Closure of Intersection of Rationals and Irrationals is Empty Set

Theorem

Let $\struct {\R, \tau}$ denote the real number line with 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.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $30 \text { - } 31$. The Rational and Irrational Numbers: $3$