Union of Local Bases is Basis
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Theorem
Let $T = \struct {X, \tau}$ be a topological space.
For each $x \in X$, let $\BB_x$ be a local basis at $x$ which consists entirely of open sets.
Then $\ds \BB = \bigcup_{x \mathop \in X} \BB_x$ is a basis for the topology $\tau$.
Proof
Let $U \in \tau$ be any open set of $T$.
Consider any $x \in U$.
Then, by definition of local basis, there exists a $B_x \in \BB_x$ such that $B_x \subseteq U$.
Also by definition of local basis:
- $x \in B_x$
So by Set Union Preserves Subsets:
- $U = \ds \bigcup_{x \mathop \in U} \set x \subseteq \bigcup_{x \mathop \in U} B_x$
Also, since $B_x \subseteq U$ for all $x$, by Union is Smallest Superset: General Result:
- $\ds \bigcup_{x \mathop \in U} B_x \subseteq U$
Thus by definition of set equality:
- $\ds U = \bigcup_{x \mathop \in U} B_x$
So, by definition, $\ds \BB = \bigcup_{x \mathop \in X} \BB_x$ is a basis for the topology $\tau$.
$\blacksquare$