Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

What is the cardinality of the following set:

$$\mathbb{A}:=\{A \ : A\subseteq\mathbb{R} \ \ \text{dense and countable}\}$$

(Is $\mathbb{A}$ a separable space?)

Thank You!

share|cite|improve this question
The question whether something is a separable space presupposes that a topology is given on that something. Did you have a particular topology on $\mathbb A$ in mind? – Andreas Blass Dec 6 '12 at 2:09
up vote 11 down vote accepted

The cardinality of $\Bbb R$ is $2^\omega$ (or if you prefer, $2^{\aleph_0}$), so $\Bbb R$ has $\left(2^\omega\right)^\omega=2^{\omega\cdot\omega}=2^\omega$ countable subsets. Thus, the cardinality of $\Bbb A$ is at most $2^\omega$, since not all countable subsets of $\Bbb R$ are dense in $\Bbb R$. However, it turns out that even though not all of these $2^\omega$ countable subsets are dense, there are $2^\omega$ of them that are dense. Here’s one way to show this.

For $x,y\in\Bbb R$ write $x\sim y$ if and only if $x-y\in\Bbb Q$.

  1. Prove that $\sim$ is an equivalence relation on $\Bbb R$.

  2. Prove that for each $x\in\Bbb R$ the $\sim$-equivalence class of $x$ is $x+\Bbb Q=\{x+q:q\in\Bbb Q\}$ and is therefore countable.

  3. Conclude that $\sim$ has $2^\omega$ equivalence classes.

  4. Show that each $\sim$-equivalence class is dense in $\Bbb R$.

The set $\Bbb A$ isn’t a space at all until you give it some topology: it’s just a collection of subsets of $\Bbb R$.

share|cite|improve this answer
Mr Brian M. Scott, Thank You! – 17SI.34SA Dec 6 '12 at 2:15

$\mathbb{Q}$ is countable. So $\mathbb{R}\setminus \mathbb{Q}$ is uncountable. For each irrational $x$, the set $\mathbb{Q} \cup \{x\}$ is also dense in $\mathbb{R}$. Hence there are an uncountable number of dense subsets of $\mathbb{R}$.

share|cite|improve this answer
What is the cardinality of the set in question? Just saying the set is uncountable does not suffice here. – David Mitra Dec 6 '12 at 2:26

Here is a slightly more fun way to do this:

First note that there are at most $\mathbb R^\mathbb N$ countable sets of real numbers, calculate the cardinality and see that this means that there cannot be more than $2^{\aleph_0}$ countable sets of real numbers.

Secondly note that $P=\mathbb{Q\setminus N}$ is dense in $\mathbb R$. Now for every $A\subseteq\mathbb N$ we have that $P\cup A$ is dense and countable. It is easy to see that if $A\neq B$ then $P\cup A\neq P\cup B$. Therefore we actually found $2^{\aleph_0}$ dense sets of real numbers which are subsets of $\mathbb Q$.

This shows that there cannot be more than $2^{\aleph_0}$ countable dense sets. As remarked by others, separability is a topological property and it is unclear what topology you are giving this collection.

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.