# Rudin's Proof of Bolzano-Weierstrass theorem

(a) If $$\{P_n\}$$ is a sequence in a compact metric space $$X$$, then some sub­sequence of $$\{P_n\}$$ converges to a point of $$X$$.
(b) Every bounded sequence in $$\mathbb R^k$$ contains a convergent subsequence.

Rudin proves (a), then argues for (b) as follows:

"(b) This follows from (a), since Theorem 2.41 implies that every bounded subset of $$\mathbb R^k$$ lies in a compact subset of $$\mathbb R^k$$," where Theorem 2.41 is the Heine-Borel Theorem.

But doesn't this require the set to be bounded AND closed? From what I can see Rudin makes no argument about that the sequence in $$\mathbb R^k$$ is closed.

• – hardmath Nov 8 '18 at 6:07
• Thank you. Will check this out – 51n84d Nov 8 '18 at 6:09
• – Lord Shark the Unknown Nov 8 '18 at 6:31

Note than (a) asks only that the sequence is "in a compact metric space". So it doesn't require the sequence to be a closed set, only that it be contained in a compact set. The points in $$\mathbb R^k$$ with norm $$\le M$$ are a closed and bounded (ergo compact) subset of $$\mathbb R^k$$.