Irrational Numbers form Metric Space

Theorem

Let $\mathbb I = \R \setminus \Q$ be the set of all irrational numbers.

Let $d: \mathbb I \times \mathbb I \to \R$ be defined as:

$\map d {x_1, x_2} = \size {x_1 - x_2}$

where $\size x$ is the absolute value of $x$.

Then $d$ is a metric on $\mathbb I$ and so $\struct {\mathbb I, d}$ is a metric space.

Proof

From the definition of absolute value:

$\size {x_1 - x_2} = \sqrt {\paren {x_1 - x_2}^2}$

This is the Euclidean metric.

in Euclidean Metric on Real Vector Space is Metric this is shown to be a metric.

$\blacksquare$

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: $4$