Continuity and compactness

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(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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  • #2
morphism
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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?
 
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Just to clarify - when they say 'real-valued,' they mean the image is strictly in [tex]\mathbb{R}[/tex], not simply in [tex]\mathbb{R}^n[/tex]?

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




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

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


Thanks! :)
 
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  • #4
morphism
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Just to clarify - when they say 'real-valued,' they mean the image is strictly in [tex]\mathbb{R}[/tex], not simply in [tex]\mathbb{R}^n[/tex]?
Yes, real-valued means the image lives in in [itex]\mathbb{R}[/itex].

I wanted to use the function [tex]f(x)=\frac{1}{|x-c|}[/tex] 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 [tex]1\leq i\leq n[/tex], let [tex]f_i(x_1,...,x_n)=x_i[/tex]. Then define each [tex]K_i[/tex] to be such that [tex]f_i(K)\subset K_i\subset\mathbb{R}[/tex]. Then [tex]K\subset K_1\times K_2\times...\times K_n[/tex] 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.
 
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Got it. Thanks so much! :)
 

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