Definition 4.1.5 (Base). Let $(X, \topo)$ be a topological space, then a base for $\topo$ is a family $\cb \subset \topo$ such that:
For every $x \in X$, there exists $U \in \cb$ such that $x \in U$.
For every $x \in X$ and $U \subset X$ open with $x \in U$, there exists $V \in \cb$ such that $x \in V \subset U$.
In which case,
Conversely, if $\cb \subset 2^{X}$ is a family such that:
For every $x \in X$, there exists $U \in \cb$ such that $x \in U$.
For every $x \in X$ and $U, V \in \cb$ such that $x \in U \cap V$, there exists $W \in \cb$ such that $x \in W \subset U \cap V$.
then $\topo(\cb)$ is a topology on $X$, and $\cb$ is a base for $\topo(\cb)$.
Proof. Let $U \in \topo$. If $U = \emptyset$, then $U$ is a union over an empty index set. Otherwise, for each $x \in U$, there exists $V_{x} \in \cb$ such that $x \in V_{x} \subset U$. In which case, $U = \bigcup_{x \in U}V_{x} \in \topo$ and $\topo \subset \topo(\cb)$. On the other hand, since $\cb \subset \topo$, $\topo \supset \topo(\cb)$ by (O3).
For the converse, $\emptyset \in \topo(\cb)$ as a union over an empty index set and $X = \bigcup_{U \in \cb}U \in \topo(\cb)$ by (TB1). For any $\seqi{U}, \seqj{V}\subset \cb$,
Thus to show that $\topo(\cb)$ satisfies (O2), it is sufficient to show that $U \cap V \in \topo(\cb)$ for all $U, V \in \cb$. To this end, for every $x \in U \cap V$, there exists $W_{x} \in \cb$ such that $x \in W_{x} \subset U \cap V$. Therefore $U \cap V = \bigcup_{x \in U \cap V}W_{x} \in \topo(\cb)$, and $\topo(\cb)$ satisfies (O2).
By definition, $\topo(\cb)$ satisfies (O3).$\square$