Local Basis Test

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

Let $x \in S$.

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

Let $\mathcal C$ be a set of open neighborhoods of $x$.

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

Necessary Condition
Let $\mathcal C$ be a local basis.

Let $B \in \mathcal B$.

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

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

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

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

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

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

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