Sober Space is T0 Space

Theorem

Let $\struct {S, \tau}$ be a sober space.

Then:

$\struct {S, \tau}$ is a $T_0$ (Kolmogorov) space

Proof

Let $x, y \in S : x \ne y$.

From Complement of Singleton Closure is Meet-Irreducible:

$S \setminus \set x^-$ is a meet-irreducible open set

By definition of a sober space:

$S \setminus \set x^-$ is uniquely defined by $x$

Hence:

$S \setminus \set x^- \ne S \setminus \set y^-$

From Equal Relative Complements iff Equal Subsets:

$\set x^- \ne \set y^-$

Since $x, y$ were arbitrary, it follows that:

$\forall x, y \in S : x \ne y \implies \set x^- \ne \set y^-$

From Characterization of T0 Space by Distinct Closures of Singletons:

$\struct {S, \tau}$ is a $T_0$ (Kolmogorov) space

$\blacksquare$

Also see

• T2 Space is Sober Space

Sources

• 2012: Jorge Picado and Aleš Pultr: Frames and Locales: Chapter $1$: Spaces and Lattices of Open Sets, $\S 1$ Sober spaces, Notes $1.2 (1)$