Let $X,Y$ be non-empty compact and Hausdorff topological spaces and $f:X \to Y$ be a continuous map. Take an element $y \in Y$.

Question: Is $f^{-1}(\{y\})$ closed in $X$?

Approaches and Ideas (assuming that this is true):

  • If we could show that $\{ y \}$ is closed in $Y$, then we would be done since inverse images of closed subsets under continuous maps are closed.
  • In order to show that this singleton set is closed, I must show that $Y \setminus \{ y \}$ must be open in $Y$. However, I know nothing about the topology of $Y$ except that it is Hausdorff and compact. But I can not see how these two properties could be used to proceed.

Could you please help me with this problem? Thank you in advance!

  • 1
    $\begingroup$ Note that compactness of $X$ and $Y$ and and Hausdorffness of $X$ were not used in the solution. All you need is that $Y$ is Hausdorff (and actually, $T_1$ suffices, if you know what that is). $\endgroup$ – Alex Kruckman Mar 31 at 12:35

If $X$ is Hausdorf, every singleton $\{x\}$ is closed. Let $y$ in $X-\{x\}$, there exists open neighborhoods $U$ of $x$ and $V$ of $y$ such that $U\cap V$ is empty, this implies that $V\subset X-\{x\}$ so $X-\{x\}$ is open.


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