Question

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?

Answer 1

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 $\mathbb{R} \setminus \bigcup_{i \in I} U_i \subseteq \bigcup_{i \in I_1} A_i$.