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Is the inverse image bounded

  • Thread starter bertram
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1. Homework Statement

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

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

3. The Attempt at a Solution
So my idea was to show that [tex]f^{-1}[/tex](K) was continuous, but i can't really figure that out immediately.
I just tried next to describe K and [tex]f^{-1}[/tex](K) as best I could.... We know that K is closed and compact (compact subsets of metric spaces are closed). This will imply that [tex]f^{-1}[/tex](K) is closed (Thm 4.8 corollary). So I have that K is closed and compact and that [tex]f^{-1}[/tex](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.
 

Answers and Replies

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

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