Do all subsets of $\mathbb{R}$ have a dense countable subset?

Question

Do all subsets $A \subset \mathbb{R}$ have a countable dense subset $B \subset A$? I first thought about $A \cap \mathbb{Q}$ but if $A= \mathbb{R} \setminus \mathbb{Q}$ this doesn't work

Answer 1

There is a simple construction that can be used. We shall use the definition of dense in $A$ that for any $\epsilon>0$ and any element $x \in A$ there is an element $y \in B$ such that $|x-y|<\epsilon$.

Now let us construct sets $B_n \subseteq A$ with the property that $B_n$ is countable and for each element $x \in A$ there is an element $y \in B_n$ such that $|x-y|<\frac{1}{n}$.

To construct $B_n$ we can use the axiom of choice to choose one element from each of the nonempty sets $[\frac{k}{n},\frac{k+1}{n}) \cap A$ (for $k \in \Bbb{Z}$) and let $B_n$ be the collection of all of these. Clearly $B_n$ satisfies the required properties.

Now it is clear that $B = \bigcup\limits_{n=1}^{\infty} B_n$ is countable and satisfies the definition of a dense set in $A$ and so we are done. $\square$

Hope this helps. If there are any issues let me know.

Answer 2

This follows from subspaces of separable metric spaces are separable.

For $\Bbb R$ is a separable metric space. Note this is not true in general (non-metric) topological spaces (for instance the Sorgenfrey plane).

Answer 3

For each pair of rational numbers $a$ and $b$ such that $a<b$ and $(a,b)\cap A\neq\emptyset$ let $x_{a,b}$ be any point of that intersection.

The set of all those points is countable and dense in $A$.