5
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.