On 2/25/2021 2:43 AM, Ganzhinterseher wrote:
> The intersection over a set {E(1), E(2), E(3), … E(n)}
> of endsegments E(k) = (k, k+1, k+2, ...) of ℕ is not empty:
>
> ∩{E(1), E(2), E(3), … E(n)} =/= { } (*)
>
> There are infinitely many endsegments E(n) which give infinite intersections when they appear at last position in (*). The intersection however does not depend on positions - a set has no order anyway. Since the position is irrelevant, the intersection of all of these endsegments is infinite and not empty also when they appear at not last position:
>
> ∩{E(n) | E(n) is satisfying (*)} =/= { } (**)
>
> However, it is impossible to find a natural number common to all E(n) satisfying (*).
>
> This is a contradiction, at least according to the conventional racist mathematics which does not tolerate contradictions. But knows about exclusion monotony.
>
> The only remedy is this: Although you cannot find any natural number in (**) this does not prove an empty intersection because all findable natural numbers have infinitely many dark, i.e., not findable successors.
>
> Regards, WM
>

A fucking war better not break out:

https://youtu.be/z0Og5MQycZg

On 5/10/2021 10:49 PM, Chris M. Thomasson wrote:
> On 2/25/2021 2:43 AM, Ganzhinterseher wrote:
>> The intersection over a set {E(1), E(2), E(3), … E(n)}
>> of endsegments E(k) = (k, k+1, k+2, ...) of ℕ is not empty:
>>
>> ∩{E(1), E(2), E(3), … E(n)} =/= { }                             (*)
>>
>> There are infinitely many  endsegments E(n) which give infinite
>> intersections when they appear at last position in (*). The
>> intersection however does not depend on positions - a set has no order
>> anyway. Since the position is irrelevant, the intersection of all of
>> these endsegments is infinite and not empty also when they appear at
>> not last position:
>>
>> ∩{E(n) | E(n) is satisfying (*)} =/= { }                      (**)
>>
>> However, it is impossible to find a natural number common to all E(n)
>> satisfying (*).
>>
>> This is a contradiction, at least according to the conventional racist
>> mathematics which does not tolerate contradictions. But knows about
>> exclusion monotony.
>>
>> The only remedy is this: Although you cannot find any natural number
>> in (**) this does not prove an empty intersection because all findable
>> natural numbers have infinitely many dark, i.e., not findable successors.
>>
>> Regards, WM
>>
>
> A fucking war better not break out:
>
> https://youtu.be/z0Og5MQycZg

https://youtu.be/UVHDGMJMWwI