Prove that $S \times S$ is not Lindelöf Prove that $S \times S$ is not Lindelöf, where $S$ denotes the Sorgenfrey line. I know that $S$ is Lindelöf even though it is not second countable. I would believe that the product would also be Lindelöf, but my professor told me it is not. Any idea of how I can find a cover that has no countable subcover for the product?
 A: Remember that $S\times S$ has as a basis the set of all rectangles which do contain their left and bottom sides, but not their right and top sides. Now look at the antidiagonal $A=\{(x, -x): x\in\mathbb{R}\}$. It's enough to build a cover $\mathcal{C}$ of $S\times S$ such that 


*

*for each $a\in A$, there is exactly one $R_a\in \mathcal{C}$ with $a\in R_a$; and moreover $a, b\in A$, $a\not=b$ implies $R_a\not=R_b$.


We can for instance let $R_a$ be the rectangle with lower-left point $a$ and upper right point . . . well, anything up and to the right of $a$, really. :P Now it's not hard to extend this to a cover, but this will have no countable subcover, since any countable subset will only cover countably many points on $A$.

A more explicit cover: say a rectangle $[a_0, b_0)\times [a_1, b_1)$ (allowing $b_i=\infty$) is good if 


*

*$(a_0, a_1)\in A$ and $b_0=b_1=\infty$, or

*$b_0<b_1$ (that is, the whole rectangle lies to the left/bottom of $A$).
Then the set of good rectangles is an open cover of $S\times S$ with no countable subcover.
A: Another idea: if you have already shown that $S \times S$ is not normal (which is often done in topology texts to show that $T_4$ is not a hereditary property), you're almost done already: the product of $T_3$ (regular and $T_1$) spaces is again $T_3$, and a Lindelöf $T_3$ space is normal. So if $S \times S$ were Lindelöf, it would be normal, which we already know it is not.
A: A more general argument (meaning, without making references to rectangles and etc.): Let $X$ be a topological space with a closed, discrete subset $H$ which is uncountable. Then $X$ is not Lindelöf, since the open cover given by $$\mathcal{U} = \{X \setminus (H \setminus \{x\}): x \in H\}$$ has no proper subcover. 
