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 $L=\{x \times (-x) | x \in R_l\}$ (where $R_l$ is the lower limit topology). It is easy to see that $L$ is closed in $R_l^2$.
Let us cover $R_l^2$ by the open set $R_l^2-L$ and by all basis elements of the form
$$[a,b)\times [-a,d)$$
Each of these open sets intersects $L$ in at most one point. Since $L$ is uncountable, no countable subcollection covers $R_l^2$
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?
$$[\alpha,b)\times[c,d) \text{ where } \alpha,b,c,d \in \mathbb{Q} \text{ and } \alpha \neq c$$
Thank you
Best Answer
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 $[a,b) \times [-a, d) \cap L = \{(a,-a)\}$.
A discrete space is Lindelöf iff is is countable (take the open cover by singletons). $L$ is not countable, contradiction.