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The Bolzano-Weierstrass theorem says that every bounded sequence in $\Bbb R^n$ contains a convergent subsequence. The proof in Wikipedia evidently doesn't go through for an infinite-dimensional space, and it seems to me that the theorem ought not to be true in general: there should be some metric in which $\langle1,0,0,0,\ldots\rangle, \langle0,1,0,0,\ldots\rangle, \langle0,0,1,0,\ldots\rangle, \ldots $ is bounded but fails to contain a convergent subsequence.

Let $M$ be a general metric space. What conditions on $M$ are necessary and sufficient for every bounded sequence of elements of $M$ to contain a convergent subsequence?

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The keyword here is "Bolzano-Weierstrass property." An easy example of a space which lacks this property is $(0, 1)$. (Note that $(0, 1)$ is homeomorphic but not bi-Lipschitz to $\mathbb{R}$ and that convergence is a topological notion but boundedness is a metric one.) –  Qiaochu Yuan Sep 25 '12 at 5:11
    
Thanks. Is there anything interesting one can say to distinguish the failure of $(0,1)$, which seems to be because of the "missing" endpoints, from the seemingly different failure of my infinite-sequence example in the post? Or conversely, is there a way to complete the space of the infinite-sequence example? –  MJD Sep 25 '12 at 5:14
    
Your question is equivalent to asking, "When is every closed bounded subspace of a metric space compact?" Notice that one necessary condition is the completeness of the metric space – but this is not enough, because the space of sequences with the $\ell^\infty$-norm is complete. –  Zhen Lin Sep 25 '12 at 5:15
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Note that if $\langle X,d\rangle$ is any metric, space there is a bounded metric $d_1$ on $X$ that generates the same topology. Thus, boundedness isn’t a topological property, but rather a metric property. The Bolzano-Weierstrass theorem is taking advantage of a very specific property of the usual metrics on $\Bbb R^n$. –  Brian M. Scott Sep 25 '12 at 8:02
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3 Answers

up vote 1 down vote accepted

The closest analogue of the Heine-Borel theorem in arbitrary metric spaces is that a subset is compact iff it's closed and totally bounded. But totally bounded sets pretty obviously have compact closure, being those which have finite covers by $\varepsilon$-balls for every $\varepsilon$, so this isn't much of an improvement. Anyway, it's easy to see that your example sequence in $\ell^2$, or whichever norm you prefer, isn't totally bounded since its elements are pairwise $\sqrt{2}$ apart, so that no $\varepsilon$-ball will cover more than one of them for $\varepsilon<\sqrt{2}$.

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I don't think your generalisation of the Heine-Borel is true. Totally boundedness is equivalent with every sequence having a Cauchy subsequence, which clearly with closedness is not enough to obtain compactness. The correct generalisation is that in arbitrary metric spaces a subset is compact iff it is complete and totally bounded. –  Thomas E. Jan 4 '13 at 7:13
    
Yes, you're right, I didn't consider the importance of the fact that the spaces for basic Heine-Borel are complete. –  Kevin Carlson Jan 6 '13 at 18:08
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A metric space is sequentially compact $\iff$ it has Bolzano Weierstrass property. And, for a metric space, compactness $\iff$ sequential compactness, and hence, the metric space should be compact for the property to hold.
And then, a metric space is compact if and only if it is complete and totally bounded. So, now for an arbitrary metric space, you must decided whether it is complete and it is totally bounded. I guess, the methods would differ from space to space and metric to metric.

Since you were talking about infnite-dimensional space, a closed unit ball in $R^{\infty}$ is not totally bounded.

Then there is another theorem (which I have not yet studied till now. Simmons has described it before, and said we will prove it later) which says that a Banach space is finite dimensional $\iff$ every bounded subspace is totally bounded. So, bounded subspaces of finite dimensional banach spaces are not the place to look for counterexamples.I do not know which other spaces are there to look then.

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The main thrust of the question regards how to decide in an arbitrary metric space whether closed & bounded implies compact-your proof doesn't address this. –  Kevin Carlson Sep 25 '12 at 5:37
    
@KevinCarlson Okay. If I say that total boundedness of a closed subspace of a complete metric space implies compactness and vice-versa. Will that address the question enough? –  Jayesh Badwaik Sep 25 '12 at 5:42
    
I guess not. So, basically, we want to find conditions for total boundedness and completeness from the metric of the space and the set of points? Is this true? –  Jayesh Badwaik Sep 25 '12 at 5:49
    
Oops, sorry that I just posted on total boundedness-I hadn't updated to see your comment. I think the OP would appreciate what you suggest, yes. –  Kevin Carlson Sep 25 '12 at 5:51
    
@KevinCarlson Its okay. I do not mind. –  Jayesh Badwaik Sep 25 '12 at 5:54
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An important example of an infinite dimensional space with this propeerty is the space of holomorphic functions, say on an open subset of the complex plane, regarded as a Fréchet space with the topology of compact convergence. This is essentially Montel's theorem and locally convex spaces in which bounded subsets are relatively compact are called Montel spaces. Many of the important locally convex spaces of analysis are, if they are not Banach spaces, Montel spaces, e.g., spaces of test functions or distributions. Most standard texts on locally convex spaces contain extensive sections on Montel spaces and their refinements (Fréchet-Schwartz spaces, Silva spaces, Fréchet nuclear spaces and so on).

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