Definition: A space for which every open covering contains a countable subcovering is called a Lindelöf space

In the book Topology written by Munkres it is said that the Sorgenfrey plane is not Lindelöf using the next argument:

Consider the subspace L={xIII for Top Bar Macy's Printed Collage Created Cutout Back Surplice ×(x)|xRl} (where Rl is the lower limit topology). It is easy to see that L is closed in R2l.

Let us cover R2l by the open set R2lL and by all basis elements of the form

[a,b)×[a,d)

Each of these open sets intersects L in at most one point. Since L is uncountable, no countable subcollection covers R2l

My question is, why you can't choose another form of all basis elements in order to intersect L with this basis elements such that the intersection have more than one point?

For example, why you can't choose elements of the basis with the following form?

[α,b)×[c,d) where α,b,c,dQ and αc

Thank you

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up vote 1 down vote Macy's for Back Bar Cutout Top Surplice III Collage Printed Created accepted

A closed subspace of a Lindelöf space is Lindelöf too, so if the Sorgenfrey plane is, so is L. But L in the subspace topology is discrete, as every singleton is open as [a,b)×[a,d)L={(a,a)}.

A discrete space is Lindelöf iff is is countable (take the open cover by singletons). L is not countable, contradiction.

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