4.4 Neighbourhoods
Definition 4.4.1 (Neighbourhood). Let $(X, \topo)$ be a topological space and $A \subset X$. A neighbourhood of $A$ is a set $V \supset X$ where there exists $U \in \topo$ such that $A \subset U \subset V \subset X$.
The set $\cn_{X, \topo}(A) = \cn_{X}(A) = \cn_{\topo}(A) = \cn(A)$ denotes the collection of all neighbourhoods of $A$, and $\cn^{o}(A)$ denotes the set of open neighbourhoods of $A$.
If $A = \bracs{x}$ is a single point, the above definition and notation applies to $x$ directly.
Definition 4.4.2 (Fundamental System of Neighbourhoods). Let $X$ be a topological space and $A \subset X$. A family $\fB \subset \cn(A)$ is a fundamental system of neighbourhoods/neighbourhood base at $A$ if for every $U \in \cn(A)$, there exists $V \in \fB$ such that $V \subset U$.
Lemma 4.4.3. Let $(X, \topo)$ be a topological space, then $U \subset X$ is open if and only if $U \in \cn_{\topo}(x)$ for all $x \in U$.
Proof. Suppose that $U \in \cn_{\topo}(x)$ for all $x \in U$. For each $x \in U$, there exists $V_{x} \in \topo$ with $x \in V_{x} \subset U$. Thus $U = \bigcup_{x \in U}V_{x} \in \topo$.$\square$
Proposition 4.4.4. Let $(X, \topo)$ be a topological space, then for each $x \in X$,
For every $V \in \cn_{\topo}(x)$, $W \in \cn_{\topo}(x)$ for all $X \supset W \supset V$.
For any $A, B \in \cn_{\topo}(x)$, $A \cap B \in \cn_{\topo}(x)$.
For every $A \in \cn_{\topo}(x)$, $x \in A$.
For every $V \in \cn_{\topo}(x)$, there exists $W \in \cn_{\topo}(x)$ such that $V \in \cn_{\topo}(y)$ for all $y \in W$.
Conversely, if $\cn: X \to 2^{X}$ is a mapping such that
$\cn(x) \ne \emptyset$ for all $x \in X$.
$\cn(x)$ satisfies (F1), (F2), (V1), and (V2).
then there exists a unique topology $\topo \subset 2^{X}$ such that $\cn = \cn_{\topo}$.
Proof. Existence: Suppose that $\cn: X \to 2^{X}$ is a mapping such that (F1), (F2), (V1), and (V2) holds for all $x \in X$. Let
Firstly, $\emptyset$ satisfies the condition vacuously, so $\emptyset \in \topo$. For any $x \in X$, $\cn(x)$ is non-empty and there exists $V \in \cn(x)$. Since $X \supset V$, $X \in \topo$ by (F1).
Let $U, V \in \topo$. If $U \cap V \ne \emptyset$, then there exists $x \in U \cap V$, and $U, V \in \cn(x)$. For any $y \in U \cap V$, $U, V \in \cn(y)$ by (F2). Hence $U \cap V \in \cn(y)$ as well. Thus $U \cap V \in \topo$.
Now, let $\seqi{U}\subset \topo$ and $U = \bigcup_{i \in I}U_{i}$. For any $x \in U$, there exists $i \in I$ such that $x \in U_{i}$. In which case, $U \in \cn(x)$ by (F1), and $U \in \cn(x)$. Therefore $U \in \topo$, and $\topo$ forms a topology on $X$.
Agreement with $\cn_{\topo}$: Fix $x \in X$. Let $V \in \cn_{\topo}(x)$, then there exists $U \in \topo$ such that $x \in U \subset V$. Since $U \in \topo$ and $x \in U$, $U \in \cn(x)$. By (F1), $V \in \cn(x)$ as well, so $\cn_{\topo}(x) \subset \cn(x)$.
Conversely, let $V \in \cn(x)$. By (V4), there exists $U_{0} \in \cn(x)$ such that $V \in \cn(y)$ for all $y \in U_{0}$. Define
then $U \supset U_{0}$ and $U \in \cn(x)$ by (V1). Let $y \in U$, then (V4) implies that there exists $W \in \cn(y)$ such that $V \in \cn(z)$ for all $z \in W$. Thus $W \subset U$ and $U \in \cn(y)$ by (F1).
Uniqueness: Let $\mathcal{R}$ be a topology such that $\cn_{\mathcal{R}}= \cn$, then by Lemma 4.4.3,
so $\topo$ is unique.$\square$