Set of Rational Numbers is not Closed in Reals

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Theorem

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

Let $\left({\R, \tau}\right)$ be the real number line under the usual (Euclidean) topology.


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


Proof

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

Let $I := \left({a \,.\,.\, b}\right)$ be an open interval in $\R$ such that $\alpha \in I$.

By elementary results, $\exists \beta \in \Q: \beta \in I$.

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