TITLE: Is the power set of the natural numbers countable?
QUESTION [62 upvotes]: Some explanations:
A set S is countable if there exists an injective function $f$ from $S$ to the natural numbers ($f:S \rightarrow \mathbb{N}$).
$\{1,2,3,4\}, \mathbb{N},\mathbb{Z}, \mathbb{Q}$ are all countable.
$\mathbb{R}$ is not countable.
The power set $\mathcal P(A) $ is defined as a set of all possible subsets of A, including the empty set and the whole set.
$\mathcal P (\{\})=\{\{\}\}, \mathcal P (\mathcal P(\{\}))=\{\{\}, \{\{\}\}\} $
$\mathcal P(\{1,2\})=\{\{\}, \{1\},\{2\},\{1,2\}\}$
My question is:
Is $\mathcal P(\mathbb{N})$ countable? How would an injective function $f:S \rightarrow \mathbb{N}$ look like?

REPLY [2 votes]: Here is my favorite proof that
$$|\mathcal{P}(M)| > |M|, \quad\forall M.$$
Proof
$$\big[\exists\phi:M \stackrel{\mathrm{on}}{\to}\mathcal{P}(M)\big]$$
$$\Downarrow$$
$$\Big[\big[A_{\not\phi} := \{m|m\in M, m\notin\phi(m)\}\big]\Rightarrow A_{\not\phi}\in\mathcal{P}(M)\Rightarrow\exists p\in M:\phi(p) = A_{\not\phi}\Big]$$
$$\Downarrow$$
$$\Big[\big[p\in A_{\not\phi}\Rightarrow p\notin A_{\not\phi}\big]\quad \mathrm{and}\quad\big[p\notin A_{\not\phi}\Rightarrow p\in A_{\not\phi}\big]\Big]$$
$$Q.E.D.$$