Way Above Closure is Open

Theorem

Let $L = \struct {S, \preceq, \tau}$ be a complete continuous topological lattice with Scott topology.

Let $x \in S$.

Then $x^\gg$ is open

where $x^\gg$ denotes the way above closure of $x$.

Proof

By Way Above Closure is Upper:

$x^\gg$ is upper.

We will prove that

$x^\gg$ is inaccessible by directed suprema.

Let $D$ be a directed subset of $S$ such that

$\sup D \in x^\gg$

By definition of way above closure:

$x \ll \sup D$

By Way Below iff Second Operand Preceding Supremum of Directed Set There Exists Element of Directed Set First Operand Way Below Element:

$\exists d \in D: x \ll d$

By definition of way above closure:

$d \in x^\gg$

By definitions of intersection and non-empty set:

$x^\gg \cap D \ne \O$

$\Box$

Thus by definition of Scott topology:

$x^\gg$ is open.

$\blacksquare$

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 WAYBEL11:36