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  "body": "\n## Proof\n\nWe prove each direction separately.\n\n1. Let $(X, d)$ be a [[metric space|063W5JMY85]], let $S \\subseteq X$, let $x \\in X$, and let $\\overline{S}$ be the [[closure|0DRKT1QJR7]] of $S$.\n2. First let $x \\in \\overline{S}$. By the [[description of the closure in a metric space|068P0P32DW]], each $n \\ge 1$ gives a point $s_n \\in S$ with $d(x, s_n) \u003c 1/n$.\n3. We show that $s_n \\to x$, in the sense of [[convergence|06JYPJ4WHH]]. Let $\\varepsilon \u003e 0$.\n4. The [[reals are Archimedean|0A3GWZ86AY]], so there is a natural number $N$ with $N \\varepsilon \u003e 1$. Then $N \\ge 1$, and $1/N \u003c \\varepsilon$.\n5. For $n \\ge N$, symmetry of $d$ gives $d(s_n, x) = d(x, s_n) \u003c 1/n \\le 1/N \u003c \\varepsilon$. So $s_n \\to x$.\n6. Conversely, let $(s_n)$ be a sequence of points of $S$ with $s_n \\to x$.\n7. By [[closure is the smallest closed superset|0SHC6Z304Z]], $S \\subseteq \\overline{S}$ and $\\overline{S}$ is closed. So $(s_n)$ is a sequence of points of the closed set $\\overline{S}$.\n8. Since [[closed sets contain their limits|09GQ3ZTQJ9]], we get $x \\in \\overline{S}$.\n\n## Idea\n\nA point of the closure has points of $S$ at distance less than $1/n$ for every $n$, and these form a sequence that tends to it. Conversely, the closure is a closed set that contains $S$. So it keeps every limit of points of $S$.\n",
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