Local Basis Test

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

Let $x \in S$.

Let $\BB$ be a local basis for $x$ in $\struct {S, \tau}$.

Let $\CC$ be a set of open neighborhoods of $x$.

Then:
 * $\CC$ is a local basis :
 * $\forall B \in \BB \implies \exists C \in \CC: C \subseteq B$

Necessary Condition
Let $\CC$ be a local basis.

Let $B \in \BB$.

Since $\BB$ is a local basis, by the definition of a local basis then $B$ is open.

Since $\CC$ is a local basis, by the definition of a local basis then:
 * $\exists C \in \CC : C\subseteq B$.

Sufficient Condition
Let $\CC$ satisfy:
 * $\forall B \in \BB \implies \exists C \in \CC: C\subseteq B$

Let $U \in \tau$ and $x \in U$.

By the definition of a local basis then:
 * $\exists B \in \BB : B\subseteq U$

By the assumption then:
 * $\exists C \in \CC: C\subseteq B \subseteq U$

By the definition of a local basis then $\CC$ is a local basis.