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On 09.01.2025 21:17, joes wrote:...for different cases. There is no empty segment, each is infinite.Am Thu, 09 Jan 2025 19:25:19 +0100 schrieb WM:Loss of all numbers is proven by the empty intersection.Losing all numbers but keeping infinitely many is impossible inThis case doesn't occur.
inclusion-monotonic sequences.
Keeping infinitely many is poved by Fritsche.
You may have noticed that every segment is different.They remain infinite. But infinitely many endsegments require allIf all endsegments remain infinite, we have aNo, they are subsets of the same cardinality. There is no
contradiction.
contradiction.
natnumbers as indices. What makes up their infinite content?
Untrue. The sequence is, unfathomably, infinite.Then not all natnumbers are outside of content and inside of the set ofNo term of the sequence is empty, if you mean that.In the sequence of end.segments of ℕ there is no number which emptiesThen there cannot exist a sequence of endsegments obeying ∀k ∈ ℕ:
an infinite set to a finite set.
E(k+1) = E(k) \ {k+1} for all k ∈ ℕ and getting empty.
indices.
Contradiction. There are inf. many.The endsegment E(n) loses its element n+1 ad becomes E(n+1).WDYM "become"? There is no point at which all naturals would be countedand there is no number which is in common with all its endsegments.Therefore all numbers get lost from the content and become indices.
- N being infinite.
>Then there are only finitely many indices.Huh? No. Then not all numbers would be "indices".ℕ has only infinite endsegments.Then it has only finitely many, because not all numbers get lost from
the content.
omega is not an element of N.What element of ℕ does not become an index?There is no such endsegment.The intersection of all (infinite) end.segments of ℕ is empty.What is the content if all elements of ℕ have become indices?
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