Closure of Intersection of Rationals and Irrationals is Empty Set

Theorem
Let $\struct {\R, \tau}$ denote the real number line under the usual (Euclidean) topology.

Let $\Q$ be the set of rational numbers.

Then:
 * $\paren {\Q \cap \paren {\R \setminus \Q} }^- = \O$

where:
 * $\R \setminus \Q$ denotes the set of irrational numbers
 * $\paren {\Q \cap \paren {\R \setminus \Q} }^-$ denotes the closure of $\Q \cap \paren {\R \setminus \Q}$.

Proof
From Set Difference Intersection with Second Set is Empty Set:
 * $\Q \cap \paren {\R \setminus \Q} = \O$

By Empty Set is Closed in Topological Space, $\O$ is closed in $\R$.

From Closed Set Equals its Closure:
 * $\O^- = \O$

Hence the result.