Is the power set of the natural numbers countable?

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?

Cantor's Theorem states that for any set $A$ there is no surjective function $A\to\mathcal P(A)$. With $A=\mathbb N$ this implies that $\mathcal P(\mathbb N)$ is not countable.

(But where on earth did you find those nice explanations of countability and power sets that didn't also tell you this?)

Power set of natural numbers has the same cardinality with the real numbers. So, it is uncountable.

In order to be rigorous, here's a proof of this.

Here is my favorite proof that $$|\mathcal{P}(M)| > |M|, \quad\forall M.$$

Proof

$$\big[\exists\phi:M \stackrel{\mathrm{on}}{\to}\mathcal{P}(M)\big]$$ $$\Downarrow$$ $$\Big[\big[A_{\not\phi} := \{m|m\in M, m\notin\phi(m)\}\big]\Rightarrow A_{\not\phi}\in\mathcal{P}(M)\Rightarrow\exists p\in M:\phi(p) = A_{\not\phi}\Big]$$ $$\Downarrow$$ $$\Big[\big[p\in A_{\not\phi}\Rightarrow p\notin A_{\not\phi}\big]\quad \mathrm{and}\quad\big[p\notin A_{\not\phi}\Rightarrow p\in A_{\not\phi}\big]\Big]$$ $$Q.E.D.$$