Let $(X,\tau)$ be an infinite topological space with the property that every subspace is compact. Prove that $(X,\tau)$ is not a Hausdorff space.

A space is Husdorff if for all $a,b\in X$ then there exists two open sets $V,U\in\tau$ such that $a\in U,b\in V$ and $U\cap V=\emptyset$.

Since all the subspaces are compact means all the elements of X are closed and then opened at the same time by the compliment so we are dealing with the discrete topology.

If I assumed the space was Hausdorff and all subspaces were compact I believe I would get a contradiction, but I do not see how.


Can someone help me see the contradictions?

Thanks in advance!


Hint: if $X$ is Hausdorff, then every subspace of $X$ is compact thus closed. What can you say about $\tau$ then?

  • $\begingroup$ I have pointed out in my post $\tau$ is the discrete topology. Is this what you wanted me to say? Thanks for your answer! $\endgroup$ – Pedro Gomes Dec 17 '18 at 13:05
  • $\begingroup$ Yes. So what can you say about a discrete compact topological space? $\endgroup$ – Mindlack Dec 17 '18 at 13:09
  • $\begingroup$ The discrete topology is Hausdorff, since for any two singletons are open sets of their own. Are you claiming the discrete space is not compact? $\endgroup$ – Pedro Gomes Dec 17 '18 at 13:51
  • $\begingroup$ I am claiming a discrete infinite space is not compact. $\endgroup$ – Mindlack Dec 17 '18 at 15:39
  • If $X$ were Hausdorff then $(X,\tau)$ is discrete (as you rightly noted).
  • If $X$ is discrete and compact (as it must be, as all subspaces are) it is finite. (Consider the open cover by singletons).
  • Contradiction, as $X$ is supposed to be infinite.

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