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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 $\mathbb{R}$?

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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
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@Dylan: The Sorgenfrey topology’s normal too! :-) –  Brian M. Scott Sep 6 '11 at 1:06

2 Answers 2

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.

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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, $q \in [a_0, b_0) \subseteq U$.

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