# Show that a subspace $X$ of the Euclidean space $\Bbb{R}^n$ is compact if and only if any sequence of elements of $X$ has a converging subsequence.

Remark: this statement holds in the considerably greater generality of any metric space but the proof of this more general result is quite involved.

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The answer to this greatly depends on how you defined compactness. Also I believe it is bad form to put the whole question in the title and not in the body of the post, and to put questions in an imperative form ("Show that...": why would I show it?). If it is homework, tag it as such. Finally, what have you tried? – Najib Idrissi Mar 7 '12 at 13:50
If your definition of compactness is closed and bounded, the result is very easy, and you should be able to show some progress. If your definition is the general one $-$ a set is compact iff every open cover of it has a finite subcover $-$ then the result isn’t much easier in $\Bbb{R}^n$ than in arbitrary metric spaces. – Brian M. Scott Mar 7 '12 at 13:55
"Any" is a potentially ambiguous word. If you mean "every", you should say so. "Prove that any A is B" can mean: pick any A and prove that it is B. But "Any A is B", with no further context, could mean every A is B. So "Prove that any A is B" can be read in two different ways. Changing "any" to "every" in cases where "every" is meant will remove the ambiguity. – Michael Hardy Mar 7 '12 at 17:18
@Michael: While I would prefer every, in my variety of English the title can only have the intended meaning, and any A is B cannot mean some A is B. – Brian M. Scott Mar 8 '12 at 5:28
@BrianM.Scott : "If anyone knows the answers, Dr. Xmith does." Surely that doesn't mean "If everyone knows...."? – Michael Hardy Mar 8 '12 at 13:45