Show that half-closed rational intervals dont form a basis for the Lower Limit Topology

Question

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.

Answer 1

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).

Answer 2

Let $C$ be any base for the lower-limit topology.

For each $r\in \mathbb R$ let $r\in f(r)\in C$ with $f(r)\subset [r,\infty).$ For real $r,r',$ if $r\ne r'$ then $f(r)\ne f(r')$ because $\min f(r)=r\ne r'=\min f(r').$

So $f:\mathbb R\to C$ is $1$-to-$1,$ and $\mathbb R$ is uncountable,so $C$ is uncountable.