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enter image description here This is an exam question that i am having trouble with. I tried to show that $f$ is a closed map, but without the assumption of surjectivity i could not prove my claim of closedness. Also i wanted to use the Lebesgue Number lemma to get the $\delta >0$ in an open cover of $Y$, but this lemma only hold for a compact metric space. How can i approach the problem? Any hint(s) will be highly appreciated. Thank you.

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Suppose that $C$ is a closed subset of $X$. Then $C$ is compact, and $f$ is continuous, so $f[C]$ is compact. Finally, $Y$ is Hausdorff (since it’s metric), so compact subsets of $Y$ are closed, and $f[C]$ is therefore closed. Thus, $f$ is a closed map.

Note that you can replace $Y$ by $f[X]$, which is a compact metric space.

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  • $\begingroup$ In the last line of your answer, you mean that we do not require surjectivity of $f$? $\endgroup$ – User101 Mar 31 '15 at 2:30
  • $\begingroup$ @User101: That’s right: no part of my argument requires surjectivity of $f$. $\endgroup$ – Brian M. Scott Mar 31 '15 at 2:38

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