Open Set Not in System of Open Neighborhoods Iff Subset of 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 for all $U \in \tau$:

$U \notin \map \UU x$ if and only if $U \subseteq S \setminus \set x^-$

Proof

First note that by definition of closed set:

$S \setminus U$ is a closed set

We have:

\(\ds U \notin \map \UU x\)

\(\leadstoandfrom\)

\(\ds x \notin U\)

Definition of System of Open Neighborhoods

\(\ds \)

\(\leadstoandfrom\)

\(\ds \set x \subseteq S \setminus U\)

Definition of Relative Complement

\(\ds \)

\(\leadstoandfrom\)

\(\ds \set x^- \subseteq S \setminus U\)

Closure of Subset of Closed Set of Topological Space is Subset

\(\ds \)

\(\leadstoandfrom\)

\(\ds U \subseteq S \setminus \set x^-\)

Relative Complement inverts Subsets of Relative Complement

$\blacksquare$