This question is from Munkres' Topology, section 13, exercise 5. I ask for verification and/or comments upon mistakes and inaccuracies.
Let $\mathcal{A}$ be a basis for a topology on $X$. We are to show that the topology generated by $\mathcal{A}$ is the intersection of all topologies on $X$ containing $\mathcal{A}$.
Here is my claimed solution:
Let $\{\tau_{\alpha}\}$ be the family of topologies containing $\mathcal{A}$. Consider the incersection $\bigcap_{a \in \alpha}\tau_{a}$. It is clear that this intersection contains $\mathcal{A}$, and moreover, is a topology, since for any two elements that are shared, intersections and unions must be shared, by the definition of a topology.
We proceed to show that $\bigcap_{a\in\alpha}\tau_a \subset \tau_{\mathcal{A}}$. Let $U$ be an arbitrary element of $\bigcap_{a\in\alpha}\tau_a$. We see that $U$ must either be in $\mathcal{A}$ or be a union or intersection of elements of $\mathcal{A}$, since $\bigcap_{a\in\alpha}\tau_a$ is clearly the largest topology containing every topology containing $\mathcal{A}$, and by the definition of a topology, this must be exactly the elements of $\mathcal{A}$ and unions and intersections therein. $\bigcap_{a\in\alpha}\tau_a \supset \tau_{\mathcal{A}}$ follows directly.
What I am not sure about here is "…by the definition of a topology, this must be exactly the elements of $\mathcal{A}$ and unions and intersections therein." Since the intersection is the largest topology containing every $\tau_\alpha$, any other element could imply new elements by unions and intersections, creating a new topology not necessarily contained in every $\tau_\alpha$. While I am somewhat confident that my intuition here is correct, I feel that my argument lacks a certain precision.
Best Answer
The argument is fine, though it is phrased a little bit unclearly. Here is a way to make it more precise. If $U\in \tau_{\mathcal{A}}$, then it is a union of basis elements, say $U=\bigcup_{i\in I}A_i$, where $A_i \in \mathcal{A}$ for all $i$. Since $\bigcap_{a\in A}\tau_a$ contains $\mathcal{A}$, we have $A_i\in \bigcap_{a\in A}\tau_a$ for all $i$ and since $\bigcap_{a\in A}\tau_a$ is a topology, we also have $U=\bigcup_{i\in I}A_i\in \bigcap_{a\in A}\tau_a$.