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?


2 Answers 2


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.

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


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