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.

  • 1
    $\begingroup$ Try and relate countable covers to rational numbers, which are a countable set. $\endgroup$
    – Dzoooks
    Nov 23, 2019 at 2:54
  • $\begingroup$ 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). $\endgroup$
    – Math1000
    Nov 23, 2019 at 3:32

1 Answer 1


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 $x<y$ and $q_x=q_y$. This implies $y<b_i$, so that $x<y<q_y=q_x<b_i$, whichh gives $y\in (x,b_i)\subset V$, which is a contradiction.


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