System of Open Neighborhoods are Equal Iff Singleton Closures are Equal

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Theorem

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

Let $x, y \in S$ such that $x \ne y$.

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

Let $\set x^-$ and $\set y^-$ denote the topological closures of $\set x$ and $\set y$ respectively.

Then:

$\map \UU x = \map \UU y$ if and only if $\set x^- = \set y^-$

Proof

We have

\(\ds \map \UU x = \map \UU y\)

\(\leadstoandfrom\)

\(\ds \tau \setminus \map \UU x = \tau \setminus \map \UU y\)

Equal Relative Complements iff Equal Subsets

\(\ds \)

\(\leadsto\)

\(\ds \bigcup \paren{\tau \setminus \map \UU x} = \bigcup \paren{\tau \setminus \map \UU y}\)

Definition of Set Union

\(\ds \)

\(\leadstoandfrom\)

\(\ds S \setminus \set x^- = S \setminus \set y^-\)

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

\(\ds \)

\(\leadstoandfrom\)

\(\ds \set x^- = \set y^-\)

Equal Relative Complements iff Equal Subsets

It remains to show, if:

$S \setminus \set x^- = S \setminus \set y^-$

then:

$\tau \setminus \map \UU x = \tau \setminus \map \UU y$

Let:

$S \setminus \set x^- = S \setminus \set y^-$

We have

\(\ds \forall U \in \tau: \, \)

\(\ds U \in \tau \setminus \map \UU x\)

\(\leadstoandfrom\)

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

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

\(\ds \)

\(\leadstoandfrom\)

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

by hypothesis $S \setminus \set x^- = S \setminus \set y^-$

\(\ds \)

\(\leadstoandfrom\)

\(\ds U \in \tau \setminus \map \UU y\)

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

By definition of set equality:

$\tau \setminus \map \UU x = \tau \setminus \map \UU y$

The result follows.

$\blacksquare$