0568 Q62 proof

Forum for the GRE subject test in mathematics.
yoyostein
Posts: 36
Joined: Tue Feb 28, 2012 12:14 am

0568 Q62 proof

Postby yoyostein » Tue Mar 20, 2012 1:32 am

Hi, could anyone give a proof of why Q62 II is true?

II. If every continuous real-valued function defined on K is bounded, then K is compact.
(K nonempty subset of R^n)

Thank you very much!

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: 0568 Q62 proof

Postby blitzer6266 » Tue Mar 20, 2012 2:03 am

First of all, in euclidean space, compact=closed and bounded

Suppose K is not compact, then either it is not closed or not bounded

Not bounded- consider the function ||x|| which will not be bounded
Not closed-there exists a limit point A such that x_n goes to A but A is not in K
Then you can look at the function 1/||x - A||. This will be continuous on K but unbounded since it goes to infinity as x goes to A

There are probably better proofs but this is at least the picture that should come to mind.

yoyostein
Posts: 36
Joined: Tue Feb 28, 2012 12:14 am

Re: 0568 Q62 proof

Postby yoyostein » Tue Mar 20, 2012 7:16 am

thanks!

this is crystal clear explanation




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