A Union of all Neighborhood Bases is a Topological Basis

Last updated Nov 1, 2022

Suppose $(X, \tau)$ is a Topological Space and for each $x \in X$, $\mathcal{B}{x}$ is a Neighborhood Basis for $x$. Then $$\mathcal{B} = \bigcup\limits{x \in X} \mathcal{B}_{x}$$ is a Topological Basis for $X$.

Suppose $U \subset X$ is Open. Then for each $x \in U$, there exists some $B_{x} \in \mathcal{B}{x}$ so that $B{x} \subset U$ (by definition of Neighborhood Basis). Thus

$$\bigcup\limits_{x \in U} B_{x} \subset U$$

On the other hand, for each $x \in U$, we have $x \in B_{x}$ since $B_{x}$ is in a Neighborhood Basis for $x$. Thus

$$U \subset \bigcup\limits_{x \in U} B_{x}$$

Therefore $U = \bigcup\limits_{x \in U} B_{x}$, completing the proof. $\blacksquare$