# Infinite topological space with cofinite topology is not Hausdorff

I found a proof to the question, but mine is completely different (sort of). Is this correct?

If $X$ were Hausdorff, then consider $u,v \in X$ with disjoint neighbourhoods $U, V$ that separates the points. But this would mean $(U \cap V)^c = \emptyset^c \iff U^c \cup V^c = X$, implying that the union of two finite sets equals $X$, an infinite set. So $X$ cannot possibly be Hausdorff.

The proof I read strategies by showing one of the disjoint open set cannot be in the complement of the other, which is a finite set. But this contradicts the other one being in the topology.

• Your proof is perfectly correct, except for misspelling "Hausdorff" twice. – bof Aug 23 '14 at 0:07
• Not twice.... Thrice.... As you yourself corrected three times. – Pranita Gupta Feb 1 at 3:53