# Inverse image of singleton sets under continuous map between compact Hausdorff topological spaces

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.

• 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). – 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.