# Is the set $A = [0,1]\setminus\mathbb{Q}$ countable or not?

Is the set $$A = [0,1]\setminus\mathbb{Q}$$ countable or not?

What I am thinking is $$A$$ consist of irrational numbers in the interval $$[0,1]$$ hence it is subset of irrational numbers. As set of irrational numbers is uncountable so I think set $$A$$ is also uncountable.

• not every subset of irrational numbers is uncountable, but if $A$ and $\mathbb Q$ were both countable then so would be $A\cup \mathbb Q$ and therefore $[0,1]\subset A\cup \mathbb Q$ – J. W. Tanner Jan 10 at 15:55
• $\varnothing$ is also a subset of the irrational numbers. – Asaf Karagila Jan 10 at 16:24

## 3 Answers

Yes, it is uncountable, but not for that reason. For instance, $$\left\{\sqrt2+n\,\middle|\, n\in\Bbb N\right\}$$ is also a set of irrational numbers, but it is countable.

However, if $$[0,1]\setminus\Bbb Q$$ was countable, then, since $$\Bbb Q\cap[0,1]$$ is countable, $$[0,1]$$ would be countable too, since it's the union of them.

• technically, $[0,1]$ is contained in the union of $A$ and $\mathbb Q$ – J. W. Tanner Jan 10 at 16:08
• @J.W.Tanner I've edited my answer. Thank you. – José Carlos Santos Jan 10 at 16:23

All countable set of $$R$$ have Lebesgue measure equal to $$0$$. So Lebesgue measure of $$[0, 1] - \mathbb{Q}$$ is $$1$$. Eventually by contraposition, $$[0, 1] - \mathbb{Q}$$ is uncoutable.

What you are thinking does not work.

For example, $$\{n\pi\mid n\in\mathbb N\setminus\{0\}\}$$ is a subset of irrational numbers but countable.

Here's an argument that works. If $$A$$ were countable, then, since $$\mathbb Q$$ is countable,

$$A\cup \mathbb Q$$ would be countable,

and therefore $$[0,1]$$, which is a subset of $$A\cup \mathbb Q$$, would be countable,

and that is a contradiction.