I am trying to show the following: every countable subset of $\Bbb R$ with at least two points is disconnected.
My attempt: let $D$ be such subset. Then take $a \in D$ and define $A=\{ a\}$ and $B = D – \{ a\}$. I want to use the result that a set is disconnected if and only if there exists mutually separated sets whose union is the set. I tried to show that $A$ and $B$ are mutually separated sets using $\overline{A} = \{ a \}$ and $\overline{B}$ but $\overline{B} = \overline{D – \{a \}} = D – \{a\}^{\circ} = D$. We have $\overline{A} \cap B = \emptyset$ but $A \cap \overline{B} = \{a\} \neq \emptyset$.
I'm not seeing another line of attack.
Best Answer
Your approach seems to imply that every point in a countable set is isolated. Namely that a countable is synonymous with "discrete" in the case of the real numbers.
This is not true, of course. The rational numbers are countable but not discrete.
Instead, let me give you the following hint:
If $D$ is a countable subset of $\Bbb R$, show that there is some $x\in\Bbb R$ such that $x\notin D$ and neither $D\cap(-\infty,x)$ nor $D\cap(x,\infty)$ are empty.