In a metric space, a subset is called compact if every sequence in the space contains a convergent subsequence (In a more general setting this is called “sequentially compact,” and compactness refers to a set where every open cover has a finite subcover, but in the case of a metric space these two definitions are equivalent.) As it turns out, every compact subset of a metric space will be both closed and bounded.

Suppose is a compact subset of some metric space . Suppose that were not bounded, then for a fixed we could find a sequence such that for each , . Note that such a sequence has no convergent subsequence. Since unboundedness implies a set can’t be compact, every compact set must be bounded. For closedness, let . There exists a sequence in , such that . By compactness of , this sequence has a convergent subsequence which converges to an element of . However, every subsequence of a convergent sequence must have the same limit as the “original” sequence, so , and this implies , so and is closed.

In a finite dimensional normed space, this becomes boundedness and closedness are no longer simply necessary conditions for compactness, but they are in fact sufficient conditions. That is, in a finite dimensional normed space, a set is compact if and only if it is both closed and bounded.

We’ve already show that compactness implies closed and bounded, so we now show the converse. Suppose is an n-dimensional normed space with basis vectors . Let and suppose is both closed and bounded. Let for any sequence in . Since is bounded, there exists a such that for each we have . We know (from an earlier lemma) there exists a such that

This tells us for any fixed , that is bounded, so it must have a convergent subsequence. Call the limit of this subsequence . We now have that must have a convergent subsequence with limit . Note this limit is in , so is compact.

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