Hi everyone I have troubles with the following proposition:

Definition: We say a metric space $(X,d)$ is an A-space iff every Hausdorff image of $X$ under a closed continuous map is metrizable.

Claim: $X$ is an A-space iff the frontier of any closed set in $X$ is compact.

(*) Assuming the Hanai-Morita theorem: Let $f:X \to Y$ closed, continuous, onto and $X$ metric. Then the following are logically equivalent

  1. $Y$ is metric
  2. $Y$ is first countable
  3. for all $p\in Y$, $\operatorname{fr}(f^{-1}(p))$ is compact

One side is pretty easy, since the range is a Hausdorff space, in particular the points are closed then $f^{-1}(p)$ is closed in $X$, but for the other I have serious problems.

Suppose $F$ is a closed set in $X$ and $X$ is an A-space, then by the Hanai-Morita theorem we know that for any map satisfying (*) we have $\forall p\in Y$ the frontier of the preimage of $p$ is a compact. Suppose to the contrary that $\operatorname{fr} (F)$ is not compact in $X$, then there is a sequence $\{x_n: n\in {\bf {N}}\}\subset \operatorname{fr}(F)$ having no cluster points in the frontier of $F$...

I was trying to repeat the proof of the Morita's theorem but I can't

I really appreciate any help.


To show that every A-space satisfies the given property, let $(X,d)$ is an A-space, let $F \subseteq X$ be closed. Consider the quotient space $Y = X / F$ and the natural quotient mapping $q : X \to Y$. Let $* \in Y$ denote the point corresponding to the collapsed closed set $F$.

Note that $Y$ is clearly Hausdorff, and $q$ is a closed, continuous, onto mapping. Therefore $Y$ is metrizable, and so by the Hanai-Morita Theorem it follows that $\operatorname{fr} ( q^{-1} [ \{ * \} ] ) = \operatorname{fr} ( F )$ is compact.


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