Sober space

Remarks

For each point $x$, the closure $\overline{\set{x}}$ is irreducible and closed, with generic point $x$. So the uniqueness part says that $\overline{\set{x}} = \overline{\set{y}}$ implies $x = y$, and a sober space is T0. Every Hausdorff space is sober, because its irreducible subsets are single points.

Example: for a commutative ring $R$, the spectrum $\operatorname{Spec} R$ with the Zariski topology is sober. An irreducible closed set has the form $V(\mathfrak p) = \set{\mathfrak q : \mathfrak q \supseteq \mathfrak p}$ for a unique prime ideal $\mathfrak p$, and $\mathfrak p$ is its generic point.

Non-example: an infinite set $X$ with the cofinite topology is not sober. $X$ is irreducible, since two proper closed sets are finite and their union is not $X$. But each point is closed, so $\overline{\set{x}} = \set{x} \ne X$, and $X$ has no generic point.