Problem 5, sec. 13, in Munkres' Topology, 2nd ed.: How to prove the assertion if $\mathscr{A}$ is a subbasis? Show that if $\mathscr{A}$ is a basis for a topology on $X$, then the topology generated by $\mathscr{A}$ equals the intersection of all topologies on $X$ that contain $\mathscr{A}$.
This is what I've managed. However, the following stumps me:
Prove the same if $\mathscr{A}$ is a subbasis.
That is, how to show that, the topology generated by $\mathscr{A}$ as a subbasis is the same as the intersection of all topologies on $X$ that contain $\mathscr{A}$?
Let $\mathscr{T}$ be the topology that consists of all unions of finite intersections of sets in $\mathscr{A}$, and let $\mathscr{T}^\prime$ be the topology that is the intersection of all topologies that contain $\mathscr{A}$.
Now $\mathscr{A}$ is a collection of subsets of $X$ whose union is all of $X$, and each set $A \in \mathscr{A}$ is in both $\mathscr{T}$ and $\mathscr{T}^\prime$.
Am I right so far?
And if so, then what next?

Since $\scr A$ is contained in $\scr T$ and $\scr T$ is a topology, it follows that $\scr T^\prime\subseteq \scr T$ (as $\scr T^\prime$ is the intersection of all topologies on $X$ that contains $\scr A$). For the other direction, if $V\notin \scr T^\prime$ then there exist a topology $\cal T$ on $X$ with $\scr A\subseteq \cal T$ such that $V\notin \cal T$. Since any arbitrary union of finite intersections of sets in $\scr A$ belongs to $\cal T$ (because $\cal T$ is a topology and $A\in \scr A\implies A\in \cal T$), it follows that $V$ cannot be a union of finite intersections of sets in $\scr A$ and hence $V\notin \scr T$. Hence $\scr T\subseteq \scr T^\prime$. So both the topologies are equal.
