Separable Subset of the Real Line Consider $\mathbb{R}$ with the usual topology. Let $A$ be an infinite subset of $\mathbb{R}$. Prove that $A$ is separable.
So, if $A$ is to be separable, it must contain a countably dense subset $H$. I need to show that this $H$ exists to prove that $A$ is separable. If I don't know more information about $A$, I am not sure how to do this.
 A: For each $q \in \mathbb{Q}$ and each $n \in \mathbb{N}$ let $I_{q,n} = (q-\frac{1}{n}, q+\frac{1}{n})$. For each $q,n$, if $A \cap I_{q,n} \ne \varnothing$ then fix some $a_{q,n} \in A \cap I_{q,n}$, or otherwise leave $a_{q,n}$ undefined.
Then certainly $D=\{ a_{q,n} : q \in \mathbb{Q}, n \in \mathbb{N} \}$ is a countable subset of $A$. It remains to prove that $D$ is dense in $A$.
I don't want to spoil it for you by giving you a proof of density, but if you want more help then let me know in the comments and I'll modify my answer appropriately.
A: Since $\mathbb{R}$ is second countable, it has a countable basis.  Intersecting each such basis element with $A$ is a basis of $A$ itself, so $A$ is also second countable.  The standard proof applies: for each basis element $B$, choose some $x_B\in B\cap A$, and let $D$ be the set of all such $x_B$.  This $D$ is countable, dense, and a subset of $A$.  To check that it is dense, for any open set $U\subset A$, there is some basis element $B$ inside it, and hence the point $x_B$, so $U\cap D$ is nonempty.
