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I was reading a note on the Sorgenfrey line at this web page.

In the first proof (proof A), the Sorgenfrey line is Lindelof. So my question is, with the same structure, is $[0,1)$ with the sorgenfrey topology also Lindelof?

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Yes, $[0,1)$ with the Sorgenfrey topology is Lindelöf, and the proof in Dan Ma’s Topology Blog works just as well. In fact, the Sorgenfrey line $S$ is hereditarily Lindelöf, and the same proof works; this is result B of the blog post.

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When he says the open cover $\mathscr{A}$ of the Sorgenfrey line S open intervals of the form $[a, b)$, does that apply to every number in the reals? –  Akaichan May 3 '13 at 15:16
    
@IvordesGreenleaf: Not necessarily. The point is that if $\mathscr{U}$ is any open cover of $S$, it has a refinement $\mathscr{R}$ consisting entirely of intervals of the form $[a,b)$, and if $\mathscr{R}$ has a countable subcover, then trivially so has $\mathscr{U}$. –  Brian M. Scott May 3 '13 at 15:18
    
Then $[0,1)$ works only because it is a countable subspace of $S$? –  Akaichan May 3 '13 at 15:22
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@IvordesGreenleaf: ??? $[0,1)$ is certainly not countable. And every subspace of $S$ is Lindelöf, by exactly the same proof. –  Brian M. Scott May 3 '13 at 15:28
    
That's what I thought, $[0,1]$ is uncountable. Just a little confused, sorry. And we just know that [0,1) is trivially a subspace of S? –  Akaichan May 3 '13 at 15:37
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