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  "body": "\n## Proof\n\nWe use the [[diagonal argument|0XCH8QVDCQ]].\n\n1. Let $A$ be a set and let $f \\colon A \\to \\mathcal{P}(A)$ be a function to the [[power set|0BC7KX51Z8]].\n2. Let $D = \\{a \\in A : a \\notin f(a)\\}$. Then $D \\subseteq A$, so $D \\in \\mathcal{P}(A)$.\n3. Suppose that $D = f(d)$ for some $d \\in A$.\n4. If $d \\in D$, then $d \\notin f(d)$ by the definition of $D$. Since $f(d) = D$, this says $d \\notin D$, a contradiction.\n5. If $d \\notin D$, then $d \\notin f(d)$, since $f(d) = D$. By the definition of $D$, this means $d \\in D$, a contradiction.\n6. So no $d \\in A$ has $f(d) = D$. Hence $f$ is not [[surjective|0XKH881BZ7]].\n7. The map $s \\colon A \\to \\mathcal{P}(A)$, $s(a) = \\{a\\}$, is [[injective|0PVG1CRAJN]], since $\\{a\\} = \\{a'\\}$ implies $a = a'$. So $|A| \\le |\\mathcal{P}(A)|$ in the [[order of cardinals|01GTP8M5RG]].\n8. Suppose that $|A| = |\\mathcal{P}(A)|$. By the definition of [[cardinality|0JTGRV4WT4]], $A$ and $\\mathcal{P}(A)$ are [[equinumerous|08EJHHMB5R]], so there is a [[bijection|0SNQEF7Z53]] $A \\to \\mathcal{P}(A)$.\n9. A bijection is surjective, which contradicts step 6. So $|A| \\neq |\\mathcal{P}(A)|$.\n10. By steps 7 and 9, $|A| \u003c |\\mathcal{P}(A)|$.\n\n## Idea\n\nThe set $D$ is built to differ from each $f(a)$ at the element $a$ itself. So $D$ cannot be any of the sets $f(a)$. This is the same diagonal trick that shows the real numbers are uncountable.\n",
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