The power set of $\mathbb{N}$, $\mathcal{P}(\mathbb{N})$, is not countable as well as the sets $\mathbb{R}$, and $\{0, 1\}^{\mathbb{N}}$ the set of all sequences which takes values 0 or 1. Use this to show that the set of all irrationals is not countable.

We have the set $\mathcal{E} = (\mathcal{E}_{n}) \in \{0, 1\}^{\mathbb{R}}$ defined by $\mathcal{E}_{n} = 1 - f(n)_{n}$, where $f(n)_{n}$ is the n-term in the sequence $f(n) \in \{ 0, 1\}^{\mathbb{N}}$

We can find a subset of $\mathbb{R}$ to show that $\mathbb{R}$ is not countable:

$$ C = \left \{ \sum_{n=0}^{\infty} \frac{\mathcal{E}_{n}}{3^n}; (\mathcal{E}_{n}) \in \{ 0, 1 \}^{\mathbb{R}} \right \} $$

$C$ is the Cantor triadic set. So, $C$ is a subset of $[0,1] \in \mathbb{R}$, and it is bijection with $\{ 0, 1\}^{\mathbb{N}}$. Hence $\mathbb{R}$ is not countable. Since $\mathbb{R}$ is not countable and $\mathbb{Q}$ is countable, then $\mathbb{R}$ \ $\mathbb{Q}$ is the set of all irrationals and it's not countable since the union of two countable sets is countable.

Now, I'm looking for other solutions to prove it, please comment on

  • Since you know $\mathbb R$ is not countable, and $\mathbb Q$ is, then doesn't it follow rather trivially that $\mathbb R \backslash \mathbb Q$ is not countable? – Gregory Grant Mar 10 '15 at 16:29
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    The question is rather oddly stated, since if you know that $\Bbb Q$ is uncountable and $\Bbb R$ is not, the rest of the information is superfluous. A much harder alternative is to show that there is a bijection between the irrationals and $\Bbb N^{\Bbb N}$; you can then observe that there is an obvious injection from $\{0,1\}^{\Bbb N}$ into $\Bbb N^{\Bbb N}$ and conclude that the irrationals are uncountable. – Brian M. Scott Mar 10 '15 at 22:42

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