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  "body": "\n## Proof\n\nWe give a map with a two-sided inverse.\n\n1. Let $A$ be a set, and let $\\{0, 1\\}^A$ be the set of functions from $A$ to $\\{0, 1\\}$.\n2. Define $\\Phi \\colon \\mathcal{P}(A) \\to \\{0, 1\\}^A$ on the [[power set|0BC7KX51Z8]] by $\\Phi(S) = \\mathbf{1}_S$, the [[indicator function|0VSB0XF4G7]] of $S$. So $\\mathbf{1}_S(a) = 1$ if $a \\in S$, and $\\mathbf{1}_S(a) = 0$ if $a \\notin S$.\n3. Define $\\Psi \\colon \\{0, 1\\}^A \\to \\mathcal{P}(A)$ by $\\Psi(\\chi) = \\{a \\in A : \\chi(a) = 1\\}$.\n4. Let $S \\subseteq A$. Then $\\Psi(\\Phi(S)) = \\{a \\in A : \\mathbf{1}_S(a) = 1\\} = S$. So $\\Psi \\circ \\Phi$ is the identity of $\\mathcal{P}(A)$.\n5. Let $\\chi \\in \\{0, 1\\}^A$ and put $S = \\Psi(\\chi)$. For $a \\in A$, $\\mathbf{1}_S(a) = 1$ exactly when $a \\in S$, that is, when $\\chi(a) = 1$.\n6. Both functions take only the values $0$ and $1$. So step 5 gives $\\mathbf{1}_S(a) = \\chi(a)$ for all $a$, and $\\Phi(\\Psi(\\chi)) = \\chi$.\n7. By steps 4 and 6, $\\Phi$ has a two-sided inverse, so $\\Phi$ is a bijection, since [[a map with a two-sided inverse is bijective|01VG2A0BHW]]. Thus $\\mathcal{P}(A)$ and $\\{0, 1\\}^A$ are [[equinumerous|08EJHHMB5R]].\n8. The cardinal $2^{|A|}$ is by definition the [[cardinality|0JTGRV4WT4]] of $\\{0, 1\\}^A$. So step 7 gives $|\\mathcal{P}(A)| = |\\{0, 1\\}^A| = 2^{|A|}$.\n\n## Idea\n\nA subset of $A$ is the same information as a yes-or-no answer for each element of $A$. The indicator function records these answers, and reading off the \"yes\" answers gives the subset back.\n",
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