# Give an example: $X, Y$ metric space, $X$ not compact, there is no $V$ for which $f^{-1}(V) \subset U$

Give an example of a non compact $X$ and sets $C$ and $U$ such that there is no $V$ satisfying the following

Let $X$ and $Y$ be metric spaces, with $X$ compact, and $f: X \to Y$ continuous. Let $C$ be a closed subset of $Y$. Then 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 is a sequel of this question.

When $X$ is not compact, $f(U^c)$ may be open.
I can't find a concret example.

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HINT: Let $X=\left\{\langle x,y\rangle\in\Bbb R^2:y>0\right\}$ and $Y=\Bbb R$, and define $f:\Bbb R^2\to\Bbb R$ by $f\big(\langle x,y\rangle\big)=x$. Let $C=\{0\}$. Clearly $f^{-1}[C]$ is the positive $y$-axis in the plane. If $V$ is an open nbhd of $0$ in $\Bbb R$, $f^{-1}[V]$ contains a set of the form $(a,b)\times(0,\to)$, where $a<0<b$. Can you find an open nbhd of the positive $y$-axis that does not contain any set of the form $(a,b)\times(0,\to)$ with $a<0<b$? (Think about the graph of $y=\frac1{x^2}$.)