$X$ and $Y$ are metric spaces, $A\subseteq X$, $A$ is bounded. map $f:X\to Y$ is continuous.


  1. Is $A$ necessarily compact?
  2. Is $f(A)$ uniformly continuous?
  3. If given that $f$ is a bijection and $f(A)$ is compact, is $f^{-1}$ necessarily continuous?

Tried: If $A$ is closed and bounded, then $A$ must be compact hence compactness of $f(A)$ follows. So the question reduces to "is boundedness of $A$ sufficient for compactness?" Also, I think 3 is correct because bijection and compact domain and codomain should be enough for homeomorphism...

I am new to topology so really appreciate any help!

  • 1
    $\begingroup$ To $1:$ Where is said that $A$ is closed? To $2:$ Do you mean $f_{|A}?$ $\endgroup$ – mfl Oct 24 '14 at 16:51
  • $\begingroup$ @mfl $A$ is not given as closed. I'm saying that if it is closed then it must be compact, since $A$ is bounded. And what is $f_{|A}$? $\endgroup$ – MatheMagic Oct 24 '14 at 17:23

For 1,2 no. Consider $f(x)=\frac{1}{x}$ on $(0,1]$

For 3, no. Consider identity map from $[0,1]$ to $[0,1]$ , where the first one with discrete metric and the second one with standard metric.

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  • $\begingroup$ nice counterexamples! $\endgroup$ – MatheMagic Oct 24 '14 at 17:25

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