Topology with Set Inclusion Forms a Frame

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Theorem

Let $T = \struct{S, \tau}$ be a topological space.

Let $\map \Omega T = \struct{\tau, \subseteq}$ be the topology $\tau$ with set inclusion.

Then:

$\map \Omega T$ is a frame

Corollary

$\map \Omega T$ is a locale

Proof

From Topology forms Complete Lattice:

$\map \Omega T$ is a complete lattice

Furthermore:

$\forall \TT \subseteq \tau : \sup \TT = \bigcup \TT, \inf \TT = \paren{\bigcap \TT}^\circ$

where $\paren{\bigcap \TT}^\circ$ denotes the interior of $\bigcap \TT$

From Interior of Open Set and Open Set Axiom $\paren {\text O 2 }$: Pairwise Intersection of Open Sets:

$\forall U, V \in \tau : \paren{U \cap V}^\circ = U \cap V$

Hence:

$\forall U, V \in \tau : U \wedge V = U \cap V$

From Intersection Distributes over Union (General Result):

$\map \Omega T$ satisfies the infinite join distributive law

Hence $\map \Omega T$ is a frame by definition.

$\blacksquare$

Also see

• Continuous Mapping Induces Frame Homomorphism

Sources

• 1982: Peter T. Johnstone: Stone Spaces: Chapter II: Introduction to Locales, $\S1.1$