Talk:Bolzano–Weierstrass theorem

Proof incomplete

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)[reply]

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)[reply]

Generalization to metric spaces

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)[reply]

That'd be useful ideed: the current formulation seems to imply it only holds for $\mathbb {R} ^{n}$ . Ceacy (talk) 13:27, 4 November 2014 (UTC)[reply]

Spelling

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)

Agree. Oleg Alexandrov 21:16, 6 May 2005 (UTC)[reply]

What's up with the weird dash? Shouldn't it be a hyphen? 67.164.12.169 07:17, 22 October 2005 (UTC)[reply]

No, the en dash is correct. See Dash#En dash. —Caesura ^(t) 19:10, 6 December 2005 (UTC)[reply]

Equivalent theorems

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)[reply]

Alternative Theorems

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)[reply]

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)[reply]

Applications to Economics

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)[reply]

Mathematical analysis has practical application..? --131.111.248.243 (talk) 13:25, 21 October 2011 (UTC)[reply]

Correctness of Theorem

I could be confused, but I think this page has a somewhat serious issue. The Heine-Borel theorem is that a set $A\subset \mathbb {R} ^{n}$ is compact if and only if it is closed and bounded. This is equivalent to what is written here. My understanding is that the Bolzano-Weierstrass theorem is much more general and deeper, that is, that a subset of a metric space is compact if and only if it is sequentially compact. Any thoughts here? If there are no arguments, I think I will update the page and try to include a proof. This could be very confusing for a beginning student in analysis. John 06:27, 23 October 2007 (UTC)[reply]

The Bolzano–Weierstrass theorem article states that:

A subset A of Rⁿ is sequentially compact if and only if it is both closed and bounded.

So I guess you are referring to the theorem being generalized to an arbitrary metric space (with "bounded and closed" being replaced by "compact"). I would very much prefer that instead of rewriting the whole article from the more general view of metric spaces, you'd rather work on expanding the "Generalizations" section in the article. How would that sound? Oleg Alexandrov (talk) 03:17, 28 October 2007 (UTC)[reply]

Yeah, that sounds good. I will spend some time doing that when I have some time. John 04:27, 28 October 2007 (UTC)[reply]

Beginning of some edits

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)[reply]

I reviewed the article and did some minor cleanup. I think that more is necessary though.

Topology Expert (talk) 03:28, 23 November 2008 (UTC)[reply]

Assessment comment

The comment(s) below were originally left at Talk:Bolzano–Weierstrass theorem/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Please add useful comments here--Cronholm144 09:25, 22 May 2007 (UTC)[reply]

Article is 'half B-class' so I left it at C. Improvement necessary:

More references

More important content (lacks important results)

Topology Expert (talk) 03:33, 23 November 2008 (UTC)[reply]

Last edited at 03:33, 23 November 2008 (UTC). Substituted at 01:49, 5 May 2016 (UTC)

Definition question

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)[reply]

A sequence as used in this branch of mathematics is, by definition, over the natural numbers. Pmokeefe (talk) 02:38, 3 September 2023 (UTC)[reply]

Notation of the general case is ambiguous

In the proof of the case for $\mathbb {R} ^{n}$ , $n$ is also used as the index of the sequence $\left(x_{n}\right)_{n=1}^{\infty }$ . This results in the misleading suggestion that the two are the same, especially in cases like " $x_{nn}$ ". One of these should be changed to a different index like $m$ . DÆS (talk) 18:25, 18 December 2024 (UTC)[reply]