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First time I ask a question here. I recently tried to prove the Bolzano-Weierstrass theorem by myself without using the nested intervals technique.

What I did is the following:

Let $(u_n)_n$ be a bounded sequence of real numbers. Then there exist $M, M'\in \mathbb R$, such that:
$$\forall n \in \mathbb N, M\le u_n\le M'.$$
Let $(u_{\phi(n)})_n$ be a subsequence of $(u_n)_n$. Then
$$\forall n \in \mathbb N, M\le u_{\phi(n)}\le M'.$$
We can then find a sequence $(u_{\phi(n+1)}-u_{\phi(n)})_n$ that is also bounded and such that $(u_{\phi(n+1)}-u_{\phi(n)})_n\ge0$ (or $(u_{\phi(n+1)}-u_{\phi(n)})_n\le0$) for every natural number $n$. It means that the subsequence $(u_{\phi(n)})_n$ is non-decreasing (or non-increasing), while bounded, and therefore is convergent.
I would be very glad to read some remarks about this attempt, thank you.

Grégoire O.
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  • "We can then find a sequence $(u_{\phi (n + 1)} - u_{\phi (n)})n$ that is also bounded and such that $(u{\phi (n + 1)} - u_{\phi (n)})_n \ge 0$... for every natural number $n$"... How do you justify that such a sequence exists? – K. Jiang Jul 30 '24 at 22:54
  • You will have to explain how you construct this (non-)decreasing sequence. – amsmath Jul 30 '24 at 22:58
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    To construct such a subsequence the concept of a peak is useful. We say $u_k$ is a peak if $u_k\geq u_n$ for all $n\geq k$ (a peak $u_k$ is larger than all the following values in the sequence). A sequence can either have an infinite number of peaks (show that this gives you an decreasing subsequence) or it can have a finite number of peaks in which you can go to the right of the last peak and iteratively construct an increasing subsequence. – Winther Jul 30 '24 at 23:11
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    Replying to K. Jiang, I got your point, thank you. – Grégoire O. Jul 31 '24 at 05:59
  • Winther, so what I must say is that a sequence is either monotonic, in which case it is obvious that there is a monotonic subsequence, or it is varying with peaks, in which case we can a subsequence defined by the set of peaks of the main sequence, to get a monotonic subsequence. – Grégoire O. Jul 31 '24 at 06:07
  • Every sequence has a monotone subsequence, see for example https://math.stackexchange.com/questions/716461/proof-verification-every-sequence-in-bbb-r-contains-a-monotone-sub-sequence so your idea works. – Steen82 Aug 01 '24 at 01:11

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