3
$\begingroup$

(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.

$\endgroup$
3
$\begingroup$

He is not saying that the bounded set itself is compact. What he is saying is that that bounded set is contained in some compact set. As a bounded set is, by definition, contained in an open ball, it is also contained in the closure of the ball, which is compact.

$\endgroup$
2
$\begingroup$

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 $\mathbb R^k$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.