Show that half-closed rational intervals dont form a basis for the Lower Limit Topology The Lower Limit Topology is the topology generated by the basis $\mathcal{B}$.  $\mathcal{B} := \{[a,b) \subseteq \mathbb{R} : a,b \in \mathbb{R}, a < b\}$ 
I need to show that: 
$\mathcal{B}_{\mathbb{Q}} := \{[a,b) \subseteq \mathbb{R} : a,b \in \mathbb{Q}, a < b\}$ is not a basis for the Lower Limit Topology.
I had shown earlier that the Lower Limit Topology contains the interval $(a,b), \forall a, b \in \mathbb{R}$ by the following proof: 

We take the collection $\mathcal{C} = \{[a+\frac{1}{n},b): n \in$
  $\mathbb{N}, n > \frac{1}{b-a}\}$. Clearly $\mathcal{C} \in \mathcal{B}$ and
  $\bigcup \mathcal{C} = (a,b)$.

In $\mathcal{B}_{\mathbb{Q}}$, for any $a,b$ that are irrational, this proof does not work (since $a+\frac{1}{n}$ and $b$ would not be rational). Initially, I thought I could show that these irrational intervals could not exist in the topology generated by $\mathcal{B}_{\mathbb{Q}}$. But, I'm no longer convinced that's true and am stuck on how to proceed.

Let $a$ be an irrational number, suppose $[a,b)$ can be written by union of the sets in $B_\mathbb{Q}$. So $a$ belongs to some set of $B_\mathbb{Q}$, say $A=[c,d)$, $c,d$ are rationals.
So clearly, $c<a$(as $a$ is irrational), so $c$ belongs to the union of those sets in $B_\mathbb{Q}$ that makes $[a,b)$, hence the union cannot equal $[a,b)$.
(Note the infimum of the numbers contained in any union of sets of $B_\mathbb{Q}$ either does not exist or equals a rational, which gives us the required contradiction).
