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.
$\blacksquare$