Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\tau$ be the topology generated by half-open intervals of the form $[a,b)$ where $a$ is a rational number and $b$ is a real number. Let $C$ denote the space endowed with the previously described topology.

Prove/disprove: $C \times C$ is a Lindelof space.

How do you proceed: here $C$ is not equal to the Sorgenfrey line (because of the rational endpoint). Do we have to use Jones lemma like when showing $\mathbb{R}_{l} \times \mathbb{R}_{l}$ is not Lindelof?

share|cite|improve this question

Hint: You can show that $C$ is second countable.

Update after the first 2 comments: Once you have that, what can you say about $C\times C$? What can you say about the relationship between second countability and Lindelöfness?

share|cite|improve this answer
@Jonas Meyer: Would $\{ [c,d]: c,d rational\}$ be a countable basis for $C$ because given a basic open set $[a,b)$ with $a$ rational and $b$ real, let $x \in [a,b)$ and choose rationals $w,z$ such that $a<z<x<w<b$ then $x \in [z,w] \subseteq [a,b)$. Correct? – user6494 Feb 1 '11 at 17:57
@Tow: Not quite, because $[c,d]$ is not open. However, $[c,d)$ is. There's one other slight problem with your argument: $x$ might be $a$. That's OK, because you can just take $z=a$ (instead of $z\gt a$). Modulo this detail, your reasoning directly translates to showing that the set of intervals $[c,d)$ with $c$ and $d$ rational is a basis. – Jonas Meyer Feb 1 '11 at 18:01
@Jonas Meyer: Thank you very much. – user6494 Feb 1 '11 at 18:18
@Tow: You're welcome. – Jonas Meyer Feb 1 '11 at 18:24
Can we show $\tau$ is regular? – Henno Brandsma Feb 3 '11 at 18:26

Your Answer


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