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

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.

share|cite|improve this question
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

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 $\clsl (A) \supseteq Y$.

share|cite|improve this answer

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 $D$.

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.