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Function invariant under Hilbert transform

I know that there is a topological space $X$ which is not compact but such that , for any Hausdorff topological space $Y$ , any continuous function $f:X \to Y$ carries closed sets to closed sets . I would like to ask , Does there exist any non-compact metric space $X$ such that , for any Hausdorff topological space $Y$ , any continuous function $f:X \to Y$ carries closed sets to closed sets ?

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We have the following result in Set Topology:

**Prop.** Let $X$ be a space with the two properties that:

(i) each point has a neighborhood basis of closed neighborhoods ($X$ is *regular*), and

(ii) every continuous image of it in a Hausdorff space is closed.

Then $X$ is compact.

**Proof.** Suppose first $X$ is Hausdorff (the case of metric spaces). Let $\{U_i\}$ an open cover of $X$. By (i), every $x\in X$ has an open neighborhood $V^x$

with $\overline{V^x}\subset U_{i(x)}$. Define a Hausdorff space à la Alexandroff: $X^*=X\cup\{\omega\}$ adding to the topology of $X$ the open nbdhs for $\omega$: $$

U^\omega=X^*\setminus\bigcup_{\text{finite}}\overline{V^x}.

$$

Then $X^*$ is Hausdorff and by (ii) $X$ is closed in $X^*$. Thus $\omega$ is open, that is

$$

\{\omega\}=X^*\setminus\bigcup_{\text{finite}}\overline{V^x},

\quad

\text{so that}

\quad

X=\bigcup_{\text{finite}}\overline{V^x}\subset\bigcup_{\text{finite}}U_{i(x)}.

$$

This completes the argument in the compact case.

In the general case, this hint: identify points in $X$ that cannot be separated in this sense: $x\equiv y$ iff every neighborhood of $x$ contains $y$ and every nbhd of $y$ contains $x$. The quotient $\widetilde X$is Hausdorff and all open sets are saturated. Then we can construct $\widetilde X^*$ to conclude. $\ \square$

In our case, $X$ is metric, hence Hausdorff and regular, and (ii) is weaker than *every continuous mapping into a Hausdorff space is closed*. Hence the answer to your question is *no, there is no such a metric space*.

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