# Countable subsets are disconnected

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.

• If the set has at least two points and is connected it must contain the corresponding interval. Hence it is uncountable. – user90041 Jun 25 '14 at 10:58
• @user90041 Thank you! – Mark Fantini Jun 25 '14 at 11:29

## 2 Answers

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.

• I've always been confused by this: what divides countable from discrete? – Mark Fantini Jun 25 '14 at 11:19
• Countable is a condition on the cardinality of the set; discrete is a topological condition on the set (every point is isolated). In the real numbers discrete implies countable, but not vice versa. – Asaf Karagila Jun 25 '14 at 11:21
• Using your hint as guidance I've tried this: since $D$ has at least two points, namely $a,b$, we can assume $a<b$. By the real numbers there exists $a<x<b$ such that $x \notin D$. Therefore $a \in D \cap (- \infty,x)$ and $b \in D \cap (x, +\infty)$. However, I'm having trouble showing that they are mutually separated. – Mark Fantini Jun 25 '14 at 11:31
• What do you know about this partition of $D$? What topological property does the parts have? – Asaf Karagila Jun 25 '14 at 11:35
• No, you are just guessing now. Look at the sets, what sort of sets did you intersect with $D$? – Asaf Karagila Jun 25 '14 at 14:46

Now let $D\subset\mathbb{R}$ be atmost countable with at least two elements, supposing it to be connected will exhibit a contradiction, since an interval has uncountably infinite elements (except $\space \{a\}\space \forall a\in \mathbb{R}$)

Warning:$(a,b)_Q$ for $a,b\in\mathbb{Q}$ denoting the induced topology on $\mathbb{Q}\cap(a,b)$ by $\mathbb{R}$ is not an interval of $\mathbb{R}$, in most books, it's not even considered to be an interval at all!