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In course of showing that the Sorgenfrey line $(\mathbb R$ endowed with the lower limit topology $\tau_l)$ is lindelöf I've made the following attempt:

I've picked up a cover $\mathcal U$ of $\mathbb R$ by the elements chosen arbitrarily from $\tau_l.$ Next for $q\in\mathbb Q$ I've choosed $G_q\in \mathcal U$ such that $q\in G_q$ to form the set $K=\{G_q:q\in\mathbb Q\}.$ My intuition says that this $K$ must cover $\mathbb R$ since no matter what irrational I choose, on its left there exists (due to denseness of $\mathbb R$) arbitrarily close rational which would get enclosed in some member of $K.$ Now here where I got stuck since I can't convert my intuition to the mathematical language. Can someone suggest me any way out?

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up vote 4 down vote accepted

This proof will not work. If it would, in general, we could prove that a separable space is always Lindelöf (pick a covering element for each member of a countable dense set). And there are lots of counterexamples to this idea (the square of this space is one of them).

Hint for a better proof: consider a cover by basic open sets $[a,b)$. Then consider the collection of the corresponding $(a,b)$ in $\mathbb{R}$ in the usual topology. This need not be a cover anymore, but as $\mathbb{R}$ is second countable, one can show that countably many of these open intervals have the exact same union as all these $(a,b)$. Then we only need to cover the countably many left end points that we might have missed...

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''If it would, in general, we could prove that a separable space is always Lindelöf...'' - Loved it !! You've identified my problem with pinpoint accuracy. Thanks. – Sugata Adhya Feb 24 '13 at 15:26
@SugataAdhya glad to be able to help – Henno Brandsma Feb 24 '13 at 15:29
excellent proof @HennoBrandsma +1 – learnmore Mar 6 '15 at 6:38

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