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Let $X$ and $Y$ be metric spaces, with $X$ compact, and $f: X \to Y$ continuous.
Let $C$ be a closed subset of $Y$.
Show that for any open neighboorhood $U$ of $f^{-1}(C)$ there is an open neighborhood $V$ of $C$ such that $$ f^{-1}(V) \subset U. $$

This question is answered here but I don't understand the answer.
Specifically, I don't see why $$ U^c \cap f^{-1}(C) = \emptyset, $$ with $U^c$ the complement of $U$ in $X$.

Here is my attempt.
Since $U$ is open and $f^{-1}(C)$ is closed then the $f^{-1}(C)$ complement of $U$ is closed. Let $U^c$ be this complement.
since $f$ is continuous, then $f(U^c)$ is closed.
Let $V=C\setminus f(U^c)$, then $V$ is open, but I don't see why $$ f^{-1}(V) \subset U. $$

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    If $A\supseteq B$, then $A^c\cap B=\varnothing$; draw a Venn diagram if you don’t see it otherwise. Here $A=U$, and $B=f^{-1}[C]$. – Brian M. Scott Jan 14 '13 at 22:03
  • What do you mean by ‘the $f^{-1}(C)$ complement of $U$’? – Brian M. Scott Jan 14 '13 at 22:05
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    $f^{-1}(C)\subset U\Rightarrow f^{-1}(C)\cap U^c=\emptyset$. – Sigur Jan 14 '13 at 22:11
  • @BrianM.Scott The question says $U$ is an open neighborhood of $f^{-1}(C)$, so this means $U \subset f^{-1}(C)$, right?. - By the $f^{-1}(C)$ complement of $U$, I mean $\left[ f^{-1}(C) \setminus U \right] \cap f^{-1}(C)$. –  Jan 14 '13 at 22:21
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    No, it means that $U\supseteq f^{-1}[C]$ (and $U$ is open). – Brian M. Scott Jan 14 '13 at 22:24
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    @BrianM.Scott, notation q: $f^{-1}[C] = f^{-1}(C)$? – alancalvitti Jan 14 '13 at 22:37
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    @alancalvitti: I am one of those who make a careful distinction between $f^{-1}[C]={x\in X:f(x)\in C}$ and $f^{-1}(C)$, the latter making sense only when $f$ is an injection from some set with $C$ in its range. – Brian M. Scott Jan 14 '13 at 22:43
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    @BrianM.Scott, in other words you distinguish inverse image from image under the inverse map and the latter doesn't necessarily exist here. But since there's no injectivity assumption the OP and commentators mean $f^{-1}[C] = f^{-1}(C)$ default? - Is there a risk of notational confusion in using $f^{-1}(C)$ here? – alancalvitti Jan 15 '13 at 00:59
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    @alancalvitti: In most contexts, including the present one, there’s no real risk of confusion and hence no harm in writing $f^{-1}(C)$ for my $f^{-1}[C]$, and you’ll see the former more often than the latter. I just prefer to maintain the distinction. – Brian M. Scott Jan 15 '13 at 01:06

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We know that $U^{c}$ is closed, and hence, compact in $X$. Thus, $f(U^{c})$ is compact, and hence, closed in $Y$ since $f$ is continuous and $Y$ is a metric space.

Let $V=Y\setminus f(U^{c})$. Then, $V$ is an open set. We'll show that $f^{-1}(V)\subset U$. Pick $x\in f^{-1}(V)$. Then, $y:=f(x)\in V$, so that $y\notin f(U^{c})$. Therefore, $x\notin U^{c}$, which means that $x\in U$. Hence, $f^{-1}(V)\subset U$, as desired.

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