Local Basis Test

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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 if and only if:
$\forall B \in \BB \implies \exists C \in \CC: C \subseteq B$


Proof

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:

$\exists C \in \CC : C\subseteq B$.

$\Box$


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.

$\blacksquare$