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It's a very basic (may be a trivial) question but what is the exact difference, if any, between Heine Borel Theorem and Bolzano Weierstrass Theorem. It is true that one (Heine Borel) can be proved from another (Bolzano Weierstrass ).

Heine Borel Theorem: Subspace of $\mathbb{R}^n$ is compact iff it is closed and bounded.

Bolzano Weierstrass Theorem: Every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence.

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Could you explain a bit more clearly just what you’re asking? The theorems very obviously say different things. –  Brian M. Scott Jan 11 '13 at 18:35

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One answer is that the Bolzano-Weierstrass theorem says that every closed, bounded set in $\Bbb R^n$ is sequentially compact, while the Heine-Borel theorem says that every closed, bounded set in $\Bbb R^n$ is compact. (The Heine-Borel theorem also asserts the converse, of course.)

In general the notions of compactness and sequential compactness are distinct. Here is an example (with proof) of a compact Hausdorff space that is not sequentially compact, and here, also with proof, is an example of a sequentially compact Hausdorff space that is not compact. However, in metric spaces the two notions of compactness coincide, so in $\Bbb R^n$ the Bolzano-Weierstrass theorem can be thought of as one direction of the Heine-Borel theorem.

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Oh, it seems that I lacked to see the difference between sequentially compact and compact. A few sources I checked seemed to imply that they are equivalent but without clarification that they meant "sequentially". Thank You. –  007resu Jan 11 '13 at 18:52
    
@Freddy: You’re welcome. –  Brian M. Scott Jan 11 '13 at 18:53

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