1
$\begingroup$

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!

$\endgroup$
  • 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.