Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
2  
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". –  Martin Feb 20 '13 at 8:08
    
@Martin: Thanks Martin. However it is a little difficult for me. So I posted here for help. –  Paul Feb 20 '13 at 8:10
3  
The very best idea would have been to post at the same place, really. –  Mariano Suárez-Alvarez Feb 20 '13 at 9:58
add comment

1 Answer

up vote 3 down vote accepted

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

share|improve this answer
    
Why could the first note be true? –  Paul Feb 20 '13 at 8:43
1  
@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.) –  Arthur Fischer Feb 20 '13 at 9:57
    
@Arthur Fischer: Thanks. I get it. –  Paul Feb 20 '13 at 10:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.