Scott Topology equals to Scott Sigma

Theorem

Let $\struct {T, \preceq, \tau}$ be a up-complete topological lattice with Scott topology.

Then $\tau = \map \sigma {T, \preceq}$

where $\map \sigma L$ denotes the Scott sigma of $L$.

Proof

This follows by Open iff Upper and with Property (S) in Scott Topological Lattice and definition Scott sigma.

$\blacksquare$

Sources

• Mizar article WAYBEL14:23