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Let us consider the lower limit topology τ={G⊂R:∀x∈G,∃ϵ>0 such that [x,x+ϵ)⊂G} on R. I am trying to show that any subspace of (R,τ) is separable, but couldn't find the countable subset in a subspace X of R which is dense. Any hint will be appreciated.

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What sets have you tried? Do you know what standard set in the lower limit topology makes it a separable space? –  Devin Murray Oct 14 '13 at 12:14
Yeah $\mathbb Q$ makes $\mathbb R$ separable, but how to find a set with the help of $\mathbb Q$? –  Anupam Oct 14 '13 at 12:20

1 Answer 1

Longish Hint: First, to show that a subspace $ \newcommand{\clrl}{\mathrm{cl}_{\mathbb{R}}} \newcommand{\clsl}{\mathrm{cl}_{\text{S}}} Y$ of a topological space $X$ is separable, it suffices to find a countable $A \subseteq Y$ such that $\overline{A} = \overline{Y}$ (and actually, $\overline{A} \supseteq Y$ will suffice).

I'll let $\clrl$ and $\clsl$ denote the closure operators on the real line (usual metric topology) and the lower-limit topology, respectively.

Since the lower-limit topology is finer than the metric topology, then $\clrl ( A ) \supseteq \clsl ( A )$ for any $A \subseteq \mathbb{R}$. The trick will be to show that the difference $\clrl ( A ) \setminus \clsl ( A )$ is always countable. In broad outline, you will want to associate to each $x \in \clrl (A) \setminus \clsl (A)$ an open interval $(a_x , b_x)$ in such a manner that given distinct $x,y \in \clrl (A) \setminus \clsl (A)$ we have $(a_x , b_x) \cap (a_y,b_y) = \varnothing$. Since the real line is separable, there can be no uncountable family of pairwise disjoint (nonempty) open sets, which implies that the difference $\clrl (A) \setminus \clsl (A)$ must be countable. (This is not entirely dissimilar to the ideas in this previous answer of mine.)

(To see that this helps, note that since the metric topology is second-countable (and therefore hereditarily separable) there is a countable $A_0 \subseteq Y$ such that $\clrl ( A_0 ) = \clrl ( Y )$. Then $Y \setminus \clsl ( A_0 ) \subseteq \clrl ( A_0 ) \setminus \clsl ( A_0 )$ is countable, and taking $A = A_0 \cup ( Y \setminus \clsl ( A_0 ) )$ we get a countable subset of $Y$ with $\clsl ( A ) \supseteq Y$.)

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