System of Open Neighborhoods is a Completely Prime Filter

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Theorem

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

Let $x \in S$.

Let $\map \UU x$ denote the system of open neighborhoods of $x$ in $\struct{S, \tau}$.


Then:

$\map \UU x$ is a completely prime filter in the complete lattice $\struct{\tau, \subseteq}$


Proof

$\map \UU x$ satisfies Filter Axiom $\paren{1}$

We have $x \in S$.

By Open Set Axiom $\paren {\text O 3 }$: Underlying Set is Element of Topology:

$S \in \tau$

Hence:

$S \in \map \UU x$


It follows that $\map \UU x$ satisfies Filter Axiom $\paren{1}$.

$\Box$

$\map \UU x$ satisfies Filter Axiom $\paren{2}$

Let $U, V \in \map \UU x$.


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

$U \cap V\in \tau$

By definition of set intersection:

$x \in U \cap V$

Hence:

$U \cap V \in \map \UU x$

From Intersection is Subset:

$U \cap V \subseteq U, U \cap V \subseteq V$


It follows that $\map \UU x$ satisfies Filter Axiom $\paren{1}$.

$\Box$

$\map \UU x$ satisfies Filter Axiom $\paren{3}$

Let $U \in \map \UU x$.

Let $v \in \tau$ such that $U \subseteq V$.


By definition of subset:

$x \in V$

Hence:

$V \in \map \UU x$


It follows that $\map \UU x$ satisfies Filter Axiom $\paren{1}$.

$\Box$

$\map \UU x$ is Proper

By definition of system of open neighborhoods:

$\map \UU x \subseteq \tau$


By empty set:

$x \notin \O$

Hence:

$\O \notin \map \UU x$

From Empty Set is Element of Topology:

$\map \UU x \ne \tau$


It follows that $\map \UU x$ is a proper subset of $\tau$.

$\Box$

$\map \UU x$ satisfies Complete Primeness

Let $\VV \subseteq \tau$:

$\bigcup \VV \in \map \UU x$


By definition of system of open neighborhoods:

$x \in \bigcup \VV$

By definition of set union:

$\exists V \in \VV : x \in V$

Hence:

$\exists V \in \VV : V \in \map \UU x$

$\Box$


It follows that $\map \UU x$ is a completely prime filter by definition.

$\blacksquare$


Also see


Sources

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