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Is the inverse image bounded

  1. Apr 16, 2008 #1
    1. The problem statement, all variables and given/known data

    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. Relevant 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.
  2. jcsd
  3. Apr 16, 2008 #2


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    Homework Helper

    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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