Let $X=\{(x,y)\in {\Bbb R}^2:y>0\}$ is the subspace of ${\Bbb R}^2$ with the usual topology, then it is still Lindelöf?
If not, with which topology can $X$ be made to be Lindelöf?
Let $X=\{(x,y)\in {\Bbb R}^2:y>0\}$ is the subspace of ${\Bbb R}^2$ with the usual topology, then it is still Lindelöf?
If not, with which topology can $X$ be made to be Lindelöf?
Yes, $X$ is Lindelöf: this is immediate from the fact that it is second countable. Every second countable space is hereditarily second countable and therefore hereditarily Lindelöf.
Added: As Mariano points out in the comments, you can also use the fact that $X$ is homeomorphic to $\Bbb R^2$, which is itself Lindelöf.