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  "body": "\n## Proof\n\nWe iterate $f$ from an arbitrary point and show that the iterates form a Cauchy sequence.\n\n1. Let $(X, d)$ be a nonempty [[complete metric space|08EWT7J0SC]], and let $f \\colon X \\to X$ be a [[contraction|0MT2S0ZX4Y]] with constant $k \\in [0, 1)$.\n2. Let $x_0 \\in X$, which exists since $X$ is nonempty. Put $x_{n+1} = f(x_n)$ for $n \\ge 0$, and let $c = d(x_1, x_0)$.\n3. We show $d(x_{n+1}, x_n) \\le k^n c$ for all $n \\ge 0$, by induction on $n$. For $n = 0$ this reads $c \\le c$.\n4. If it holds for $n$, then $d(x_{n+2}, x_{n+1}) = d(f(x_{n+1}), f(x_n)) \\le k \\, d(x_{n+1}, x_n) \\le k^{n+1} c$.\n5. Let $m \u003e n \\ge 0$. By the [[triangle inequality|01M6CW9VM6]] applied $m - n - 1$ times, $d(x_m, x_n) \\le \\sum_{j=n}^{m-1} d(x_{j+1}, x_j)$.\n6. By step 3 and the [[sum of a finite geometric series|05APWSX53S]] with ratio $k \\ne 1$, this sum is at most $k^n c \\sum_{i=0}^{m-n-1} k^i = k^n c \\, (1 - k^{m-n}) / (1 - k)$.\n7. Since $0 \\le k^{m-n}$, we get $d(x_m, x_n) \\le k^n c / (1 - k)$ for all $m \u003e n \\ge 0$.\n8. Since $0 \\le k \u003c 1$, [[$k^n \\to 0$|0Q7K4AXE53]]. Let $\\varepsilon \u003e 0$, and choose $N$ with $k^N c / (1 - k) \u003c \\varepsilon$.\n9. For $m \u003e n \\ge N$, step 7 gives $d(x_m, x_n) \\le k^n c / (1 - k) \\le k^N c / (1 - k) \u003c \\varepsilon$, since $k^n \\le k^N$.\n10. So $(x_n)$ is a [[Cauchy sequence|0MCK8K2D7Z]]. Since $X$ is complete, it [[converges|06JYPJ4WHH]] to some $x^* \\in X$.\n11. For each $n$, $d(f(x^*), x^*) \\le d(f(x^*), f(x_n)) + d(x_{n+1}, x^*) \\le k \\, d(x^*, x_n) + d(x_{n+1}, x^*)$.\n12. The right side of step 11 tends to $0$ as $n \\to \\infty$. So $d(f(x^*), x^*) \u003c \\varepsilon$ for every $\\varepsilon \u003e 0$, hence $d(f(x^*), x^*) = 0$.\n13. A metric is positive on distinct points, so $f(x^*) = x^*$. Thus $x^*$ is a fixed point.\n14. Let $y$ be a fixed point. Then $d(x^*, y) = d(f(x^*), f(y)) \\le k \\, d(x^*, y)$, so $(1 - k) \\, d(x^*, y) \\le 0$.\n15. Since $1 - k \u003e 0$ and $d(x^*, y) \\ge 0$, we get $d(x^*, y) = 0$, so $y = x^*$. The fixed point is unique.\n16. By steps 10 to 15, for each start $x_0$ the iterates converge to a fixed point, which is the unique fixed point $x^*$.\n17. Fix $n \\ge 0$ and let $m \u003e n$. By the triangle inequality and step 7, $d(x_n, x^*) \\le k^n c / (1 - k) + d(x_m, x^*)$.\n18. Since $d(x_m, x^*) \\to 0$ as $m \\to \\infty$, step 17 gives $d(x_n, x^*) \\le k^n c / (1 - k) + \\varepsilon$ for each $\\varepsilon \u003e 0$.\n19. So $d(x_n, x^*) \\le \\dfrac{k^n}{1 - k} \\, d(x_1, x_0)$ for all $n \\ge 0$.\n\n## Idea\n\nEach application of $f$ shrinks the step between consecutive iterates by the factor $k$.\nThe steps add up to a geometric series, so the iterates form a Cauchy sequence. Completeness gives a limit, and the contraction property makes it a fixed point and the only one.\n",
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