Closure of Intersection of Rationals and Irrationals is Empty Set
Jump to navigation
Jump to search
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: $3$