The simplest way of proving that $|\mathcal{P}(\mathbb{N})| = |\mathbb{R}| = c$ What is the simplest way of proving (to a non-mathematician) that the power set of the set of natural numbers has the same cardinality as the set of the real numbers, i.e. how to construct a bijection from $\mathcal{P}(\mathbb{N})$ to $\mathbb{R}$?

I will prove every reasonable "equinumerosity" involved. There is a key lemma (Schröder–Bernstein theorem) that states that $|A|=|B|$ iff there is an injective map from $A$ to $B$ and an injective map from $B$ to $A$.


*

*$|\mathbb{R}|=|(0,1)|$ is trivial: $f(x)=\frac{1}{\pi}\left(\frac{\pi}{2}+\arctan(x)\right)$ is an increasing function that maps $\mathbb{R}$ to $(0,1)$;

*$|(0,1)|=|[0,1]|$ is less trivial: take an enumeration of the rational numbers in $[0,1]$ such that $q_0=0,q_1=1$, then map $q_n$ into $q_{n+2}$ (Hilbert's hotel);

*$|(0,1)|=|2^{\mathbb{N}}|$ is tricky: every number in $(0,1)$ has a unique canonical binary representation, where canonical means that no tails of $11111\ldots$ are allowed: that gives an injective map from $(0,1)$ to $2^\mathbb{N}$. On the other hand, if $\{a_n\}_{n\geq 0}$ is a sequence such that $a_i\in\{0,1\}$, the map
$$ \{a_n\}_{n\geq 0} \mapsto \sum_{n\geq 0}\frac{2a_n+1}{5^{n+1}} $$
sends $2^{\mathbb{N}}$ into a Cantor subset of $(0,1)$ in a injective way.

*$|\mathcal{P}(\mathbb{N})|=|2^{\mathbb{N}}|\,$ is trivial again: every  non-empty subset $A\subseteq \mathbb{N}$ can be associated with the sequence $\{a_n\}_{n\geq 0}$ in which $a_n=1$ if $n\in A$ and $a_n=0$ otherwise.


By $(4)+(3)+(2)+(1)$, $$\left|\mathcal{P}(\mathbb{N})\right|=\left|\mathbb{R}\right|$$
as wanted. The trickiest part comes from noticing that this approach shows that a bijection exists without an explicit construction: the "i.e." in OP's question is not really an "i.e.", unless we are constructivists.
