MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

All topological spaces considered here are Hausdorff.

It is a well-known consequence of the minimality of a compact topology that an injective continuous map

$f\colon X\to Y$

where $X$ is compact, must be automatically a homeomorphism onto its range. I am interested in possibly non-compact spaces which share this property. I would like to kindly ask whether there is a characterisation of this class of spaces.

share|cite|improve this question
There is an easy case: Let $f$ be proper, meaning the inverse image of every compact subset of $Y$ is compact in $X$. Then $f$ would be a homeomorphism onto its image. – Vahid Shirbisheh Dec 22 '12 at 18:33
$Y$ also should be compact in the above! – Vahid Shirbisheh Dec 22 '12 at 18:47
This is a similar question. – Joseph Van Name Dec 22 '12 at 20:53
up vote 8 down vote accepted

The spaces that you are looking for are precisely the minimal Hausdorff spaces. i.e. A Hausdorff space $X$ is a minimal Hausdorff if every injective continuous map from $X$ to a Hausdorff space is an embedding. See the Book General Topology by Stephen Willard problems 17M for more information about these spaces. The Minimal Hausdorff spaces are precisely the Hausdorff spaces where every open filter with a unique accumulation point converges.

We say that a topological space is semiregular if the regular open sets form a basis for the topology. A Hausdorff space is said to be $H$-closed if it is closed in every Hausdorff extension. Every minimal Hausdorff space is $H$-closed, and a Hausdorff space is a minimal Hausdorff space if and only if it is semiregular and $H$-closed.

While there are minimal Hausdorff spaces which are not compact, the notions of minimal Hausdorff and compactness coincide for regular spaces.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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