# Scott Topological Lattice is T0 Space

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## 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