Talk:Bolzano–Weierstrass theorem

message: The proof of the lemma in the case n = 1 is not complete: it assumes that every bounded real sequence has a "peak". This is not true – for example, the bounded sequence \left(x_n\right)_{n=1}^\infty where a_n=\frac n{n+1} has no "peak". I suggest the following.
Let a=\sup_{n\in\mathbb N}\{x_n\}. (This exists because the sequence is bounded.) Let n_1=1 and consider x_{n_1}=x_1. Either x_1=a or x_1 < a. In the latter case, there exists n_2 such that x_1 < x_{n_2}\leq a. Clearly n_2 > 1 = n_1. Choose n_2 to be as small as possible. Now either x_{n_2}=a or else x_{n_2} < x_{n_3}\leq a for some n_3. By minimality of n_2, n_3 > n_2. Again take n_3 to be as small as possible.
Continuing the process, we get one of two eventualities: either x_{n_r}=a for some n_r, or no such n_r exists. In the first case, n_r is a "peak" as defined in the article, and the rest of the proof can proceed as in the article. In the latter case, we have x_{n_1} < x_{n_2} < x_{n_3} < \cdots with n_1 < n_2 < n_3 < \cdots, which is a monotone increasing subsequence as desired.
SophieAthena (talk) 13:53, 6 November 2012 (UTC)
replies: text: In the proof of the lemma, let N=0 when there is no peak, which appears to be handled correctly in the rest of the proof of the lemma.  I edited the article to make the "no peak" case explicit Pmokeefe (talk) 15:47, 3 September 2023 (UTC)
thread_title: Proof incomplete

message: Shouldn't the Theorem's generalization to metric spaces be included in the article?
In a compact metric space (S,d), every infinite set included in S admits at least an accumulation (limit) point. Or in a compact metric space, every sequence admits a convergent subsequence.  —Preceding unsigned comment added by Hearth (talk • contribs) 01:58, 24 April 2008 (UTC)
thread_title: Generalization to metric spaces

message: Weierstrass/Weierstraß
It is not really my buisness, but I think in English it should be spelled Weierstrass, you know, all these books, and everything. Tosha 07:35, 23 Feb 2005 (UTC)
replies: text: Agree. Oleg Alexandrov 21:16, 6 May 2005 (UTC)
thread_title: Spelling

message: Regarding this edit: What does it mean for theorems to be equivalent? I understand what it means that propositions A and B are equivalent (namely, A implies B and B implies A), but that does not make sense for theorems which are basically tautologies. -- Jitse Niesen (talk) 23:45, 8 November 2006 (UTC)
thread_title: Equivalent theorems

message: In my real analysis class, I had two Bolzano-Weierstrass theorems:
one for sets, and one (as is here) for sequences
Is there somewhere where the sets one is taken care of?
Peter Stalin 18:21, 27 March 2007 (UTC)
replies: text: I'm not exactly sure what you are asking (but I'm not that smart, so its probably my fault)? What book did you use (if you have a page reference, that would be awesome)?  Are you thinking of this formulation?  Smmurphy(Talk) 18:09, 29 May 2007 (UTC)
thread_title:  Alternative Theorems 

message: Hi.  While the application to economics listed here is interesting, it makes use of technical language that seems out of place in this article.  If we had a nice list of a few of the many applications of this theorem in Mathematics, it would be a lot better.  We could then include this application as one example.  Regardless, this little section needs rewritten to be more understandable by someone with more knowledge of economics than I. Grokmoo 18:24, 13 July 2007 (UTC)
thread_title:  Applications to Economics 

message: I removed a short section on generalizations, because neither of the results discussed in this section was in fact a generalization of Bolzano-Weierstrass.  I will be making some further edits in the near future, including some content-related additions.  Plclark (talk) 06:26, 21 September 2008 (UTC)
thread_title:  Beginning of some edits 

message: Substituted at 01:49, 5 May 2016 (UTC)
thread_title: Assessment comment

message: In the definition "The theorem states that each bounded sequence in R^n has a convergent subsequence", should it only be for bounded infinite sequences, since it doesn't seem to hold for finite sequences? Nikolaih☎️📖 02:26, 11 May 2021 (UTC)
replies: text: A sequence as used in this branch of mathematics is, by definition, over the natural numbers. Pmokeefe (talk) 02:38, 3 September 2023 (UTC)
thread_title:  Definition question