Characterization of Analytic Basis by Local Bases

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

Let $P$ be a set of subsets of $S$ such that
 * $P \subseteq \tau$

and
 * for all $p \in S$: there exists local basis $B$ at $p: B \subseteq P$

Then $P$ is basis of $T$.

Proof
By assumption:
 * $P \subseteq \tau$

Let $U$ be an open subset of $S$.

Define:
 * $X := \set {V \in P: V \subseteq U}$

By definition of subset:
 * $X \subseteq P$

We will prove that:
 * $\forall u \in S: u \in U \iff \exists Z \in X: u \in Z$

Let $u \in S$.

We will prove that:
 * $u \in U \implies \exists Z \in X: u \in Z$

Assume that:
 * $u \in U$

By assumption:
 * there exists local basis $B$ at $u: B \subseteq P$.

By definition of local basis:
 * $\exists V \in B: V \subseteq U$

Thus by definitions of subset and $X$:
 * $V \in X$

Thus by definition of local basis:
 * $u \in V$

Assume that:
 * $\exists Z \in X: u \in Z$

By definition of $X$:
 * $Z \subseteq U$

Thus by definition of subset:
 * $u \in U$

Thus by definition of union:
 * $U = \bigcup X$

Hence $P$ is basis of $L$.