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It is known that the continuous Hausdorff(!) image of a compact metrizable space is (compact and) metrizable. In general, the continuous image of a compact (pseudo-)metrizable space need not be pseudometrizable. Example: Take $K = [0,1]$ with the usual topology and $L = [0,1]$ with the cofinite topology. Then $L$ carries its minimal $T_1$-topology (thus not Hausdorff) and $L$ is not first-countable, thus not pseudometrizable (however, $L$ is Fréchet-Urysohn). But the identity map $id : K \to L$ is continuous.

So let $K$ be compact pseudometrizable, $L$ a general topological space and $f : K \to L$ continuous and surjective.

  1. What are the "minimal" additional requirements on $f$ which imply that $L$ is pseudometrizable?
  2. If we assume that $L$ is also completely regular, is then $L = f(K)$ pseudometrizable? If not, what are the minimal additional requirements on $f$ in such a setup? (This second question is for applications in topological vector spaces. The space $L$ is not (completely) regular, since $T_1$+regular $\Rightarrow$ $T_3$ $\Rightarrow$ $T_2$, but $L$ is not $T_2$).
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In general topology, it is often the case that theorems that rely on Hausdorffness can be sometimes generalized to the non-Hausdorff setting by imposing regularity as a separation axiom. Here is a theorem that solves Question 2. Note also that this solves Question 1, since then necessarily $Y$ must be completely regular (pseudometrizable spaces are completely regular).

Theorem: If $X$ is compact pseudometrizable, $Y$ is regular and $f : X \to Y$ is continuous and surjective then $Y$ is pseudometrizable.

I follow the proof in [Aliprantis, Border, "Infinite Dimensional Analysis: A Hitchhiker's Guide", Corollary 3.43] for the statement "$X$ compact metrizable, $Y$ Hausdorff, $f$ continuous surjective $\Rightarrow$ $Y$ metrizable" with the obvious modifications replacing non-closed compact sets with their closures.

Proof: Since $X$ is compact and $f$ is a continuous it follows that $Y = f(X)$ is compact. Since $Y$ is compact and regular it follows that $Y$ is also completely regular and normal. By the Urysohn metrization theorem in the non-Hausdorff setting, every regular second-countable topological space is pseudometrizable. Thus it is enough to show that $Y$ is second-countable. If $G \subseteq X$ is open then $X \setminus G$ is closed and thus compact (as a closed subset of a compact space). Therefore, $f(X \setminus G)$ is compact, but it need not be closed in $Y$. However, its closure $\overline{f(X \setminus G)}$ in $Y$ is compact since $Y$ is regular (see here). Thus, if $G$ is open in $X$ then $Y \setminus \overline{f(X \setminus G)}$ is open in $Y$. Since $X$ is compact and pseudometrizable it follows that $X$ is second-countable. So let $\mathcal{B}$ be a countable base for $X$ and we can assume that $\mathcal{B}$ is closed under finite unions. Define $\mathcal{C} := \{ Y \setminus \overline{f(X \setminus G)} \mid G \in \mathcal{B} \}$. Then $\mathcal{C}$ is countable an we claim that $\mathcal{C}$ is a base for $Y$. Let $W \subseteq Y$ open, $W \neq \emptyset$. Take $y \in W$. Then $\{ y \}$ is compact and again by regularity of $Y$, $W$ is also an open neighborhood of its closure $\overline{ \{ y \} }$ (see again here). Then $f^{-1}(\overline{ \{ y \} })$ is closed in $X$, thus compact and we have $f^{-1}(\overline{ \{ y \} }) \subseteq f^{-1}(W)$. Now since $X$ is second-countable the open set $f^{-1}(W)$ is a union of sets in $\mathcal{B}$ which then forms an open cover of $f^{-1}(\overline{ \{ y \} })$ and thus there are finitely many such sets $B_1, \dots, B_n \in \mathcal{B}$ with $f^{-1}(\overline{ \{ y \} }) \subseteq G \subseteq f^{-1}(W)$ where $G := \bigcup_{i=1}^n B_i$. Since $\mathcal{B}$ is closed under finite unions we have $G \in \mathcal{B}$. It follows that $Y \setminus \overline{f(X \setminus G)} \subseteq f(G) \subseteq W$ and we are done.

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