$\mathcal B_{\mathbb Q}$ = = { [p, q) ⊆ R : p, q ∈ Q, p < q } is not a bases for the Lower Limit Topology I'm having a bit of trouble proving this:
The definition of Lower Limit Topology I am working with: $ \{[a, b) \subseteq \mathbb R \ \text s.t \  a < b\}$. 
The only thing I can think of is that since the rationals are a countable set in the Reals, there will only be countable many intervals and their union can not equal the entire lower limit topology, or in other words there are not enough intervals B in $\mathcal B_{\mathbb Q}$ s.t $\forall x \in$ every open set U of the lower limit topology $\exists B \in \mathcal B_{\mathbb Q} $ s.t $x \in B \in U$.
I can't figure out how to rigorously formulate this. 
 A: What does it mean to say that a collection $\mathcal B$ of open sets is a base? It means that, given any point $a$ and any neighborhood $U$ of $a,$ we can find a set $B\in\mathcal B$ with $a\in B\subseteq U.$
To show that $\mathcal B_\mathbb Q$ is not a base, all we have to do is find a point $a$ and a neighborhood $U$ of $a$ such that there is no set $B\in\mathcal B_\mathbb Q$ with $a\in B\subseteq U.$
Looking at the definition of $\mathcal B_\mathbb Q,$ I think I'd try taking a irrational number for $a,$ say $a=\pi.$
Looking at the definition of the Sorgenfrey topology, oops, I mean the "lower limit topology", I'd try taking a neighborhood of $\pi$ which is not a neighborhood in the usual topology, say, the interval $[\pi,\infty).$
There we have it. Is there a set $B\in\mathcal B_\mathbb Q$ such that $\pi\in B\subseteq[\pi,\infty)?$ If we can prove that there is no such $B,$ then everything is fine.
Hmm. Suppose $B=[p,q)\in\mathcal B_\mathbb Q.$ Now there are three cases: either $p\lt\pi,$ or $p=\pi,$ or $p\gt\pi$ . . .
