Let $X$ be a set and $S \subset \mathcal{P}(X)$. Consider the topology induced by $S$, $\tau(S)$ (defined to be the collection of unions of finite intersections of elements of $S$). We know that $S$ is a subbase for $\tau(S)$. But under what conditions would $S$ actually also be a base for $\tau(S)$?
The professor seem to have hinted that if the intersection of any two elements of $S$ is also in $S$, then $S$ will also be a base for $\tau(S)$, and not just a subbase for $\tau(S)$. Is this true? Why is this true? Under what other conditions will $S$ also be a base for $\tau(S)$?
The usual ones to recognise a base:
$$\bigcup S = X$$ and $$\forall S_1,S_2 \in S: \forall x \in S_1 \cap S_2: \exists S_x \in S: x \in S_x \subseteq S_1 \cap S_2$$
The last condition is saying that the intersection of subbasic elements is a union of subbasic elements.