Closure of Intersection of Rationals and Irrationals is Empty Set

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

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

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

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

Proof
From Set Difference Intersection with Second Set is Empty Set:
 * $\Q \cap \left({\R \setminus \Q}\right) = \varnothing$

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

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

Hence the result.