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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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Do you mean $\Bbb R^2$ where you’ve written $R$? – Brian M. Scott May 14 '12 at 3:29
I assumed that was the case; I hope I was correct! – MJD May 14 '12 at 3:29
Yes. I'm sorry:) – Paul May 14 '12 at 3:31
up vote 5 down vote accepted

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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The space is even homeomorphic to $\mathbb R^2$ so, assuming the «still» in the question means that the OP knows $\mathbb R^2$ is Lindelöf, then he also knows $X$ is Lindelöf :) – Mariano Suárez-Alvarez May 14 '12 at 3:31
@Mariano: Good point; I’ll add it to the answer. – Brian M. Scott May 14 '12 at 3:32

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