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Am Wed, 09 Oct 2024 18:56:29 +0200 schrieb WM:Impossible. Infinite endsegments contain almost all numbers. Therefore it is not possible that almost all numbers are used as indices.Am 09.10.2024 um 18:12 schrieb Alan Mackenzie:Especially since both all segments are infinite,WM <wolfgang.mueckenheim@tha.de> wrote:>
and there are infinitely many of them.
The intersection of all endsegments containing at least 3 numbers with just these endsegments contains 3 numbers.You should really be more careful with your phrasing. Intersection withThe completion of the above sets does not change the principle:But the terms of the sequence do. Here is a simple finite example:Note: The shrinking endsegments cannot acquire new numbers.An end segment is what it is. It doesn't change.
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
{2, 3, 4, 5, 6, 7, 8, 9, 10}
{3, 4, 5, 6, 7, 8, 9, 10}
{4, 5, 6, 7, 8, 9, 10}
{5, 6, 7, 8, 9, 10}
{6, 7, 8, 9, 10}
{7, 8, 9, 10}
{8, 9, 10}
{9, 10}
{10}
{ } .
Non-empty inclusion-monotonic sets like infinite endsegments have a
non-empty intersection. All endsegments have an empty intersection.
what?
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