0
$\begingroup$

Let $X$ be a compact space such that for any continuous map $f\colon C\rightarrow X$ with $C$ compact space, if $f$ is a continuous bijection then $f$ is an homeomorphisms.

Prove that all compact subspace of $X$ is closed. Is interesting because it sounds that $X$ has to be Hausdorff, since any compact in a Hausdorff space is closed. So one result that may help is that a space si Hausdorff when the diagonal map is closed. Also, if a continuous bijection is closed then it is a homeomorphism.

$\endgroup$
1
  • 2
    $\begingroup$ See here or here. $\endgroup$ Commented Sep 27 at 19:04

1 Answer 1

Reset to default
1
$\begingroup$

We have that $1_X\colon (X,\sigma)\rightarrow (X,\tau)$ is continuous if and only if $\tau\subseteq \sigma$.

If we take $(X,\tau)$ the one of the question and $(X, \sigma)$ a compact space. Then $\tau=\sigma$. Which means that $\tau$ is a maximal compact topology. Then the result follows from the following question.

Maximal compact topology iff compact sets are closed

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.