Equivalence of Definitions of Local Basis/Local Basis for Open Sets Implies Neighborhood Basis of Open Sets

Theorem

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

Let $x$ be an element of $S$.

Let $\BB$ be a set of open neighborhoods of $x$ such that:

$\forall U \in \tau: x \in U \implies \exists H \in \BB: H \subseteq U$

Then $\BB$ satisfies:

every neighborhood of $x$ contains a set in $\BB$.

Proof

Let $N$ be a neighborhood of $x$.

Then there exists $U \in \tau$ such that $x \in U$ and $U \subseteq N$ by definition.

By assumption, there exists $H \in \BB$ such that $H \subseteq U$.

From Subset Relation is Transitive, $H \subseteq N$.

The result follows.

$\blacksquare$