Let $X$ be a set, and $S \subset P(X)$. Let $T(S)$ be the topology generated by $S$, i.e. the smallest topology containing $S$. Is $S$ a subbase for $T(S)$?
Definition of subbase: $B$ is a subbase if the collection of open subsets consisting of all finite intersections of elements of $B$ together with $X$ forms a basis.
It does not seem obvious to me. Let $U \in T(S)$ be an arbitrary open. If $S$ is indeed a subbase, then $U$ is a union of (finite) intersections of elements in $S$. I.e. we only need one iterarion of taking intersections and unions of $S$.
But to form $T(S)$ we need at least infinite iterations of taking (finite) intersections and unions of $S$.
Note: there seemed to be two definitions of subbase: subbase. With the one definition the question is true obviuous, by the other it is not.
Best Answer
If $X$ is a set and $\mathcal S\subseteq\wp(X)$ then it always generates a topology in the sense that a smallest topology that contains $\mathcal S$ as subcollection exists.
This merely because the intersection of all topologies on $X$ that contain $\mathcal S$ is again a topology that contains $\mathcal S$.
In that situation $\mathcal S$ is by definition a subbase of the generated topology $\tau(\mathcal S)$.
So a proof that $S$ is a subbase of $\tau(\mathcal S)$ is actually not needed.
What can be proved is that in this situation the topology $\tau(\mathcal S)$ can be described as the collection of sets that can be written as a union of finite intersections of $\mathcal S$.
For that it is enough to show that the latest collection is indeed a topology that contains $\mathcal S$ and secondly that every topology that contains $\mathcal S$ as a subcollection will also contain this collection as a subcollection.
edit:
Let $\tau(\mathcal S)$ be the smallest topology that contains $\mathcal S$ as a subcollection, and let $\mathcal B$ denote the collection of finite intersections of elements of $\mathcal B$. Then it is obvious that $\mathcal B$ is closed under finite intersection. Further it contains the empty intersection which is by convention the set $X$, so $\mathcal B$ covers $X$. These two conditions are exactly what is needed to ensure that the set of all unions of subsets of $\mathcal B$ is a topology on $X$, and $\mathcal B$ will serve as a base for that topology. The sets of this topology will be the unions of elements of $\mathcal B$.
Now since $\tau(\mathcal S)$ is a topology that contains $\mathcal S$ it will also contain $\mathcal B$ as a subcollection and will also contain the unions of sets that are in $\mathcal B$ as elements. Then as smallest topology that contains $\mathcal S$ it must coincide with that topology.
As said the collection of finite intersections of $S$ serves as a base for it, so also according to the alternative defintion $\mathcal S$ is a subbase for $\tau(\mathcal S)$.