# Interior, closure and boundary of the sets of all rational and irrational numbers.

Consider $\mathbb Q$, the set of rational numbers, and its complement $\mathbb R\setminus \mathbb Q$, the set of irrational numbers.

I noticed that their interiors, closures and boundaries are the same, that is:

• Interior: $\varnothing$
• Closure: $\Bbb R$
• Boundary: $\Bbb R$

Why does this happen? Is this a part of some general pattern?

• Yes, those answers are correct. Can you make any stab at an explanation, at least of why the closures are both $\Bbb R$? That’s very straightforward. – Brian M. Scott Jan 22 '13 at 4:02

Whenever $A$ and $B$ are dense disjoint subsets of a topological space $X$, we have $\overline A=X=\overline B$ by the definition of being dense. Since $B\subset A^c$ and $A\subset B^c$, it follows that $\overline{A^c}=X=\overline{B^c}$. The closure of complement is the complement of interior; hence, both interiors are empty. The conclusion about boundary follows by removing interior from the closure: $X\setminus \varnothing=X$.