Scott Topological Lattice is T0 Space

Theorem

Let $T = \left({S, \preceq, \tau}\right)$ be a complete topological lattice with Scott topology.

Then $T$ is a $T_0$ space.

Proof

Let $x, y \in S$ such that

$x \ne y$

By Closure of Singleton is Lower Closure of Element in Scott Topological Lattice:

$\left\{ {x}\right\}^- = x^\preceq$ and $\left\{ {y}\right\}^- = y^\preceq$

Thus by Lower Closures are Equal implies Elements are Equal:

$\left\{ {x}\right\}^- \ne \left\{ {y}\right\}^-$

Hence by Characterization of T0 Space by Distinct Closures of Singletons:

$T$ is $T_0$ space.

$\blacksquare$

Sources

• 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott: A Compendium of Continuous Lattices

• Mizar article WAYBEL11:10