Sujet : Re: How many different unit fractions are lessorequal than all unit fractions? (infinitary)
De : wolfgang.mueckenheim (at) *nospam* tha.de (WM)
Groupes : sci.mathDate : 10. Oct 2024, 19:18:48
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <ve95q7$38c74$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
User-Agent : Mozilla Thunderbird
On 09.10.2024 19:47, joes wrote:
Am Wed, 09 Oct 2024 18:56:29 +0200 schrieb WM:
Am 09.10.2024 um 18:12 schrieb Alan Mackenzie:
WM <wolfgang.mueckenheim@tha.de> wrote:
>
Especially since both all segments are infinite,
and there are infinitely many of them.
Impossible. Infinite endsegments contain almost all numbers. Therefore it is not possible that almost all numbers are used as indices.
Note: The shrinking endsegments cannot acquire new numbers.
An end segment is what it is. It doesn't change.
But the terms of the sequence do. Here is a simple finite example:
{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}
{ } .
The completion of the above sets does not change the principle:
Non-empty inclusion-monotonic sets like infinite endsegments have a
non-empty intersection. All endsegments have an empty intersection.
You should really be more careful with your phrasing. Intersection with
what?
The intersection of all endsegments containing at least 3 numbers with just these endsegments contains 3 numbers.
The intersection of all endsegments with all endsegments is empty.
Regards, WM