If $V$ is a liniar space and $dimV=n$ finite, then the set $ S=\{x \in V;\|x\|_1=1\} $ is bounded and closed .And from Heine Borel theorem $S$ will be compact.$$ $$ My question: Is there a way to show a sequence from $S$ have a convergent subsequence? $$$$ If not how to proof $S$ is closed and bounded?
2026-05-04 17:04:43.1777914283
Unit sphere is sequentially compact \ compact
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Suppose we have a sequence in S with no convergent subsequence in S. Then for each x in S there is a open neighborhood containing no points in the sequence except x. S is in the union of all these open neighborhoods. Because of S being compact, there exists a finite subcover. However at least one of these finite sets covering S has to contain infinite elements of the sequence in contradiction to our choice of the open neighborhoods.