1
$\begingroup$

I found this: Let $X$ be a compact space and $f:X \rightarrow Y$ a continuous injection. Let $f(X)$ be dense in $Y$. Prove that $f$ is a homeomorphism.

So, my question is: is it possible to prove that? I tried proving it and I couldn't, since Y is not necessarily a Hausdorff space.

The Hausdorff-condition is needed when proving that f is a closed mapping, but I guess you could do it some other way.

$\endgroup$
  • 3
    $\begingroup$ Hausdorff is needed. Take $Y=X$ with the trivial topology and consider the identity map from $X$ to $Y$. $\endgroup$ – David Mitra Jul 24 '14 at 9:45
  • 1
    $\begingroup$ Please consider changing the title, to be more informative. $\endgroup$ – Elimination Jul 24 '14 at 9:49
  • $\begingroup$ Note that $f(X)$ is compact and thus closed in $Y$ under the assumption that $Y$ is Hausdorff. Together with being dense it implies that $f(X)=Y$. With that you apply the classical "continuous bijection from a compact space" argument. $\endgroup$ – freakish Nov 8 '18 at 18:46
1
$\begingroup$

From the comments above by @DavidMitra.


It is necessary that $Y$ is Hausdorff. For a counterexample when $Y$ is not Hausdorff, let $Y$ be the set $X$ equipped with the trivial topology. Assume that the topology on $X$ is not the trivial topology. Then, the identity map from $X$ to $Y$ is injective, continuous, and the image is dense in $Y$ (in fact it is all of $Y$). However, $f$ is not a homeomorphism.

$\endgroup$

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.