# Property of continuous map from compact space onto Hausdorff space

I am self studying Topology, and I am having difficulty in proving the following statement.

Let $$f$$ be a continuous mapping defined on a compact space $$X$$ onto a Hausdorff space $$Y$$. Suppose $$y\in Y$$ and $$U$$ is an open subset of $$X$$ that contains $$f^{-1}[y]$$. Show that there exists an open neighborhood $$V$$ of $$y$$ such that $$f^{-1}[V] \subset U$$.

I know that

1. $$Y=f[X]$$ is compact, and thus, being Hausdorff, it is also normal;
2. $$f$$ is a closed map;
3. $$f^{-1}[y]$$ is closed and compact.

Can you give me some hints on how to use these facts (or other consequences of the hypothesis) to prove the theorem?

Because $$U$$ is open $$X\backslash U$$ is closed and therefore compact. Then $$f(X\backslash U)$$ is closed because $$Y$$ is a Hausdorff space. Also $$y\not\in f(X\backslash U)$$ because $$f^{-1}(y)\subset U\ \cap X\backslash U = \varnothing$$.

Let denote $$V=Y\backslash f(X\backslash U)$$ it is open and $$y\in V$$. $$f^{-1}(V)=f^{-1}(Y)\backslash f^{-1}(f(X\backslash U)) = X\backslash f^{-1}(f(X\backslash U))=W$$

$$W$$ is open because $$f$$ is continuous and $$f(X\backslash U)$$ is closed.

$$X\backslash U \subset f^{-1}(f(X\backslash U))$$ and therefore $$W = X\backslash f^{-1}(f(X\backslash U)) \subset X\backslash(X\backslash U) = U$$

$$W\subset U$$ as we need.

• In your last expression, are you assuming $f^{-1}[f[X\backslash U]]=X\backslash U$? Can you actually state that?
– dfnu
Jan 13, 2021 at 12:35
• You are right it is not generally true, I have edited my proof. Jan 13, 2021 at 12:42

We know $$f^{-1}[\{y\}] \subseteq U$$. We know that $$f$$ is a closed map (continuous from a compact space to a Hausdorff space) so $$f[X\setminus U]$$ is closed in $$Y$$ and so $$V:= Y\setminus f[X\setminus U]$$ is open in $$Y$$.

I claim that $$y \in V$$ and $$f^{-1}[V] \subseteq U$$.

Proof: if $$y \notin V$$ then $$y \in f[X\setminus U]$$ so $$y=f(x)$$ for some $$x \notin U$$, but this implies $$x \in f^{-1}[\{y\}] \nsubseteq U$$ a contradiction. So $$y \in V$$.

If $$x \in f^{-1}[V]$$ then $$f(x) \in V$$ by definition and if $$x \notin U$$, we'd have $$x \in X\setminus U$$ so $$f(x) \in f[X\setminus U]$$ or $$f(x) \notin V$$, a contradiction, and hence $$x \in U$$ and the inclusion holds.