Closure of Rational Numbers is Real Numbers

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

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

Then:
 * $\Q^- = \R$

where $\Q^-$ denotes the closure of $\Q$.

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

It follows by definition of everywhere dense that $\Q^- = \R$.