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  "body": "\n## Proof\n\nWe give injections in both directions and apply the Schröder–Bernstein theorem.\n\n1. Define $L \\colon \\R \\to \\mathcal{P}(\\Q)$ by $L(x) = \\{q \\in \\Q : q \u003c x\\}$, a subset of the [[power set|0BC7KX51Z8]] of $\\Q$.\n2. The map $L$ is [[injective|0PVG1CRAJN]]. Let $x \\neq y$, say $x \u003c y$. By the [[density of the rationals|0STX0WRDNG]], some $q \\in \\Q$ has $x \u003c q \u003c y$. Then $q \\in L(y)$ and $q \\notin L(x)$.\n3. By [[the rationals are countable|0DS9KWFDK2]], there is a bijection $e \\colon \\N \\to \\Q$. Its inverse $e^{-1} \\colon \\Q \\to \\N$ is also a [[bijection|01VG2A0BHW]].\n4. Define $E \\colon \\mathcal{P}(\\Q) \\to \\mathcal{P}(\\N)$ by $E(S) = e^{-1}(S)$. If $E(S) = E(T)$, then $S = e(e^{-1}(S)) = e(e^{-1}(T)) = T$. So $E$ is injective.\n5. The composite $E \\circ L \\colon \\R \\to \\mathcal{P}(\\N)$ is [[injective|01FPXMTBX1]].\n6. For $S \\subseteq \\N$ and $k \\in \\N$, let $s_k(S) = \\sum_{n \\in S,\\ n \u003c k} 10^{-(n+1)}$. The sequence $(s_k(S))_k$ is increasing, since each step adds a term $\\ge 0$.\n7. By the [[sum of a finite geometric series|05APWSX53S]], for $0 \\le m \\le M$ we have $\\sum_{n=m}^{M} 10^{-(n+1)} = \\frac{10^{-(m+1)}(1 - 10^{-(M-m+1)})}{1 - 1/10} \u003c \\frac{10^{-m}}{9}$.\n8. By step 7 with $m = 0$, each $s_k(S) \u003c 1/9$. By the [[monotone convergence theorem|0FM6N4R0VB]], $(s_k(S))_k$ converges. Let $F(S) \\in \\R$ be its limit.\n9. We show that $F \\colon \\mathcal{P}(\\N) \\to \\R$ is injective. Let $S \\neq T$. By the [[well-ordering principle|01BX15HSRR]], there is a least $k$ in exactly one of $S$ and $T$. Say $k \\in S$ and $k \\notin T$.\n10. Let $N \u003e k + 1$. The sets $S$ and $T$ agree below $k$, so the terms with $n \u003c k$ cancel in $s_N(S) - s_N(T)$. The term for $n = k$ is $10^{-(k+1)}$.\n11. Each term with $k \u003c n \u003c N$ lies between $-10^{-(n+1)}$ and $10^{-(n+1)}$. So step 7 with $m = k + 1$ gives $s_N(S) - s_N(T) \\ge 10^{-(k+1)} - \\frac{10^{-(k+1)}}{9} = \\frac{8}{9} \\cdot 10^{-(k+1)}$.\n12. By the [[algebra of limits|07XBM3A9J5]], $s_N(S) - s_N(T) \\to F(S) - F(T)$. Since [[limits preserve non-strict inequalities|0N32V0E3XJ]], $F(S) - F(T) \\ge \\frac{8}{9} \\cdot 10^{-(k+1)} \u003e 0$. So $F(S) \\neq F(T)$.\n13. By the [[Schröder–Bernstein theorem|0A9V6RKQG6]], steps 5 and 12 give a bijection $\\R \\to \\mathcal{P}(\\N)$. So $\\R$ and $\\mathcal{P}(\\N)$ are [[equinumerous|08EJHHMB5R]], and $|\\R| = |\\mathcal{P}(\\N)|$ by the definition of [[cardinality|0JTGRV4WT4]].\n14. By the [[cardinality of the power set|0GPFKS96T7]], $|\\mathcal{P}(\\N)| = 2^{|\\N|} = 2^{\\aleph_0}$, and this is the [[cardinality of the continuum|0WC28W9DNM]] $\\mathfrak{c}$. Hence $|\\R| = \\mathfrak{c} = |\\mathcal{P}(\\N)|$.\n\n## Idea\n\nA real number is determined by the rationals below it, and there are only as many sets of rationals as sets of natural numbers. Conversely, a set $S$ of natural numbers gives the decimal $0.d_0 d_1 d_2 \\ldots$ with $d_n = 1$ for $n \\in S$ and $d_n = 0$ otherwise. Digits $0$ and $1$ never produce two expansions of the same number, so this map is injective.\n",
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