Regular space which is not Hausdorff I know that normality in the absence of $T_{1}$ does not imply regularity (Sierpinski space being a counterexample as it is vacuously normal but not regular). I have the feeling that similarly regularity in the absence of $T_{1}$ does not imply Hausdorff. I tried thinking of a counter example but obviously every such counter example mustn't be $T_{1}$ and I'm not familiar with many spaces which are not $T_{1}$ (the only ones that comes to mind are Trivial topologies and the Sierpinski space). 
Help would be appreciated :)
 A: Let $X=\{0,1,2,3\}$, and endow $X$ with the topology
$$\tau=\big\{\varnothing,\{0,1\},\{2,3\},X\big\}\;;$$
then $\langle X,\tau\rangle$ is regular but not Hausdorff. ($X$ is homeomorphic to the product of the discrete two-point space with the indiscrete two-point space.)
Added: Given a space $\langle X,\tau\rangle$, we can define an equivalence relation $\sim$ on $X$ by setting $x\sim y$ iff $x$ and $y$ have the same open nbhds. If we identify equivalent points (i.e., take the quotient $X/\sim$), we always get a $T_0$-space. An $R_0$-space is one in which we get a $T_1$-space. As you can see, the example above is $R_0$: the quotient $X/\sim$ is just a discrete two-point space. You can start with any $T_3$-space and ‘fatten up’ some points to get an $R_0$, regular space that is not Hausdorff.
The Wikipedia article on separation axioms has definitions of some of the more obscure ones, including $R_0$, as well as of the familiar ones.
A: Depending on the textbook's definition of regular or normal, it may be required that the space is $T_1$ to begin with.  That would imply that for each $x\in X$, then $\{x\}$ must be closed...which would not be true in this specific example.  It truly depends on how regular or normal are defined.
