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  "body": "\n## Proof\n\nWe show that $A$ equals its closure.\n\n1. Let $(X, d)$ be a [[metric space|063W5JMY85]] and let $A \\subseteq X$. Assume $A$ with the [[subspace metric|0T2WYDTRP2]] $d_A$ is [[complete|08EWT7J0SC]].\n2. Let $x$ be a point of the [[closure|0DRKT1QJR7]] $\\overline{A}$.\n3. Since the [[closure is the set of limits of sequences|003QP7A8HG]], there is a sequence $(a_n)$ of points of $A$ that [[converges|06JYPJ4WHH]] to $x$ in $X$.\n4. Let $\\varepsilon \u003e 0$. There is $N$ with $d(a_n, x) \u003c \\varepsilon / 2$ for $n \\ge N$.\n5. For $m, n \\ge N$, the [[triangle inequality|01M6CW9VM6]] gives $d_A(a_m, a_n) = d(a_m, a_n) \\le d(a_m, x) + d(x, a_n) \u003c \\varepsilon$.\n6. So $(a_n)$ is a [[Cauchy sequence|0MCK8K2D7Z]] in $(A, d_A)$. Since $A$ is complete, $(a_n)$ converges in $A$ to some $a \\in A$.\n7. Since $d(a_n, a) = d_A(a_n, a)$, the sequence $(a_n)$ also converges to $a$ in $X$.\n8. For each $n$, $d(x, a) \\le d(x, a_n) + d(a_n, a)$. The right side tends to $0$, so $d(x, a) \u003c \\varepsilon$ for every real $\\varepsilon \u003e 0$.\n9. So $d(x, a) = 0$. A metric is positive on distinct points, so $x = a$, and thus $x \\in A$.\n10. By steps 2 to 9, $\\overline{A} \\subseteq A$. Also $A \\subseteq \\overline{A}$, because the [[closure is the smallest closed superset|0SHC6Z304Z]].\n11. So $\\overline{A} = A$. Since [[a set is closed iff it equals its closure|0D2W7FS52T]], $A$ is [[closed|0G5AV9Q4RR]] in $X$.\n\n## Idea\n\nA point of the closure is a limit of points of $A$. Such a sequence is Cauchy, so completeness of $A$ gives it a limit inside $A$.\nLimits are unique, so the original point is that limit and lies in $A$.\n",
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