Rational Number Space is Topological Space/Proof 2

Theorem

Let $\struct {\Q, \tau_d}$ be the rational number space formed by the rational numbers $\Q$ under the usual (Euclidean) topology $\tau_d$.

Then $\tau_d$ forms a topology.

Proof

Let $\left({\R, \tau_d}\right)$ be the real number space $\R$ under the Euclidean topology $\tau_d$.

By definition of rational numbers, $\Q \subseteq \R$.

From Topological Subspace is Topological Space we have that $\left({\Q, \tau_d}\right)$ is a topology.

$\blacksquare$