The theorem states that, for the metric space , a subset is compact if and only if it is closed and bounded.
We will prove this theorem for first, then we show that if is compact and is compact, then is compact.
For any metric space , a compact subset is bounded, i.e there exists a real number , such that for any , . Suppose not, then for any , and , is not empty. Consider the open cover , then it has no finite sub-cover.
Hence, we have proven that, compactness implies closed and bounded.
The converse in general is not true. Consider any infinite set with discrete metric, . Then, is closed and bounded, but not compact, since , and are open sets.
It is true in .
Proposition Any closed interval is compact. Let be an open cover of . Suppose there is no finite subcover, then we can divide the interval in half and find out which half is causing the trouble. Let , let be either or , that does not admit finite subcover. Repeating this process, we get a nested sequence of closed intervals , the diameter . Let , then is a bounded increasing sequence, and is a bounded decreasing sequence. We have a limit . But is counted in certain , and there is a , such that . Hence, if , we have , contradicting with does not admit finite subcover.
Corollary Any closed bounded subset in is compact. Since it is a closed subset, and contained in a compact subset for large enough.
Proposition If is a collection of compact sets, such that any finite subcollection of them has non-empty intersection, then .
Proof: Suppose not, then . Choose any , then . By compactness, we have a finite sub-cover of This means get contradiction.
Proposition If is an infinite subset, then has a limit point in .
If not, then every point in has an open ball containing at most one point in . By compactness of , is covered by finitely many such , hence is also covered by such finitely many , which contradicts with being infinite.
Corollary If in is a sequence, then there exists a convergent subsequence. (Compactness implies sequential compactness.)
If the value set is finite, then there is a value which appears infinite many times in the sequence, and we can take the constant subsequence. If the value set is infinite, then we by proposition, there exists a , such that for any , is an infinite set, hence is infinite, thus is a subsequential limit of .