Way Above Closures that Way Below Form Local Basis

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Theorem


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

Let $p \in S$.


Then $\set {q^\gg: q \in S \land q \ll p}$ is a local basis at $p$.


Proof

Define $B := \set {q^\gg: q \in S \land q \ll p}$

By Way Above Closure is Open:

$B \subseteq \tau$

By definition of way above closure:

$\forall X \in B: p \in X$

Thus by definition:

$B$ is set of open neighborhoods.


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Let $U$ be an open subset of $S$ such that

$p \in U$

By Open implies There Exists Way Below Element:

$\exists u \in U: u \ll p$

Thus by definition of $B$:

$u^\gg \in B$

By definition of Scott topology:

$U$ is upper.

We will prove that

$u^\gg \subseteq U$

Let $z \in u^\gg$

By definition of way above closure:

$u \ll z$

By Way Below implies Preceding:

$u \preceq z$

Thus by definition of upper section:

$z \in U$

$\blacksquare$


Sources

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