Definition 3.1.5 (Base of Topology).label Let $X$ be a set then $\cali{B}\suf\cali{P}(X)$ is a base if it satisfies:
- (B1)
For each $x\in X$ there exists $B_{x}\in\cali{B}$ such that $x\in B_{x}$ i.e. $\cali{B}$ is a covering of $X$
- (B2)
If $x\in B_{1},B_{2}\in\cali{B}$ then there exists $B_{3}\in\cali{B}$ such that $x\in B_{3}\suf B_{1}\cap B_{2}$
then the topology generated by $\cali{B}$ defined as
is a topology on $X$. Suppose $X$ has a topology $\tau$ then we say $\cali{B}$ is a base for $\tau$ if $\cali{B}\suf \tau$ and
- (B’1)
$\cali{B}$ satisfies (B1)
- (B’2)
For each $x\in X, U\in\tau$ such that $x\in U$, there exists $B\in\cali{B}$ such that $x\in B\suf U$
in which case $\tau=\tau(\cali{B})$.
Proof. If $\cali{B}$ is a base then $\emp=\cups{}{\emp}\in\tau(\cali{B})$ and by (B1) $X=\cups{}{x\in X}B_{x}\in \tau(\cali{B})$ showing (O1). Observe that for any $\curl{B_i}_{i\in I},\curl{B_j}_{j\in J}\suf\cali{B}$ and $*\in \curl{\cap,\cup}$ we have
by (B2) this observation shows (O3) and
shows (O2).
Suppose now that $\cali{B}$ is a base for $\tau$ then by (O3) and $\cali{B}\suf \tau$ we have $\tau(\cali{B})\suf \tau$. For any $U\in \tau$ either $U=\emp\in\tau(\cali{B})$, or $U\neq \emp$. In which case by (B’2) for each $x\in U$ pick $B_{x}\in\cali{B}$ such that $x\in B_{x}\suf U$ then $U=\cups{}{x\in U }B_{x}\in \tau(\cali{B})$ as desired.$\square$
Comments
You probably don’t need $* \in \{\cap, \cup\}$ if you only use it for $\cap$. The union for (O2) shouldn’t be over the product $I_{j} \times J$, since $I_{j}$ depends on $J$. Instead, it should be something like: Let $\{U_{j}\}_{j \in J}\subset \tau(\mathcal{B})$. For each $j \in J$, let $I_{j}$ such that $U_{j} = \bigcup_{i \in I_j}B_{i, j}$, then