# An half-open cover has a countable subcover

Let $$\mathscr{U}$$ consist of half-open interval $$[a,b)$$ such that $$\mathscr{U}$$ covers $$\mathbb{R}$$. Then does there exists a countable subcover for any such $$\mathscr{U}$$.

I know this is true if we have an open cover and I alway's assumed that this was true for half-open covers, but I can't quite figure out the details.

• Try and relate countable covers to rational numbers, which are a countable set. Nov 23, 2019 at 2:54
• Are you considering $\mathbb R$ in the usual topology, or in the topology generated by the half-open intervals $[a,b)$ (known as $\mathbb R_\ell$ or the Sorgenfrey line)? It shouldn't matter though, since both topologies are Lindelöf spaces (as suggested by @Dzoooks comment). Nov 23, 2019 at 3:32

Say $$\mathscr{U}=\{[a_i,b_i):i\in I\}$$ for some (uncountable) set $$I$$. Let $$\mathscr{V}=\{(a_i,b_i):i\in I\}$$ Put $$V:=\bigcup\mathscr{V}$$. As $$\mathscr{V}$$ is an open (in the Euclidean topology of $$V$$), there exists some countable $$\mathscr{V}_0\subset \mathscr{V}$$ such that $$V\subset \bigcup \mathscr{V}_0$$.
We'll be done if we can show that $$\mathbb{R}\setminus V$$ is countable. Let $$x\in\mathbb{R}\setminus V$$, so that $$x=a_i$$ for some $$i\in I$$. Pick some rational number $$q_x\in (x,b_i)$$. I claim that $$x\mapsto q_x$$ is injective. Indeed, suppose $$x and $$q_x=q_y$$. This implies $$y, so that $$x, whichh gives $$y\in (x,b_i)\subset V$$, which is a contradiction.