Irrational Number Space is Topological Space

Theorem

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

Then $\tau_d$ forms a topology.

Proof

Let $\struct {\R, \tau_d}$ be the real number space $\R$ under the Euclidean topology $\tau_d$.

By definition of irrational numbers, $\R \setminus \Q \subseteq \R$.

From Topological Subspace is Topological Space we have that $\struct {\R \setminus \Q, \tau_d}$ 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: $31$. The Irrational Numbers