Cardinality of all dense and countable sets of $\mathbb{R}$ 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! 
 A: 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.
A: $\mathbb{Q}$ 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 $\mathbb{R}$.
A: 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 $x-y\in\Bbb Q$.


*

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

*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.

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

*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 $\Bbb R$.
