Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 2 / 13G / a

Define $\mathcal{T}$ to be the family of all unions of members of $\mathcal{B}$, including the empty union. We will verify that this is a topology and that $\mathcal{B}$ is a basis of a topology. The two conditions play different roles: the covering condition supplies the whole space, while the local intersection condition supplies closure under finite intersections.

By the covering condition, $X={\bigcup }_{B\in \mathcal{B}}B$ belongs to $\mathcal{T}$. The empty set belongs to $\mathcal{T}$ as the empty union. An arbitrary union of sets that are themselves unions of members of $\mathcal{B}$ is still a union of members of $\mathcal{B}$. Thus $\mathcal{T}$ contains $X,\varnothing$ and is closed under arbitrary unions, as required for a topology.

Take $U,V\in \mathcal{T}$ and $x\in U\cap V$. Their representations as unions provide $B_1,B_2\in \mathcal{B}$ with $x\in B_1\subseteq U$ and $x\in B_2\subseteq V$. The intersection condition supplies $B\in \mathcal{B}$ with $x\in B\subseteq B_1\cap B_2\subseteq U\cap V$.

It follows that $U\cap V$ is the union of all members of $\mathcal{B}$ contained in it, and hence belongs to $\mathcal{T}$. Induction gives closure under every finite intersection. Together with part (i), this proves that $\mathcal{T}$ is a topology. Each $B\in \mathcal{B}$ belongs to $\mathcal{T}$ as a union with one member, and every open set is a union of such members by construction. Thus $\mathcal{B}$ is a basis for $\mathcal{T}$. Here the printed subset sign is interpreted as inclusion, allowing equality, as in the usual basis criterion.