User:Leigh.Samphier/Topology/Frame of Topological Space is Frame

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Theorem

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


Let $\map \Omega T$ denote the frame of topological space $T$.


Then:

$\map \Omega T$ is a frame


Proof

Recall that the frame of topological space $T$ is defined to be:

$\map \Omega T = \struct{\tau, \subseteq}$

where $\subseteq$ denotes the subset relation.


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

Sources

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