(adsbygoogle = window.adsbygoogle || []).push({}); (Problem 62 from practice GRE math subject exam:)Let K be a nonempty subset of [tex]\mathbb{R}^n[/tex], n>1. Which of the following must be true?

I. If K is compact, then every continuous real-valued function defined on K is bounded.

II. If every continuous real-valued function defined on K is bounded, then K is compact.

III. If K is compact, then K is connected.

I know (I) is true and (III) is not necessarily true. I'm working on (II), which the answer key says is true, but I can't seem to prove it. I tried using several versions of the definition of compactness:

Closed and bounded-

K is obviously bounded if you take the function f(x)=x. Then f(K)=K is bounded.

To show K is closed, I assumed it wasn't: there is some sequence [tex](x_n)\subseteq K[/tex] such that the sequence converges to a point c outside of K. Then the sequence [tex]f(x_n)\subseteq f(K)[/tex] must converge to some point d, not necessarily in f(K). I'm not sure where to go from there or what contradiction I am looking for.

Covers / finite subcovers-

Let [tex]\{U_i\}[/tex] be any open cover of K. Then [tex]\{f(U_i)\}[/tex] is a cover of f(K). What I'd like to do is somehow force a finite subset of [tex]\{f(U_i)\}[/tex] to be a cover of f(K) - possibly using the fact that f(K) is bounded - and thus find a finite subcover for K. The problem is that I don't know that I can find a subcover for f(K).

Any ideas?

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# Continuity and compactness

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