I have been trying to prove that the Power set of Positive Integers is uncountable using :
Is $i$ an element of the $i$-th subset
But I can not seem to prove it.
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My answer at the duplicate should be enough, I hope. – Asaf Karagila Mar 01 '18 at 07:21
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Given any set $X$, you can prove that no function $f : X \to \mathcal{P}(X)$ is surjective. To prove this this, take any such function $f$ and let $B_f = \{ a \in X \mid a \not \in f(a) \} \in \mathcal{P}(X)$; it follows that $B_f \ne f(x)$ for any $x \in X$. This result is known as Cantor's theorem.
A corollary of this is that no function $\mathbb{Z}^+ \to \mathcal{P}(\mathbb{Z}^+)$ is surjective. Can you see where to take it from here?
Clive Newstead
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So since no function $\mathbb{Z}^+ \to \mathcal{P}(\mathbb{Z}^+)$ is surjective, there are some elements in $\mathcal{P}(\mathbb{Z}^+)$ that are missed and therefor it is uncountable? – Rick Mar 01 '18 at 05:36
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@Rick: You need to connect this fact with the definition of 'countable' that you're working with. Usually, what it means for a set $X$ to be countable is that there is a bijection $\mathbb{N} \to X$, or equivalently a bijection $\mathbb{Z}^+ \to X$. If no function $\mathbb{Z}^+ \to \mathcal{P}(\mathbb{Z}^+)$ is surjective, why can $\mathcal{P}(\mathbb{Z}^+)$ not be countable? – Clive Newstead Mar 01 '18 at 05:54
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So since no function can't be surjective, it can not be countable as to be countable, the function must have a bijection, this contradicts. – Rick Mar 01 '18 at 06:04
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@Rick: Pretty much. It's not that 'the function must be a bijection', but rather that 'some function must be a bijection'; what you prove is that no function is a bijection. – Clive Newstead Mar 01 '18 at 13:53