Cardinality of the power set

Statement

Let $A$ be a set. Then the power set of $A$ is equinumerous with the set $\{0, 1\}^A$ of functions from $A$ to $\{0, 1\}$, so $$|\mathcal{P}(A)| = 2^{|A|}.$$

Remarks

The bijection sends a subset $S$ to its indicator function $\mathbf{1}_S$. For a finite set with $n$ elements, the power set has $2^n$ elements. For $A = \mathbb{N}$, $|\mathcal{P}(\mathbb{N})| = 2^{\aleph_0}$, the cardinality of the continuum.

Explanations

L1 human

A subset is a choice of "in" or "out" for each element. With $n$ elements, there are $2 \cdot 2 \cdots 2 = 2^n$ such choices.