From Wikipedia:
Given a topological space $(X, \tau)$, a base for the topology $\tau$ is a family $\mathcal{B} \subseteq \tau$ of open sets such that every open set of the topology can be represented as the union of some subfamily of $\mathcal{B}$.
Equivalently, a family $\mathcal{B}$ of subsets of $X$ is a base for the topology $\tau$ if and only if $\mathcal{B} \subseteq \tau$ and for every open set $U$ in $X$ and point $x\in U$ there is some basic open set $B \in \mathcal{B}$ such that $x \in B \subseteq U$.
Question: Why are the above definitions of a base equivalent?
Best Answer
Suppose $\mathcal{B}$ is a base under definition 1. Consider some $x \in U$ where $U$ is open. Then write $U = \bigcup Z$, where $Z$ is a subfamily of $\mathcal{B}$. Since $x \in U$, there exists some $B \in Z$ such that $x \in B$. Then $B \subseteq \bigcup Z = U$, and thus $x \in B \subseteq U$. Furthermore, $B \in Z \subseteq \mathcal{B}$, so $B \in \mathcal{B}$.
For the other direction, I claim that $U = \bigcup \{B \in \mathcal{B} \mid B \subseteq U\}$. Clearly, we have $\bigcup \{B \in \mathcal{B} \mid B \subseteq U\} \subseteq U$. Now consider an arbitrary $x \in U$. Then there exists $B \in \mathcal{B}$ such that $x \in B \subseteq U$. Then $X \in \bigcup \{B \in \mathcal{B} \mid B \subseteq U\}$. So $U \subseteq \bigcup \{B \in \mathcal{B} \mid B \subseteq U\}$.