subbasis generating usual topology I am trying to prove that  $S := \{ (-\infty, b) : b\in\Bbb R \}\cup \{ (a, ∞) : a\in\Bbb R\}$ generates the usual topology on $\Bbb R$.
I am trying to show that this set generates the usual basis. However, when I take finite intersections of sets of $S$, I get one of the following sets: $\{0\},\;(a,b),\;(-\infty, b)$ and $(a,\infty )$.  My understanding is that sets in the form of $(a,b)$ generate the basis for the usual topology and that null set is a subset of every set, but what do we do about these "extraneous"sets $(-\infty, b)$ and $(a,\infty)$?  They are not part of the usual basis that generates the usual topology on $\Bbb R$ are they? 
 A: Every open set with respect to the usual topology $\tau$ is an union of intervals of the type $(a,b)$, with $a<b$. Therefore, since $S$ contains each such interval, the topology $\tau_S$ generated by $S$ contains $\tau$. But then, since each element of $S$ belongs to $\tau$, $\tau_S\subset\tau$, and therefore $\tau_S=\tau$.
A: There can be many "generating sets" for a topology. A collection of subsets $\mathcal{A}$ of a set $X$ generates the topology $\tau$ if the smallest (in inclusion sense) topology on $X$ that contains $X$.
There are two types of generating collections, in a way: a base (where we "only" need to take unions of them to get all open sets) and a subbase (where we also take finite intersections to create a "base", so they're often more "economical"). 
In your case, sets of the form $(-\infty, a)$ are open in $\tau$, they are clearly unions of sets in the standard base: $(-\infty, a)= \bigcup_{n \in \Bbb N^+} (a-(n+2), a-n)$, e.g. and similarly for sets of the form $(a, +\infty)$. They are not members of the standard base, but all sets in the standard base are in a any topology that contains them because $(a,b)=(-\infty, b) \cap (a, +\infty)$ and open sets are closed under finite intersections.
So if we have a known generating set $\mathcal{B}$ for $\tau$ and we have another candidate $\mathcal{A} \subseteq \tau$ of which we know we can make (by finite intersections and all unions) all members of $\mathcal{B}$, we must have that $\mathcal{A}$ also generates $\tau$. (The topology generated by $\mathcal{B}$ is at most $\tau$ (because they're already open in $\tau$, and we take the minimal topology containing the family) and at least $\tau$ (because we get all members of $\mathcal{B}$ which is generating for $\tau$)).
