Proposition 3.3.4 (Characteristics of Neighbourhoods, [Proposition 1.2.2, Bou95]).label Let $(X,\Tc)$ be a topological space, then for each $x\in X$
- (V1)
For each $V\in\cali{N}_{\Tc}(x)$, if $W\supf V$ then $W\in\cali{N}_{\Tc}(x)$
- (V2)
For each $A,B\in\cali{N}_{\Tc}(x)$, $A\cap B\in\cali{N}_{\Tc}(x)$
- (V3)
For each $A\in\cali{N}_{\Tc}(x),x\in A$
- (V4)
For each $V\in\cali{N}_{\Tc}(x)$, there exists $W\in\cali{N}_{\Tc}(x)$ such that $V\in\cali{N}_{\Tc}(y)$ for all $y\in W$
Conversely, if $\cali{N}:X\to\cali{P}(X)$ is a mapping such that
- (1)
$\cali{N}(x)\neq\emp$ for all $x\in X$
- (2)
$\cali{N}(x)$ satisfies (V1-4)
then there uniquely exists a topology $\Tc\suf\cali{P}(X)$ such that $\cali{N}=\cali{N}_{\Tc}$.
Proof.
- (V1)
$\suf$ is transitive
- (V2)
$\Tc$ is closed under finite intersection
- (V3)
$\suf$ is transitive
- (V4)
Follows from Lemma 3.3.3
Conversely let $\Tc\define \curl{U\suf X:\fall x\in U, U\in\cali{N}(x)}$. Firstly $\emp\in\Tc$ vacuously. For any $x\in X$, there exists $V\in\cali{N}(x)$ and $V\suf X$ so $X\in\Tc$ by (V1). Let $U,V\in\Tc$ such that $U\cap V\neq \emp$ then for any $x\in U\cap V$ we have $U,V\in \cali{N}(x)$ so $U\cap V\in \cali{N}(x)$ by (V2) hence $U\cap V\in\Tc$. Let $\curl{U_i}_{i\in I}\suf \Tc$ then for any $x\in \cups{}{i\in I }U_{i}$, $x\in U_{i}$ for some $i$ so $U\in\cali{N}(x)$ by (V1) showing that $\Tc$ is a topology on $X$. Fix $x\in X$. Let $V\in\cali{N}_{\Tc}(x)$, then there exists $U\in\Tc$ such that $x\in U\suf V$ in particular $U\in\cali{N}(x)$ so $V\in\cali{N}(x)$ by (V1). OTOH suppose $V\in\cali{N}(x)$ then by (V4), there exists $U_{0}\in\cali{N}(x)$ such that $V\in\cali{N}(y)$ for all $y\in U_{0}$. Define
then $U\supf U_{0}$ and $U\in\cali{N}(x)$ by (V1). For each $y\in U$, by (V4) there exists $W\in\cali{N}(y)$ suhc that $V\in\cali{N}(z)$ for all $z\in W$. Thus $W\suf U$ and $U\in\cali{N}(y)$ by (V1). Lastly if $\cali{R}$ is a topology such that $\cali{N}_{\cali{R}}=\cali{N}$ then by Lemma 3.3.3
$\square$
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