Boundary of the Rational Numbers

Consider the set of rational numbers $\mathbb{Q}$ with the standard topology inherited from $\mathbb{R}$, namely the usual topology on the real line.

We want to determine the boundary of $\mathbb{Q}$.

The boundary of $\mathbb{Q}$, denoted by $\partial \mathbb{Q}$, is the set of all points that lie both in the closure of $\mathbb{Q}$ and in the closure of its complement in $\mathbb{R}$.

$$ \partial \mathbb{Q} = \operatorname{Cl}(\mathbb{Q}) \cap \operatorname{Cl}(\mathbb{R} \setminus \mathbb{Q}) $$

The complement of the rational numbers in $\mathbb{R}$, that is, $\mathbb{R} \setminus \mathbb{Q}$, is the set of irrational numbers. We denote it by $\mathbb{I}$.

$$ \partial \mathbb{Q} = \operatorname{Cl}(\mathbb{Q}) \cap \operatorname{Cl}(\mathbb{I}) $$

Since the rational numbers are dense in $\mathbb{R}$, their closure is the whole real line:

$$ \operatorname{Cl}(\mathbb{Q}) = \mathbb{R} $$

Substituting this into the expression for the boundary gives:

$$ \partial \mathbb{Q} = \mathbb{R} \cap \operatorname{Cl}(\mathbb{I}) $$

The closure of $\mathbb{Q}$ consists of all points of $\mathbb{R}$ that can be approximated arbitrarily closely by rational numbers. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, every real number is either rational or a limit point of rational numbers. Hence: $$ \operatorname{Cl}(\mathbb{Q}) = \mathbb{R} $$

The irrational numbers are dense in $\mathbb{R}$ as well. Therefore:

$$ \operatorname{Cl}(\mathbb{I}) = \mathbb{R} $$

It follows that:

$$ \partial \mathbb{Q} = \mathbb{R} \cap \mathbb{R} $$

and hence:

$$ \partial \mathbb{Q} = \mathbb{R} $$

Thus, the boundary of the rational numbers is the entire real line.

In other words, every point of $\mathbb{R}$ lies on the boundary between the rational numbers and the irrational numbers.

This is because every open interval in $\mathbb{R}$ contains both rational and irrational numbers. Consequently, every real number can be approached arbitrarily closely by points of $\mathbb{Q}$ and by points of $\mathbb{R} \setminus \mathbb{Q}$.

Alternative Solution

The boundary of a set can also be expressed as its closure minus its interior:

$$ \partial \mathbb{Q} = \operatorname{Cl}(\mathbb{Q}) \setminus \operatorname{Int}(\mathbb{Q}) $$

Since the closure of the rational numbers is the whole real line, we obtain:

$$ \partial \mathbb{Q} = \mathbb{R} \setminus \operatorname{Int}(\mathbb{Q}) $$

The interior of $\mathbb{Q}$ is empty:

$$ \operatorname{Int}(\mathbb{Q}) = \emptyset $$

Therefore:

$$ \partial \mathbb{Q} = \mathbb{R} \setminus \emptyset $$

The interior of $\mathbb{Q}$ is the set of all rational numbers that have an open neighborhood contained entirely in $\mathbb{Q}$. But every open interval in $\mathbb{R}$ contains both rational and irrational numbers. Therefore, no open interval is contained entirely in $\mathbb{Q}$, and the interior of $\mathbb{Q}$ is empty: $$ \operatorname{Int}(\mathbb{Q}) = \emptyset $$

It follows that:

$$ \partial \mathbb{Q} = \mathbb{R} $$

This again confirms that, in the usual topology on $\mathbb{R}$, the boundary of the rational numbers is the entire real line.