A space is called completely Hausdorff if for any two distinct points $x$ and $y$ there are disjoint closed neighborhoods of $x$ and $y$ . This is called the $T_{2{1\over 2}}$ axiom . So easily can be seen that every regular space is completely Hausdorff. And a completely Hausdorff space is Hausdorff.

Now for the first case I guess $\mathbb R_k$ space will serve as a counter example. This is Hausdorff and if I have not made silly mistakes in deriving, completely Hausdorff as well but not regular.

I need to find an example of a space that is Hausdorff but not completely Hausdorff.

  • $\begingroup$ @bof : Obviously typo was at the title as Completely Hausdorff is always Hausdorff, you know that right $?$ $\endgroup$ – user118494 Jan 3 '16 at 9:12
  • $\begingroup$ Engelking, General Topology,calls a $T(2\frac {1}{2})$ space an Urysohn space. In case you try to find this in,e.g, Wikipedia. $\endgroup$ – DanielWainfleet Jan 3 '16 at 9:14

The minimal Hausdorff topology from Steen and Seebach is an example.

Let $A$ be the linearly ordered set $\{0,1,2,\ldots,\omega,\ldots, -3,-2,-1\}$ in the order topology. $X = A \times \omega \cup \{a, -a\}$ where $a$ and $-a$ are new points with neighbourhood base $M^{+}_n(a) = \{a\} \cup \{(i,j) : i \lt \omega, j > n\}$ for $a$ and $M^{-}_n(-a) = \{-a\} \cup \{(i,j): i \gt \omega, j \gt n\}$ for $-a$. Then $X$ is Hausdorff and $a$ and $-a$ do not have disjoint closed neighbourhoods.


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