It is well known that if $X$ and $Y$ are topological spaces with $X$ locally compact Hausdorff and $Y$ compactly generated, then $X \times Y$ (with the ordinary product topology) is compactly generated. Does this fail if $X$ is compact, compactly generated, but not Hausdorff?


1 Answer 1


I think the space $Z=\mathbb{Q}^* \times \mathbb{Q}$ works for your Question. (Here $\mathbb{Q}^*$ is the one point compactification of the rational numbers)

At first let me recall some properties of $\mathbb{Q}^*$ which you could find Them in the Monthly Article (Between $T_1$ and $T_2$) http://www.jstor.org/discover/10.2307/2316017?uid=3738280&uid=2&uid=4&sid=47699073521117.

  • $\mathbb{Q}^*$ is a $KC$ space.(i.e. every compact subset of this space is closed)
  • It's easy to show that $\mathbb{Q}^*$ and $\mathbb{Q}$ are compactly generated.

Now I bring Two theorems from the Article "On KC and K-spaces" which are important in the sequel to show that $Z$ is the space which you needed. you could find this Article from:


Theorem 1: Let $X , Y$ be topological spaces.If $X$ is $KC$ and $Y$ is hausdorff; Then $X\times Y$ is $KC$.

Theorem 2: In the topological space $X$ the following are equivalent:

  • $X$ is locally compact and Hausdorff.
  • $X^*\times X$ is $KC$ and compactly generated.

For the sake of Theorem 1, you could see that $\mathbb{Q}^* \times \mathbb{Q}$ is $KC$. Also from part one of theorem 2 and because $\mathbb{Q}$ is not locally compact, We can conclude that $\mathbb{Q}^* \times \mathbb{Q}$ is not compactly generated.

  • $\begingroup$ What is $KC$? Do you have a counterexample for product of compactly generated spaces being compactly generated? $\endgroup$
    – Sigur
    Commented Mar 21, 2014 at 18:09
  • $\begingroup$ @Sigur what Ali_Reza has done is exactly what you asked for isn't it ? He explained what KC meant and gave you the requested counter example $Q\times Q^*$ $\endgroup$
    – brunoh
    Commented May 24, 2023 at 19:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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