Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

On page 75 of Counterexamples in Topology, the author writes that the lower limit topology on $\mathbb{R}$ is separable since $\mathbb{Q}$ is dense in $\mathbb{R}$.

Could someone offer more detail on why this is so? I know $\overline{\mathbb{Q}}=\mathbb{R}$ in the standard topology, but might there be something different going on since the closure of $\mathbb{Q}$ might be proven differently depending on the topology of $\mathbb{R}$?

share|cite|improve this question
How do you prove it for the normal (ahem: usual) topology? Why wouldn't that proof adapt? – Dylan Moreland Sep 6 '11 at 1:05
@Dylan: The Sorgenfrey topology’s normal too! :-) – Brian M. Scott Sep 6 '11 at 1:06
up vote 4 down vote accepted

A topological space is separable if there exists a countable dense subset. In our case we claim that $\mathbb Q \subset \mathbb R_l$ is a dense subset. Note that the topology on $\mathbb R_l$ is generated by the open sets $[a,b)$ where $a,b \in \mathbb R$ and $a<b$. So it suffices to show that $\mathbb Q \cap [a,b) \neq \emptyset$. This is evident because we know that $\mathbb Q \cap (a,b) \neq \emptyset$ when $a<b$.

What might have confused you is the fact that the lower limit topology isn't metrizable. So being separable doesn't imply that $\mathbb R_l$ is second countable and in fact it isn't.

share|cite|improve this answer
Thanks, that seems simple enough. – Gotye Sep 6 '11 at 2:30

Let $\mathbb{R}_l$ be the lower limit topology on $\mathbb{R}$. We can prove $\overline{\mathbb{Q}} = \mathbb{R}$ by showing that every open set of $\mathbb{R}_l$ contains an element of $\mathbb{Q}$.

A basis for $\mathbb{R}_l$ is all intervals of the form $[a, b)$ for $a, b \in \mathbb{R}$ with $a < b$. Let $U$ be any nonempty open set in $\mathbb{R}_l$. Then $U$ contains a basis element $[a_0, b_0)$, and there is some $q \in \mathbb{Q}$ so that $a_0 < q < b_0$. Therefore, $q \in [a_0, b_0) \subseteq U$.

share|cite|improve this answer

Your Answer


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.