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

2 Answers 2

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).

(The idea presented below is not entirely dissimilar to the ideas in this previous answer of mine.)

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 $A \subseteq \mathbb{R}$.

Lemma: For any $A \subseteq \mathbb{R}$ the difference $\clrl ( A ) \setminus \clsl ( A )$ is countable.

proof outline. Given $x \in \clrl ( A ) \setminus \clsl ( A )$ there must be a $b_x > x$ such that $[ x , b_x ) \cap A = \varnothing$. We can show that $[x,b_x) \cap [y,b_y) = \varnothing$ for distinct $x , y \in \clrl (A) \setminus \clsl (A)$. Thus $\{ [ x , b_x ) : x \in \clrl ( A ) \setminus \clsl ( A ) \}$ is a family of pairwise disjoint nonempty open sets in the lower-limit topology, and since the lower-limit topology is separable this family cannot be uncountable. $\dashv$

Given $Y \subseteq \mathbb{R}$, to show that $Y$ as a subspace of the lower-limit topology is separable we first note that since the real line is second-countable (and therefore hereditarily separable) there is a countable $A_0 \subseteq Y$ such that $\clrl ( A_0 ) = \clrl ( Y )$. By the lemma above $Y \setminus \clsl ( A_0 ) \subseteq \clrl ( A_0 ) \setminus \clsl ( A_0 )$ is countable, and so $A = A_0 \cup ( Y \setminus \clsl ( A_0 ) ) \subseteq Y$ is also countable. It is fairly straightforward to show that $\clsl (A) \supseteq Y$.

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Consider a set $X\subseteq\mathbb R$ with its relative topology as a subspace of the Sorgenfrey line $(\mathbb R,\tau)$. Let $\mathbb Q$ be the set of all rational numbers.

For each $r\in\mathbb Q$ such that $X\cap(-\infty,r)$ has a greatest element, let $x_r$ be the greatest element of $X\cap(-\infty,r)$.

For each rational interval $(q,r)$ such that $X\cap(q,r)\ne\emptyset$, choose an element $y_{q,r}\in X\cap(q,r)$. (Axiom of choice used here.)

Let $D=\{x_r:r\in\mathbb Q,\ X\cap(-\infty,r)\ \text{has a greatest element}\}\cup\{y_{q,r}:q,r\in\mathbb Q,\ q\lt r,\ X\cap(q,r)\ne\emptyset\}.$

Clearly $D$ is a countable subset of $X$.To see that $D$ is dense in $X$, observe that the topology of $X$ has a base consisting of all nonempty sets of the form $X\cap[a,b)$ where $a,b\in\mathbb R,\ a\lt b$, and that every such set contains an element of $D$.

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