Rational Numbers are not Connected

Theorem
The set of rational numbers $\Q$ is not a connected topological space.

Proof
Let $\alpha \in \R$ be an irrational number.

By definition, $\alpha \notin \Q$.

Consider the sets:
 * $S := \Q \cap \left({-\infty \,.\,.\, \alpha}\right)$
 * $T := \Q \cap \left({\alpha \,.\,.\, \infty}\right)$

Let $x \in S$.

Let $B_\epsilon \left({x}\right)$ be the open $\epsilon$-ball of $x$ in $\Q$.

Then:
 * $\forall x \in S: \exists \epsilon \in \R_{>0}: B_\epsilon \left({x}\right) \subseteq S$

by setting $\epsilon = \alpha - x$.

Similarly:
 * $\forall x \in T: \exists \epsilon \in \R_{>0}: B_\epsilon \left({x}\right) \subseteq T$

by setting $\epsilon = x - \alpha$.

Thus $S$ and $T$ are open sets of $\Q$.

Then:
 * $S \cup T = \Q$
 * $S \cap T = \O$
 * $S, T \ne \O$

So $S$ and $T$ fulfil the conditions for $S \mid T$ to be a separation of $\Q$.

Hence the result by definition of connected topological space.