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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 :)

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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.

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    $\begingroup$ The smallest example is the set $\{ 0, 1 \}$ with the indiscrete (trivial) topology. The space is obviously not $T_1$. It is, however, regular. $\endgroup$ – dfeuer Jun 24 '13 at 20:13
  • $\begingroup$ @dfeuer: True, but I wanted an example that wasn’t vacuously regular. (I could still have omitted the point $3$, of course, but I liked the more uniform construction.) $\endgroup$ – Brian M. Scott Jun 24 '13 at 20:15
  • $\begingroup$ Interestingly, this $T_0$-space with the quotient map is universal in the sense that every map from $X$ to a $T_0$-space $Y$ factors uniquely through it :-) $\endgroup$ – Stefan Hamcke Jun 24 '13 at 20:16
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    $\begingroup$ It's not actually vacuous, technically speaking, because there is a closed set, $\varnothing$, and a point not in it, $0$. But certainly it is an uninformatively trivial example. $\endgroup$ – dfeuer Jun 24 '13 at 20:18
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    $\begingroup$ Very nice example Brian, as usual. Somehow I completely overlooked the fact the trivial topology is regular for some reason. Thanks to all repliers. $\endgroup$ – Serpahimz Jun 24 '13 at 20:18
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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.

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