Set of Rational Numbers is not Closed in Reals

Theorem

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

Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.

Then $\Q$ is not closed in $\R$.

Proof

Let $\alpha \in \R \setminus \Q$.

Let $I := \openint a b$ be an open interval in $\R$ such that $\alpha \in I$.

By Between two Real Numbers exists Rational Number:

$\exists \beta \in \Q: \beta \in I$.

Thus $I$ contains elements of $\Q$ and so $\R \setminus \Q$ is not open in $\R$.

Thus by definition, $\Q$ is not closed in $\R$.

$\blacksquare$

Sources

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