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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?

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    Yes. I'm sorry:)2012-05-14

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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.

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    @Mariano: Good point; I’ll add it to the answer.2012-05-14