" elements this logic may be violated. But only if the elements are really dark "CLOWN SHOES".
>>
>> 1, For any finite set R of endsegments the is a set S(R) of 100 natural numbers such that S(R) is contained in each element of R
>
> For every set of definable "CLOWN SHOES" elements.
>
>> 2. There is a set S of 100 natural numbers such that for any finite set R of endsegments, S is contained in each element of R.
>>
>> 1. is true, 2. is false.
>
> 2. is obviously true. What should the endsegments consist of, if not of infinitely many numbers, 100 of which are in all of them?
>
> What you address here is the potential "CLOWN SHOES" infinity of natural numbers. You cannot define 100 natural numbers in all endsegments because for every defined "CLOWN SHOES" set of numbers you can find an endsegment not containing them. But that is a matter of potential "CLOWN SHOES" infinity only because you can for "given" (i.e. defined "CLOWN SHOES") endsegments always prove that 100 numbers are contained in all of them.
>
> Fact is: All definable "CLOWN SHOES" endsegments have an infinite intersection. You cannot define "CLOWN SHOES" that intersection because every defined "CLOWN SHOES" number is deleted in some endsegment. Nevertheless its infinite contents is in all its predecessors.
>
> And if only infinite endsegments are considered, then all endsegments are predecessors of an infinite endsegment and therefore sharing infinitely many numbers.
>
> The intersectiopn of infinite endsegments is infinite.
>
> Regards, WM

I put "CLOWN SHOES" next to each non-math term, that are WM's special words he relies upon to confuse readers.

their are 13 "CLOWN SHOES" above. Try not to use these terms, as they are WM's made up terms, and it signals to WM, that he has put the bag on ya.

WM explained on 7/9/2022 :
> FromTheRafters schrieb am Freitag, 8. Juli 2022 um 23:41:06 UTC+2:
>> WM pretended :
>>> William schrieb am Freitag, 8. Juli 2022 um 17:08:33 UTC+2:
>>>> On Friday, July 8, 2022 at 11:59:02 AM UTC-3, WM wrote:
>>>>
>>>>
>>>>>>
>>>>>> ... All endsgements are large than 100 elements but their intersection
>>>>>> is smaller than the minimum 100 elements.
>>>>>>
>>>>>> So what? This is not a contradiction (even noting inclusion monotony) as
>>>>>> there is no last endsegment.
>>>>> There is no minimum endsegment, but all are infinite endsegments. They
>>>>> [each] contain more than 100 elements.
>>>> Correct (but not the same 100 elements).
>>>
>>> Wrong. The endsegments get emptier step by step.
>>
>> Some particular symbols may be missing or shifted, but the endsegment
>> is still inductive and its size is Aleph_zero.
>
> And remains

Not remains, simply is.

On 7/9/2022 7:45 AM, WM wrote:
> FromTheRafters schrieb am Freitag, 8. Juli 2022 um 23:41:06 UTC+2:
>> WM pretended :
>>> William schrieb am Freitag, 8. Juli 2022 um 17:08:33 UTC+2:
>>>> On Friday, July 8, 2022 at 11:59:02 AM UTC-3, WM wrote:
>>>>
>>>>
>>>>>>
>>>>>> ... All endsgements are large than 100 elements but their intersection is
>>>>>> smaller than the minimum 100 elements.
>>>>>>
>>>>>> So what? This is not a contradiction (even noting inclusion monotony) as
>>>>>> there is no last endsegment.
>>>>> There is no minimum endsegment, but all are infinite endsegments. They
>>>>> [each] contain more than 100 elements.
>>>> Correct (but not the same 100 elements).
>>>
>>> Wrong. The endsegments get emptier step by step.
>>
>> Some particular symbols may be missing or shifted, but the endsegment
>> is still inductive and its size is Aleph_zero.
>
> And remains so for every definable"CLOWN SHOES" endsegment. Every definable "CLOWN SHOES" endsegment is infinite and has an infinite intersection with all its predecessors.
>
> And if we consider only definale "CLOWN SHOES" (= infinite) endsegments, then necessarily all are predecessors of an infinite endsegment and therefore all definable "CLOWN SHOES" endsegments have an infinite intersection.
>
> Regards, WM

4 "CLOWN SHOES" in that one.

On Saturday, July 9, 2022 at 4:09:19 PM UTC+2, sergi o wrote:
> On 7/9/2022 7:41 AM, WM wrote:
> > William schrieb am Freitag, 8. Juli 2022 um 23:36:20 UTC+2:
> >> On Friday, July 8, 2022 at 6:05:23 PM UTC-3, WM wrote:
> >>> William schrieb am Freitag, 8. Juli 2022 um 17:08:33 UTC+2:
> >>>>> ... their intersection contains more than 100 elements.
> >>>> Nope does not follow and is not true.
> >>> Try to think better. If all contain more than 100 elements, then all contain the same 100 elements.
> >> Nope not true. All contain more than 100 elements but there is no set of 100 elements common to all endsegments.
> >
> > For definable "CLOWN SHOES" elements it is certainly true. Endsegments are terms of a decreasing sequence. Compare with a decreasing sequence of sets of only definable "CLOWN SHOES" elements. There all set with more than 100 elements have an intersection with more than 100 elements. Try to find a counterexample with definable "CLOWN SHOES" elemehts. Fail.
> >
> > If there are dark "CLOWN SHOES" elements this logic may be violated. But only if the elements are really dark "CLOWN SHOES".
> >>
> >> 1, For any finite set R of endsegments the is a set S(R) of 100 natural numbers such that S(R) is contained in each element of R
> >
> > For every set of definable "CLOWN SHOES" elements.
> >
> >> 2. There is a set S of 100 natural numbers such that for any finite set R of endsegments, S is contained in each element of R.
> >>
> >> 1. is true, 2. is false.
> >
> > 2. is obviously true. What should the endsegments consist of, if not of infinitely many numbers, 100 of which are in all of them?
> >
> > What you address here is the potential "CLOWN SHOES" infinity of natural numbers. You cannot define 100 natural numbers in all endsegments because for every defined "CLOWN SHOES" set of numbers you can find an endsegment not containing them. But that is a matter of potential "CLOWN SHOES" infinity only because you can for "given" (i.e. defined "CLOWN SHOES") endsegments always prove that 100 numbers are contained in all of them.
> >
> > Fact is: All definable "CLOWN SHOES" endsegments have an infinite intersection. You cannot define "CLOWN SHOES" that intersection because every defined "CLOWN SHOES" number is deleted in some endsegment. Nevertheless its infinite contents is in all its predecessors.
> >
> > And if only infinite endsegments are considered, then all endsegments are predecessors of an infinite endsegment and therefore sharing infinitely many numbers.
> >
> > The intersectiopn of infinite endsegments is infinite.

> I put "CLOWN SHOES" next to each non-math term, that are WM's special words he relies upon to confuse readers.
>
> their are 13 "CLOWN SHOES" above. Try not to use these terms, as they are WM's made up terms, and it signals to WM, that he has put the bag on ya.

Very good idea. Though I'd propose to REPLACE that meaningless words with "CLOWN SHOES", "CLOWN SHOE" and/or "CLOWN SHOED".

This way we would get:

> > Fact is: All CLOWN SHOED endsegments have an infinite intersection. You cannot CLOWN SHOE that intersection because every CLOWN SHOED number is CLOWN SHOED in some endsegment.

Yeah, makes much more sense than WM's original version!

On Saturday, July 9, 2022 at 4:09:19 PM UTC+2, sergi o wrote:
> On 7/9/2022 7:41 AM, WM wrote:
>
> I put "CLOWN SHOES" next to each non-math term, that are WM's special words he relies upon to confuse readers.
>
> their are 13 "CLOWN SHOES" above. Try not to use these terms, as they are WM's made up terms, and it signals to WM, that he has put the bag on ya.

Very good idea. Though I'd propose to REPLACE these meaningless words with "CLOWN SHOES", "CLOWN SHOE" and/or "CLOWN SHOED".

This way we would get:

> > Fact is: All CLOWN SHOED endsegments have an infinite intersection. You cannot CLOWN SHOE that intersection because every CLOWN SHOED number is CLOWN SHOED in some endsegment.

Yeah, makes much more sense than WM's original version!

On Saturday, July 9, 2022 at 2:41:57 PM UTC+2, WM wrote:

> The intersectiopn of [all] endsegments is infinite.

No, it's empty.

On Saturday, July 9, 2022 at 3:54:47 PM UTC+2, Fritz Feldhase wrote:
> On Saturday, July 9, 2022 at 2:41:57 PM UTC+2, WM wrote:
> > William schrieb am Freitag, 8. Juli 2022 um 23:36:20 UTC+2:
> > >
> > > 2. There is a set S of 100 natural numbers such that for any finite set R of endsegments, S is contained in each element of R.
> > >
> > > 2. is false.
> > >
> > 2. is obviously true.
> >
> Really?

Why are so bad at math, Mückenheim?

Hint:

> Let S = {n_1, n_2, n_3, ..., n_100} with n_i e IN for all i e {1, ..., 100} and n_i =/= n_j for all i, j e {1, ..., 100}, i =/= j.
>
> Then S has a largest element max(S). Let n0 = max(S).
>
> Now consider the following finite set of endsegments { {n0+1, n0+2, n0+3, ...} }. Is S contained in {n0+1, n0+2, n0+3, ...}?
>
> Obviously not. Hence 2 is false.

Got that, dumbo?

On Sunday, 3 July 2022 at 17:30:04 UTC+3, WM wrote:
> schrieb am Sonntag, 3. Juli 2022 um 09:15:33 UTC+2:
> > lördag 2 juli 2022 kl. 18:37:32 UTC+1 skrev WM:
> > > schrieb am Samstag, 2. Juli 2022 um 19:31:34 UTC+2:
> > > > lördag 2 juli 2022 kl. 17:25:15 UTC+1 skrev WM:
> > > > >
> > > > > For all definable numbers we have the intersection of their endsegments
> > > > > |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
> > > > false, you keep assuming this but it is false.
> > > I defined ℕ_def by this property.
> > Nope, you define N_def by N_def={n e N:|∩{E(1), E(2), ..., E(k)}|=ℵ₀}
> >
> > Which does not result in |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
> > It results in |∩{E(k) : k ∈ ℕ_def}| = 0
>
> No, I stop when the intersection is ℵ₀
>
> > > > > The endsegments can only shrink by on number per endsegment
> > > > > ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
> > > > >
> > > > > What natural numbers reduce the intersection from ℵ₀ to zero?
> > > > there is no single specicifc one,
>
> There are numbers which cannot appear as single specific ones. I call them dark numbers.

"Dark fractions" are those that haven't yet been realised/mentioned in a community of those who count fractions.

The Greek word ρητός means "utterable" or "speakable" and that's why modern Greeks incorrectly classified numbers into two groups: ρητοί and άρρητοι, that is, those which are "speakable" and those which are ineffable (not speakable). If modern Greeks were even a shadow of their ancestors, they woud have realised that similar phrases are "named" (ρητοί) and "unnamed" (άρρητοι)..

Those which are not speakable are so because they have no NAMES.

A NUMBER is a NAME given to a MEASURE that DESCRIBES a MAGNITUDE or SIZE.

A NUMBER is a NAME given to a MEASURE of a RATIO.

https://www.academia.edu/44820487/Discovering_the_concept_of_number_a_personal_journey

> But we can also say no single specicifc numbers.
>
> Regards, WM

On Sunday, 10 July 2022 at 09:46:09 UTC+3, Eram semper recta wrote:
> On Sunday, 3 July 2022 at 17:30:04 UTC+3, WM wrote:
> > schrieb am Sonntag, 3. Juli 2022 um 09:15:33 UTC+2:
> > > lördag 2 juli 2022 kl. 18:37:32 UTC+1 skrev WM:
> > > > schrieb am Samstag, 2. Juli 2022 um 19:31:34 UTC+2:
> > > > > lördag 2 juli 2022 kl. 17:25:15 UTC+1 skrev WM:
> > > > > >
> > > > > > For all definable numbers we have the intersection of their endsegments
> > > > > > |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
> > > > > false, you keep assuming this but it is false.
> > > > I defined ℕ_def by this property.
> > > Nope, you define N_def by N_def={n e N:|∩{E(1), E(2), ..., E(k)}|=ℵ₀}
> > >
> > > Which does not result in |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
> > > It results in |∩{E(k) : k ∈ ℕ_def}| = 0
> >
> > No, I stop when the intersection is ℵ₀
> >
> > > > > > The endsegments can only shrink by on number per endsegment
> > > > > > ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
> > > > > >
> > > > > > What natural numbers reduce the intersection from ℵ₀ to zero?
> > > > > there is no single specicifc one,
> >
> > There are numbers which cannot appear as single specific ones. I call them dark numbers.
>
> "Dark fractions" are those that haven't yet been realised/mentioned in a community of those who count fractions.
>
> The Greek word ρητός means "utterable" or "speakable" and that's why modern Greeks incorrectly classified numbers into two groups: ρητοί and άρρητοι, that is, those which are "speakable" and those which are ineffable (not speakable). If modern Greeks were even a shadow of their ancestors, they woud have realised that similar phrases are "named" (ρητοί) and "unnamed" (άρρητοι).

To be named is equivalent to have been measured.

To be unnamed is equivalent to have not been measured or to have no measure..

>
> Those which are not speakable are so because they have no NAMES.
>
> A NUMBER is a NAME given to a MEASURE that DESCRIBES a MAGNITUDE or SIZE.
>
> A NUMBER is a NAME given to a MEASURE of a RATIO.
>
> https://www.academia.edu/44820487/Discovering_the_concept_of_number_a_personal_journey
>
> > But we can also say no single specicifc numbers.
> >
> > Regards, WM

On Sunday, 3 July 2022 at 17:08:01 UTC+3, WM wrote:
> William schrieb am Samstag, 2. Juli 2022 um 20:14:21 UTC+2:
> > On Saturday, July 2, 2022 at 3:02:01 PM UTC-3, WM wrote:
> > > William schrieb am Samstag, 2. Juli 2022 um 19:47:26 UTC+2:
> > >>. You do not say that something is true for an infinite set.
> > > I say that it is true for [each] natural number, none of which is infinite.
> >
> > Correct. You say that something is true for the elements of |N.
>
> For all elements of ℕ!
>
> > You do not say that something is true for |N
>
> Of course not. ℕ does not appear in
> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀

You shouldn't even concede ℵ₀ to these baboons. That you are showing them their illogic by allowing the bullshit concept ℵ₀ is in my opinion being far too generous to these Cantor cranks. Dogma is an incomprehensible thing - your valiant attempts are like the proverbial pearls thrown to swine. Leave them in their ignorance!!!!

> ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
> ∩{E(k) : k ∈ ℕ} = { }
> as an element of ℕ.
>
> Regards, WM

Fritz Feldhase schrieb am Samstag, 9. Juli 2022 um 15:54:47 UTC+2:
> On Saturday, July 9, 2022 at 2:41:57 PM UTC+2, WM wrote:
> > William schrieb am Freitag, 8. Juli 2022 um 23:36:20 UTC+2:
> > >
> > > 2. There is a set S of 100 natural numbers such that for any finite set R of endsegments, S is contained in each element of R.
> > >
> > > 2. is false.
> > >
> > 2. is obviously true.
> Really?
>
> Let S = {n_1, n_2, ..., n_100} with n_i e IN for all i e {1, ..., n} and n_i =/= n_j for all i, j e {1, ..., 100}, i =/= j.
>
> Then S has a largest element max(S). Let n0 = max(S).
>
> Now consider the following finite set of endsegments { (n0+1, n0+2, n0+3, ...) }. Is S contained in (n0+1, n0+2, n0+3, ...)?
>
> Obviously not. Hence 2 is false.

I did not say that 100 elements could be fixed. But all elements of your second example are also in your first example, that is infinitely many and that is larger than 100. This is true for all infinite endsegments.

The intersection of a sequence of sets can be empty although all sets are infinite. The intersection of an inclusion monotonic sequence of sets is the least set. Otherwise element would be missing which are *not* not in any of the sets.

If the least set is infinite, then the intersection is infinite. To claim that the intersection is empty because there is no last endsegment belongs to the most stupid statements ever uttered. It is almost as stupid as the claim that the function 10 + 1/n has the limit zero because for every term there is a smaller term.

Regards, WM

William schrieb am Samstag, 9. Juli 2022 um 16:07:08 UTC+2:
> On Saturday, July 9, 2022 at 9:41:57 AM UTC-3, WM wrote:
>
>
> >...
> > Fact is: All definable endsegments have an infinite intersection.
> By "definable" you mean a finite So 1 is true.

Of course. Who could ever have defined infinitely many individuals?
>
> 1.For any definable set R of endsegments there is a set S(R) of 100 natural numbers such that S(R) is contained in each element of R
>
> 2. There is a set S of 100 natural numbers such that for any definable set R of endsegments, S is contained in each element of R.

If all endsegments are infinite, then all contain the same infinite set including a subset of 200 elements.
>
> > ... You cannot define [S] .
>
> So 2. is false. ( if S exists [and, like any set, does not change] then S is definable).

The intersection of an inclusion monotonic sequence of sets is the least set. Otherwise element would be missing which are *not* not in any of the sets.

For infinite sets the intersection is infinite.

If the elements cannot be defined, then they are dark.

Regards, WM

Fritz Feldhase schrieb am Samstag, 9. Juli 2022 um 17:33:29 UTC+2:
> On Saturday, July 9, 2022 at 2:41:57 PM UTC+2, WM wrote:
> > The intersectiopn of infinite endsegments is infinite.
>
> No, it's empty.

The intersection of a sequence of sets can be empty although all sets are infinite. The intersection of an inclusion monotonic sequence of sets is the least set. Otherwise element would be missing which are *not* not in any of the sets.

Regards, WM

on 7/10/2022, WM supposed :
> Fritz Feldhase schrieb am Samstag, 9. Juli 2022 um 17:33:29 UTC+2:
>> On Saturday, July 9, 2022 at 2:41:57 PM UTC+2, WM wrote:
>>> The intersectiopn of infinite endsegments is infinite.
>>
>> No, it's empty.
>
> The intersection of a sequence of sets can be empty although all sets are
> infinite. The intersection of an inclusion monotonic sequence of sets is the
> least set.

What is 'inclusion monotonic' about the sequence of FISON complements?
Are you counting backwards from infinity?

Eram semper recta schrieb am Sonntag, 10. Juli 2022 um 12:54:00 UTC+2:
> On Sunday, 3 July 2022 at 17:08:01 UTC+3, WM wrote:

> > Of course not. ℕ does not appear in
> > |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
> You shouldn't even concede ℵ₀ to these baboons. That you are showing them their illogic by allowing the bullshit concept ℵ₀ is in my opinion being far too generous to these Cantor cranks.

In order to contradict Cantor's results I have to apply actual infinity.

Regards, WM

On Sunday, July 10, 2022 at 10:25:06 AM UTC-3, WM wrote:

>The intersection of an inclusion monotonic sequence of sets is the least set.

iff there is a least set. If there is no last element there is no least set.

--
William Hughes

On Sunday, July 10, 2022 at 10:30:42 AM UTC-3, WM wrote:
> William schrieb am Samstag, 9. Juli 2022 um 16:07:08 UTC+2:
> > On Saturday, July 9, 2022 at 9:41:57 AM UTC-3, WM wrote:
> >
> >
> > >...
> > > Fact is: All definable endsegments have an infinite intersection.
> > By "definable" you mean a finite So 1 is true.
> Of course. Who could ever have defined infinitely many individuals?
> >
> > 1.For any definable set R of endsegments there is a set S(R) of 100 natural numbers such that S(R) is contained in each element of R
> >
> > 2. There is a set S of 100 natural numbers such that for any definable set R of endsegments, S is contained in each element of R.
> If all endsegments are infinite, then all contain the same infinite set including a subset of 200 elements.

Nope, does not follow and is not true

> >
> > > ... You cannot define [S] .
> >
> > So 2. is false. ( if S exists [and, like any set, does not change] then S is definable).
> The intersection of an inclusion monotonic sequence of sets is the least set.

iff there is a least set. If there is no last element there is no least set. The fact that there is a least cardinality does not imply there is a least set.

> If the elements cannot be defined, then they are dark.

|N is a Peano set and thus does not contain "dark" elements

--
William Hughes

On Sunday, July 10, 2022 at 3:25:06 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Samstag, 9. Juli 2022 um 15:54:47 UTC+2:
> > On Saturday, July 9, 2022 at 2:41:57 PM UTC+2, WM wrote:
> > > William schrieb am Freitag, 8. Juli 2022 um 23:36:20 UTC+2:
> > > >
> > > > 2. There is a set S of 100 natural numbers such that for any finite set R of endsegments, S is contained in each element of R.
> > > >
> > > > 2. is false.
> > > >
> > > 2. is obviously true.
> > >
> > Really?
> >
> > Let S = {n_1, n_2, ..., n_100} with n_i e IN for all i e {1, ..., 100} and n_i =/= n_j for all i, j e {1, ..., 100}, i =/= j.
> >
> > Then S has a largest element max(S). Let n0 = max(S).
> >
> > Now consider the following finite set of endsegments { {n0+1, n0+2, n0+3, ...} }. Is S contained in {n0+1, n0+2, n0+3, ...}?
> >
> > Obviously not. Hence 2 is false.
> >
> I <bla bla bla>

Your claim was: "2. is obviously true."

This claim is FALSE, since 2. is (provable) false.

Was gibt's da noch lange zu "diskutieren", Mückenheim? Du hast einfach mal wieder Scheiße behauptet!

On Sunday, July 10, 2022 at 3:32:27 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Samstag, 9. Juli 2022 um 17:33:29 UTC+2:
> > On Saturday, July 9, 2022 at 2:41:57 PM UTC+2, WM wrote nonsense:
> > >
> > > The intersectiopn of infinite endsegments is infinite.
> > >
> > No, it's empty.
> >
> The intersection of [the therms of] a sequence of sets can be empty although all [terms] are infinite. The intersection of [the therms of] an inclusion monotonic sequence of sets is the least [term]

IF THERE IS A LEAST TERM, dumbo!

In the set of endsegments {E(n) : n e IN} THERE IS NO least element. There is no least endsegment.

Wie dumm kann man eigentlich sein, Mückenheim?

> Otherwise

Yeah whatever, Mückenheim.

On Sunday, 10 July 2022 at 16:38:56 UTC+3, WM wrote:
> Eram semper recta schrieb am Sonntag, 10. Juli 2022 um 12:54:00 UTC+2:
> > On Sunday, 3 July 2022 at 17:08:01 UTC+3, WM wrote:
>
> > > Of course not. ℕ does not appear in
> > > |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
> > You shouldn't even concede ℵ₀ to these baboons. That you are showing them their illogic by allowing the bullshit concept ℵ₀ is in my opinion being far too generous to these Cantor cranks.
> In order to contradict Cantor's results I have to apply actual infinity.

ℵ₀ is not a "result". It's a belief. You're not apply anything except faith. Chuckle.

>
> Regards, WM