The reals have the cardinality of the continuum

Statement

The set $\mathbb{R}$ of real numbers has cardinality $2^{\aleph_0}$: $$|\mathbb{R}| = \mathfrak{c} = |\mathcal{P}(\mathbb{N})|.$$

Remarks

One proof uses the Schröder–Bernstein theorem. A real number gives a set of rationals, the rationals less than it, so $|\mathbb{R}| \le |\mathcal{P}(\mathbb{Q})| = 2^{\aleph_0}$. Binary sequences give an injection in the other direction, for example through decimal expansions with digits $0$ and $1$. With Cantor's theorem, this gives another proof that $\mathbb{R}$ is uncountable.

Explanations

L1 human

The real numbers and the subsets of $\mathbb{N}$ have the same size. Roughly, a real number is an infinite sequence of digits, and an infinite sequence of $0$ and $1$ is a subset of $\mathbb{N}$.