# A regular, non-Hausdorff, non-normal space

I'm trying to understand the relationship between the different separation axioms:

• T2 (Hausdorff): two distinct points can be separated by open sets
• Regular: a closed set and a point not in it can be separated by open sets
• Normal: two disjoint closed sets can be separated by open sets

I know that T4 (normal + T2) implies regular, so if a space is not regular, it is either not Hausdorff or not normal. I can't find out if it's possible to have a space that is regular, but not Hausdorff and not normal.

• regular + $T_0$ already implies $T_1$ and regular + $T_1$ implies $T_2$, so an example cannot be $T_0$ either. The usual go-to example of an an indiscrete space does not work (it's normal). – Henno Brandsma May 13 '18 at 15:11
• Another "restriction" : a regular and second countable space is normal, so that already rules out all finite spaces and many more. – Henno Brandsma May 13 '18 at 20:44

Take your favourite regular non-normal space $X$, for example the Sorgenfrey plane, now consider the indiscrete space $Y=\{0,1\}$, finally $X\times Y$ is regular because product of regular spaces (you can see it here, despite he says that a regular space is also $T_1$, the same proof works perfectly without this hypotesis), it isn't Hausdorff because $Y$ is not. It isn't also non-normal because if $A,B\subset X$ are closed set that you cannot separate with open, also $A\times Y$ and $B\times Y$ will be closed set of $X\times Y$ that you cannot separate with open, because the nonempty open are of the form $A\times B$ where $A\subset X$ and necessarily $B=Y$.