Is the Inverse Image of a Compact Set Always Bounded?

[bertram]

Homework Statement

Let f be a continuous mapping from metric spaces X to Y. K \subset Yis compact. Is f^{-1}(K) bounded?

Homework Equations

Theorem 4.8 Corollary (Rudin) A mapping f of a metric space X into Y is continuous iff f^{-1}(C) is closed in X for every closed set C in Y.

The Attempt at a Solution

So my idea was to show that f^{-1}(K) was continuous, but i can't really figure that out immediately.
I just tried next to describe K and f^{-1}(K) as best I could... We know that K is closed and compact (compact subsets of metric spaces are closed). This will imply that f^{-1}(K) is closed (Thm 4.8 corollary). So I have that K is closed and compact and that f^{-1}(K) is closed. I just don't know how to make the ends meet. Maybe I'm doing this wrong, or just missing something obvious.

Thanks in advance for any help.

 

[Dick]

Think about why it might not be true before you start trying to prove it. Suppose f:R->R and f(x)=sin(x)?