Let $R$ denote the set of all real numbers. $B$ is any Bernstein set of $R$.

Bernstein Set: A subset of the real line that meets every uncountable closed subset of the real line but that contains none of them. It's from wiki.

We topologize $R$ now: the set $B$ is discrete and its complement has the usual topology. How to see the new topological space is Lindelof?

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    $\begingroup$ On math.SE you were linked to Dan Ma's topology blog dantopology.wordpress.com/2012/10/30/… where there is a proof in the first paragraph on "Non-normal product spaces". $\endgroup$ – Martin Feb 20 '13 at 8:08
  • $\begingroup$ @Martin: Thanks Martin. However it is a little difficult for me. So I posted here for help. $\endgroup$ – Paul Feb 20 '13 at 8:10
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    $\begingroup$ The very best idea would have been to post at the same place, really. $\endgroup$ – Mariano Suárez-Álvarez Feb 20 '13 at 9:58

Note that the open subsets of (what I will denote by) $\mathbb{R}_B$ are of the form $U \cup A$ where $U \subseteq \mathbb{R}$ is open in the usual topology, and $A \subseteq B$ is arbitrary.

Suppose that $\{ U_i \cup A_i : i \in I \}$ is an open cover of $\mathbb{R}\_B$. Note that there is a countable $I_0 \subseteq I$ such that $\bigcup_{i \in I_0} U_i = \bigcup_{i \in I} U_i$. Next note that $\mathbb{R} \setminus \bigcup_{i \in I} U_i \subseteq B$ is closed (in $\mathbb{R}$) and is therefore countable, so there is a countable $I_1 \subseteq I$ such that $\mathbb{R} \setminus \bigcup_{i \in I} U_i \subseteq \bigcup_{i \in I_1} A_i$.

  • $\begingroup$ Why could the first note be true? $\endgroup$ – Paul Feb 20 '13 at 8:43
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    $\begingroup$ @John: Do you mean my characterisation of the open subsets of $\mathbb{R}_B$? (Which follows from the fact that the topology generated by the usual open subsets of $\mathbb{R}$ and the singletons from $B$.) Or that there is a countable $I_0$? (Which follows from the fact that $\mathbb{R}$ is second-countable, and thus hereditary Lindelöf.) $\endgroup$ – user642796 Feb 20 '13 at 9:57

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