Talk:Heine-Borel Theorem/Real Line

I believe, by doing a little research, that the initial statement is wrong. It should be made to apply only to a closed real interval, not a general closed set in $\R$. Comments? --prime mover 03:46, 16 March 2012 (EDT)


 * Somehow, I do think that the theorem statement is true. I think that it is just that the proof doesn't prove the general case for a closed subspace of $\R^n$.


 * By the Heine-Borel Theorem (General Case), we have that a subset of $\R^n$ is compact if and only if it is complete and totally bounded (if my proof is valid).


 * Then we have that $\R^n$ is complete.


 * A subspace of a complete metric space is closed if and only if it is complete.


 * So a subspace of $\R^n$ is closed if and only if it is complete.


 * It is quite easy to show that a subspace of $\R^n$ is bounded if and only if it is totally bounded.


 * So we get the Heine-Borel Theorem (Special Case), which states that a subspace of $\R^n$ is compact if and only if it is closed and bounded.


 * (Anyway, sorry for the page changes.) Abcxyz 10:10, 16 March 2012 (EDT)


 * Also, the result that a closed real interval is sequentially compact is basically the Bolzano-Weierstrass Theorem. We have from Sequence of Implications of Metric Space Compactness Properties that a metric space is compact if and only if it is sequentially compact. So I was thinking that maybe the proof in this article could be included as a third proof of the Bolzano-Weierstrass Theorem. Comments? Abcxyz 10:20, 16 March 2012 (EDT)