Closure of Irrational Numbers is Real Numbers

Theorem
Let $\left({\R, \tau_d}\right)$ be the real numbers under the usual (Euclidean) topology.

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

Then:
 * $\left({\R \setminus \Q}\right)^- = \R$

where $\left({\R \setminus \Q}\right)-$ denotes the closure of $\R \setminus \Q$.

Proof
From Irrationals are Everywhere Dense in Reals, $\R \setminus \Q$ is everywhere dense in $\R$.

It follows by definition of everywhere dense that $\left({\R \setminus \Q}\right)^- = \R$.