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  "body": "\n## Proof\n\nWe prove each implication. The key step shows that a Cauchy sequence follows its convergent subsequences.\n\n1. Let $(X, d)$ be a [[metric space|063W5JMY85]].\n2. Assume $X$ is [[complete|08EWT7J0SC]]. Let $(x_n)$ be a [[Cauchy sequence|0MCK8K2D7Z]] in $X$. By completeness, it [[converges|06JYPJ4WHH]] to some $x \\in X$.\n3. With $n_k = k$, the sequence $(x_{n_k})$ is a [[subsequence|0R8J3ZQMJS]] of $(x_n)$, namely $(x_n)$ itself. So $(x_n)$ has a convergent subsequence.\n4. Now assume that each Cauchy sequence in $X$ has a convergent subsequence. Let $(x_n)_{n \\ge 1}$ be a Cauchy sequence in $X$.\n5. By assumption, there are integers $1 \\le n_1 \u003c n_2 \u003c \\cdots$ and a point $x \\in X$ with $x_{n_k} \\to x$ as $k \\to \\infty$.\n6. The $n_k$ are strictly increasing integers with $n_1 \\ge 1$. By induction on $k$, we get $n_k \\ge k$ for all $k \\ge 1$.\n7. Let $\\varepsilon \u003e 0$ be real. Since $(x_n)$ is Cauchy, there is an integer $N$ with $d(x_m, x_n) \u003c \\varepsilon / 2$ for all $m, n \\ge N$.\n8. Since $x_{n_k} \\to x$, there is an integer $L$ with $d(x_{n_k}, x) \u003c \\varepsilon / 2$ for all $k \\ge L$.\n9. Let $k = \\max(N, L)$. Then $k \\ge L$, and $n_k \\ge k \\ge N$ by step 6.\n10. Let $n \\ge N$. By the [[triangle inequality|01M6CW9VM6]] and steps 7 to 9, $d(x_n, x) \\le d(x_n, x_{n_k}) + d(x_{n_k}, x) \u003c \\varepsilon / 2 + \\varepsilon / 2 = \\varepsilon$.\n11. Since $\\varepsilon$ was arbitrary, $x_n \\to x$. So each Cauchy sequence in $X$ converges, and $X$ is complete.\n\n## Idea\n\nIn a complete space, the whole sequence converges, so it is its own convergent subsequence.\nConversely, late terms of a Cauchy sequence are close to each other, and some late term of the subsequence is close to the limit. The triangle inequality joins the two estimates.\n",
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