Proving $K$ is compact directly.

Question

If $K$ is a subset of metric space $\mathbb{R}^n$ and if every real valued continuous function on $K$ is bounded, then $K$ is compact.

I know a proof considering $K$ is unbounded and not closed. This is proof by contradiction.

Is there any direct way to prove?

Please help!

Does this answer your question? If every real-valued continuous function is bounded on $X$ (metric space), then $X$ is compact. — Arctic Char, Aug 02 '20 at 06:52
Yes! But is there any direct way where we can show for any open cover of $K$ there exists a finite subcover? — Learning, Aug 02 '20 at 07:08

Stephen Montgomery-Smith · Answer 1 · 2020-08-02T07:38:15.173

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Suppose that $U_n$ is a sequence of open sets that cover $(X,d)$. Let $$ f(x) = \sum_{n=1}^\infty 2^{-n} \min\{d(x,U_n^C),1\} .$$ We know $f(x) > 0$ for $x \in X$. So by hypothesis, $1/f$ is bounded above by $2^{N-1}$ for some $N \in \mathbb N$. Hence $f(x) > 2^{-N}$ for all $x\in X$, and so $X \subset \bigcup_{n=1}^N U_n$.

This shows every countable cover has a finite subcover.

Not sure how to complete it.

edited Aug 02 '20 at 07:38

answered Aug 02 '20 at 07:16

Stephen Montgomery-Smith

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It seems that your $N$ depends on $x$? And it's not clear how the assumptions on the space of continuous functions are used. – Arctic Char Aug 02 '20 at 07:25
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No it doesn't. Changed my answer to make that clearer. – Stephen Montgomery-Smith Aug 02 '20 at 07:26
Can you please clarify how $f > 2^{-N}$ implies that $X \subset U_1\cup\cdots \cup U_N$? – Arctic Char Aug 02 '20 at 08:01
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Because if $x \not\in U_1 \cup \cdots \cup U_N$, then $f(x) \le \sum_{n=N+1}^\infty 2^{-n}$. – Stephen Montgomery-Smith Aug 02 '20 at 08:02

Proving $K$ is compact directly.

1 Answers1