The definition of "normal" I use in this question is the same as on Wikipedia and as in Munkres' "Topology": any two disjoint closed sets can be separated by disjoint neighbourhoods. Similarly, T${}_1$ means any two distinct points can be separated by neighbourhoods, and T${}_0$ means any two distinct points are topologically distinguishable.
Now, it's easy to see that a normal T${}_1$ space is Hausdorff (since singletons are closed, hence normalcy on singletons reduces to Hausdorffness). I'm curious if that's the weakest additional separation axiom needed in order to guarantee Hausdorfness, or if we can go any further.
Based on my own fiddling around trying to prove it, I would guess that normal + T${}_0$ does not imply Hausdorff, but I haven't looked for a counterexample yet. Are there any normal T${}_0$, non-Hausdorff counterexamples that would disprove this claim?
Best Answer
Let $X=\{0,1\}$, and let $\mathcal{T}=\left\{\emptyset,\{0\},X\right\}$. Then $(X,\mathcal{T})$ is a $T_0$ topological space.
Let $A$ and $B$ be closed sets with $A\cap B=\emptyset$. Then it must be the case that either $A=\emptyset$ or $B=\emptyset$. In the former case, let $U=\emptyset$ and let $V=X$. In the latter case let $U=X$ and let $V=\emptyset$. It follows that $U$ and $V$ are disjoint open sets with $A\subseteq U$ and $B\subseteq V$.
Finally, $(X,\mathcal{T})$ is not $T_1$, since there is no open $U$ with $1\in U$ and $0\notin U$.