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.
$\blacksquare$