Irrational Number Space is Totally Separated

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

Then $\left({\R \setminus \Q, \tau_d}\right)$ is totally separated.

Proof
Let $x, y \in \R \setminus \Q$.

From Between two Real Numbers exists Rational Number:
 * $\exists \alpha \in \Q: x < \alpha < y$

Consider the unbounded open real intervals:
 * $A := \left({-\infty \,.\,.\, \alpha}\right), B := \left({\alpha \,.\,.\, +\infty}\right)$

Let:
 * $U := A \cap \left({\R \setminus \Q}\right), V := B \cap \left({\R \setminus \Q}\right)$

Let $\beta \in \R \setminus \Q$.

Then either:
 * $(1): \quad \beta < \alpha$

in which case:
 * $\beta \in U$

or:
 * $(2): \quad \beta > \alpha$

in which case:
 * $\beta \in V$

Thus $U \cup V = \R \setminus \Q$.

Let $a \in U$.

Then $a < \alpha$ and so $a \notin V$.

Similarly, let $b \in V$.

Then $b > \alpha$ and so $b \notin U$.

Then note that $x \in U$ and $y \in V$.

Thus $U \ne \varnothing$ and $V \ne \varnothing$.

Thus, by definition, $U$ and $V$ constitute a separation of $\R \setminus \Q$ such that $x \in U$ and $y \in V$.

Hence the result by definition of totally separated.