TITLE: How can I prove that the Sorgenfrey line is a Lindelöf space?
QUESTION [6 upvotes]: How can I prove that the Sorgenfrey line is a Lindelöf space? Now, Sorgenfrey line is $\mathbb{R}$ with the basis of $\{[a,b) \mid a,b\in\mathbb{R}, a<b\}$, and in general, a topological space is called a "Lindelöf space" iff every open cover has a countable subcover. Please show me an elegant proof.

REPLY [4 votes]: The proof given by bof is correct, but incomplete. Here is a completion.
When bof assumed ad absurdum that $\sup{C_a} = b \in \mathbb{R}$, he implicitely assumed that $b > a$, which a priori does not need to be true (as $b$ could be equal to $a$). For the sake of completeness, first remark that $C_a$ is non-empty, as the interval $[a, a]$ is trivially countably covered, so $a \in C_a$.
Now let $A$ be a Sorgenfrey-open set that covers $a$. Then, because $A$ is open, there must exist a neighbourhood $[a, a+\epsilon[$ in $A$ for some $\epsilon > 0$. Now observe that the interval $[a, a+\frac{\epsilon}{2}]$ is covered by $A$, so $a + \frac{\epsilon}{2}$ will certainly be a member of $C_a$. It follows that $b = \sup{C_a} \geq a + \frac{\epsilon}{2} > a$, which justifies bofs assumption that $b > a$.