Union of Local Bases is Basis

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Theorem

Let $T = \left({X, \tau}\right)$ be a topological space.

For each $x \in X$, let $\mathcal B_x$ be a local basis at $x$ which consists entirely of open sets.

Then $\displaystyle \mathcal B = \bigcup_{x \mathop \in X} \mathcal B_x$ 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 $B_x \in \mathcal B_x$ such that $B_x \subseteq U$.

Also by definition of local basis, $x \in B_x$.

So by Set Union Preserves Subsets:

$U = \displaystyle \bigcup_{x \mathop \in U} \left\{{x}\right\} \subseteq \bigcup_{x \mathop \in U} B_x$

Also, since $B_x \subseteq U$ for all $x$, by Union is Smallest Superset: General Result:

$\displaystyle \bigcup_{x \mathop \in U} B_x \subseteq U$

Thus by definition of set equality:

$\displaystyle U = \bigcup_{x \mathop \in U} B_x$

So, by definition, $\displaystyle \mathcal B = \bigcup_{x \mathop \in X} \mathcal B_x$ is a basis for the topology $\tau$.

$\blacksquare$


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