Forum for the GRE subject test in mathematics.
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Joined: Sun Aug 16, 2009 12:19 pm


Postby weilan8 » Wed Sep 23, 2009 7:40 pm

Hi everyone, I have a question from GR0568 and [edited away, please don't post actual exam problems]. Thank you!

Let K be a nonempty subset of Rn, where n>1. Which of the following statements 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.
(A) I only (B) II only (C) III only (D) I and II only (E) I, II, and III

[answer is D. For II, we can give counterexample such as f(x)=sin(pi*x) where x is (-1,1), which f(x) is [-1,1]. However, the number of counterexample is limited if x belongs to an open set. So, I’m not sure that if every continuous real-valued function defined on K is bounded then will K must be a closed set?]

Posts: 24
Joined: Thu Sep 24, 2009 12:17 am

Re: GR0568_#62 and some previous exam questions

Postby prong » Thu Sep 24, 2009 12:25 am

In the first question, consider your function f(x) = sin(pi*x), on K = (-1,1). Im(f(x)) is [-1,1] so f(x) is bounded. Property (II), however, doesn't depend on the existence of a single continuous real-valued function that is bounded on K. (II) says that if EVERY continuous f: K->R is bounded, then K is compact.

I think you're arguing that property (II) would imply K=(-1,1) is compact. However, the existence of a bounded continuous function f is not sufficient; to prove K is compact by using property (II), you have to show that whenever f is some ARBITRARY continuous function on K, f is bounded. In this case, sin(pi*x) is certainly bounded--but the function f(x) = 1/(1-x) is continuous with respect to K but not bounded, so it's not true that EVERY continuous f is bounded. Therefore property (II) can't show that K is compact, since you would need the set {1/(1-x) | -1 < x < 1} to be bounded.

[Edited away: Please don't post solutions to problems that shouldn't have been posted]

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