Is Every Continuous Function on a Compact Subset of \mathbb{R}^n Bounded?

[jjou]

(Problem 62 from practice GRE math subject exam:) Let K be a nonempty subset of $$\mathbb{R}^n$$, 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 $$(x_n)\subseteq K$$ such that the sequence converges to a point c outside of K. Then the sequence $$f(x_n)\subseteq f(K)$$ 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 $$\{U_i\}$$ be any open cover of K. Then $$\{f(U_i)\}$$ is a cover of f(K). What I'd like to do is somehow force a finite subset of $$\{f(U_i)\}$$ 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?

 

[morphism]

The function f(x)=x isn't 'good', because it can't possibly be defined on K - K is a subset of R^n. To show that K is bounded, use the projection maps instead.

To show that K is closed, you can continue with what you're doing. Can you use the point c=(c_1,...,c_n) to define a continuous, but unbounded, real-valued function on K?

 

[jjou]

Just to clarify - when they say 'real-valued,' they mean the image is strictly in $$\mathbb{R}$$, not simply in $$\mathbb{R}^n$$?

I wanted to use the function $$f(x)=\frac{1}{|x-c|}$$ which is defined for x in K. This function would be unbounded, but is it still continuous?




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Please check -
To use the projection maps to show boundedness of K:

For each $$1\leq i\leq n$$, let $$f_i(x_1,...,x_n)=x_i$$. Then define each $$K_i$$ to be such that $$f_i(K)\subset K_i\subset\mathbb{R}$$. Then $$K\subset K_1\times K_2\times...\times K_n$$ is bounded.


Thanks! :)

 

[morphism]

Just to clarify - when they say 'real-valued,' they mean the image is strictly in $$\mathbb{R}$$, not simply in $$\mathbb{R}^n$$?

Yes, real-valued means the image lives in in $\mathbb{R}$.

I wanted to use the function $$f(x)=\frac{1}{|x-c|}$$ which is defined for x in K. This function would be unbounded, but is it still continuous?

Yup - it's a composition of two continuous maps (x -> 1/(x-c) and x -> |x|).

To use the projection maps to show boundedness of K:

For each $$1\leq i\leq n$$, let $$f_i(x_1,...,x_n)=x_i$$. Then define each $$K_i$$ to be such that $$f_i(K)\subset K_i\subset\mathbb{R}$$. Then $$K\subset K_1\times K_2\times...\times K_n$$ is bounded.

Yes, that's fine. Although on second thought I'd just use the norm map instead of projections to prove that K is bounded.

 

[jjou]

Got it. Thanks so much! :)