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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.

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    $\begingroup$ 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 $[0,1]\subset A\cup \mathbb Q$ $\endgroup$ – J. W. Tanner Jan 10 at 15:55
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    $\begingroup$ $\varnothing$ is also a subset of the irrational numbers. $\endgroup$ – Asaf Karagila Jan 10 at 16:24
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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.

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

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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.

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