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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!

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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
    
I have to wonder where did you come up with the separability question in mind. The original question merely asks for the first part. Also, please remember to use the [homework] tag. –  Asaf Karagila Dec 6 '12 at 10:59
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3 Answers 3

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 $\Bbb R$.

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Mr Brian M. Scott, Thank You! –  17SI.34SA Dec 6 '12 at 2:15
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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.

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$\mathbb{Q}$ 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 $\mathbb{R}$.

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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
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