# Contradiction in the proof a space with every compact subspace is not Hausdorff

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.

Question:

Can someone help me see the contradictions?

Hint: if $$X$$ is Hausdorff, then every subspace of $$X$$ is compact thus closed. What can you say about $$\tau$$ then?
• I have pointed out in my post $\tau$ is the discrete topology. Is this what you wanted me to say? Thanks for your answer! – Pedro Gomes Dec 17 '18 at 13:05
• 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.