Rational Number Space is Topological Space
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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 1
From Rational Numbers form Metric Space we have that $\Q$ is a metric space under the Euclidean metric.
From Metric Induces Topology, it follows that the Euclidean topology forms a topology on $\Q$.
$\blacksquare$
Proof 2
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.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $30$. The Rational Numbers