Union of Open Sets Not in System of Open Neighborhoods is Complement of Singleton Closure

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Theorem

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

Let $x \in S$.

Let $\map \UU x$ denote the system of open neighborhoods of $x$.

Let $\set x^-$ denote the topological closure of $\set x$.

Then:

$\bigcup \set{U \in \tau : U \notin \map \UU x} = S \setminus \set x^-$

Proof

From Open Set Not in System of Open Neighborhoods Iff Subset of Complement of Singleton Closure:

$\forall U \in \tau : \paren{U \notin \map \UU x \iff U \subseteq S \setminus \set x^-}$

From Union of Subsets is Subset:

$\bigcup \set{U \in \tau : U \notin \map \UU x} \subseteq S \setminus \set x^-$

From Topological Closure is Closed:

$\set x^-$ is a closed set

From Complement of Closed Set is Open Set:

$S \setminus \set x^- \in \tau$

Since $x \notin S \setminus \set x^-$ then:

$S \setminus \set x^- \in \set{U \in \tau : U \notin \map \UU x}$

From Set is Subset of Union:

$S \setminus \set x^- \subseteq \bigcup \set{U \in \tau : U \notin \map \UU x}$

By definition of set equality:

$\bigcup \set{U \in \tau : U \notin \map \UU x} = S \setminus \set x^-$

$\blacksquare$