Urysohn space

Remarks

Equivalently, any two distinct points have disjoint closed neighbourhoods. Every Urysohn space is Hausdorff, since the interiors of $\overline{U}$ and $\overline{V}$ are disjoint open sets around $x$ and $y$. Every regular Hausdorff space is Urysohn.

Example: $\R$ is Urysohn. For $x < y$, put $r = (y - x)/3$; the closures $[x - r, x + r]$ and $[y - r, y + r]$ of the open intervals of radius $r$ are disjoint.

Non-example: an infinite set with the cofinite topology is not Urysohn. Any two nonempty open sets in it meet, so even the Hausdorff condition fails.

Notation: $T_{2\frac12}$. A space in which distinct points are separated by a continuous function $f \colon X \to [0, 1]$ is called completely Hausdorff; some authors call that an Urysohn space.