Irrational Number Space is Completely Normal

Theorem

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


Then $\struct {\R \setminus \Q, \tau_d}$ is a completely normal space.


Proof

From Euclidean Space is Complete Metric Space, a Euclidean space is a metric space.

Hence in particular $\struct {\R, \tau_d}$ is a metric space.

From Subspace of Metric Space is Metric Space, it follows that $\struct {\R \setminus \Q, \tau_d}$ is likewise a metric space.

From Metric Space fulfils all Separation Axioms it follows that $\struct {\R \setminus \Q, \tau_d}$ is a completely normal space.

$\blacksquare$


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: $4$
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