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

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## 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:

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

Then $\BB$ satisfies:

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

## Proof

Let $U \in \tau$ such that $x \in U$.

From Set is Open iff Neighborhood of all its Points then $U$ is a neighborhood of $x$.

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

The result follows.

$\blacksquare$