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Let $X$ be the real numbers, and $r \in \mathbb{R}$ some real number. We define the following topology on X; $T = \{ U \subset\mathbb{R}\:|\:r\notin U\text{ or }\mathbb{R}\setminus U\text{ finite}\} $. It can be shown that this topology is Lindelöf (i.e. every open cover has a countable subcover). But suppose that we look at the set $B$ = $X\setminus{\{r\}}$ under the subspace topology. Would this one still be Lindelöf?

I would argue that the Lindelöf property depends on the fact that $r\in X$, but I'm not sure..

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Actually, the lindelöf property is proved solely by the cofinite subsets. It would not be possible to prove the property with open intervals of $\mathbb{R}\setminus{\{0\}}$ ? – omar Jan 12 '13 at 11:50
$B$ is open, and an open subspace of a Lindelöf space need not be Lindelöf. – Z. L. Jan 12 '13 at 12:54
In fact, $B$ can't be Lindelöf since the topology induced on it is the discrete topology and $X = \mathbb{R}$ which is uncountable. – Z. L. Jan 12 '13 at 12:57
@ZangoLotino. $T$ could be seen as the union of the cofinite topology and all sets not containing $0$. The latter would result in the discrete topology with $B$. But what about the cofinite part? Is there a reason to not use the cofinite sets when constructing a countable subcover? – omar Jan 12 '13 at 13:37
up vote 1 down vote accepted

All subsets of $B$ (including the cofinite sets) are open, so it is indeed the discrete topology. And it is easily shown that any uncountable discrete space is not Lindelöf. Consider $\mathcal{O}=\{\{x\}: x \in B\}$, as an open cover and show that it has no countable subcover. And then by defintion is not Lindelöf.

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