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By definition, $\mathbb{Q}$ is closed iff $\mathbb{R}\setminus\mathbb{Q}$ -- the set of irrational real numbers, is an open set in $\mathbb{R}$.

If $\mathbb{R}\setminus\mathbb{Q}$ is open, then it has to be a union of open intervals (by definition).

Therefore, if $\mathbb{R}\setminus\mathbb{Q}$ is open, it must have some open interval $(a, b)$ as a subset.

The question is to show that every open interval $(a, b)$ must contain rational numbers.

Can anyone give me some hints on this problem? I was trying to use the definition of $\mathbb{Q}$ which is a number is rational if it can be written as $\frac{p}{q}$ for integers $p, q \in \mathbb{Z}$ and $q \neq 0$. But I don't think there's $\mathbb{Z}$ in $(a, b)$ since it's a subset of irrational real numbers? Am I thinking it wrong...

Asaf Karagila
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d0nut
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    Hint: If $q$ is a large positive integer then $(aq,bq)$ contains an integer $p$. ($q>\frac 1 {b-a}$ will do). – geetha290krm Sep 23 '22 at 09:43
  • You are right, $(a,b)$ may not contain an element of $\mathbb Z$. But maybe you can construct an integer $n\ge 1$ such that $(a,b)$ contains an element of $\frac1n \mathbb Z$? – Claudius Sep 23 '22 at 09:43
  • If you know that the rationals are dense then it's straightforward – Lelouch Sep 23 '22 at 09:54
  • You can take the decimal approximation of (a+b)/2 to get a rationals number in the interval (a, b) – Lelouch Sep 23 '22 at 09:56

2 Answers2

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If $\mathbb{Q}$ was closed, every sequence of rational numbers would converge to a rational number. You just need to show that this is not the case... Take any irrational number $z$ and denote by $x_n$ its decimal expansion up to $n$ decimal places. Although $(x_n)\subset \mathbb{Q}$, $\lim x_n = z \notin \mathbb{Q}$.

On a different direction, it is easy to show that every open set $(a,b)$ contains rational numbers. To see this, just choose an integer $m$ such that $(b-a) m > 1$ and observe that there must be some integer $n$ in the interval $(ma, mb)$, which means that $\frac nm \in (a,b)$.

PierreCarre
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Assume that $\mathbb{Q}$ is closed. Then $\mathbb{R}-\mathbb{Q}$ is open.

That implies that for each irrational a there is a neighborhood

$S(a,\epsilon)$ contained in $\mathbb{R}-\mathbb{Q}$.

But this is a clear contradiction because there are rationals in every

open neighborhood of any point in $\mathbb{R}$.

This comes from the construction of the real numbers.