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  "body": "\n## Proof\n\nWe prove each direction separately.\n\n1. Let $(X, d)$ be a [[metric space|063W5JMY85]], let $S \\subseteq X$ and let $x \\in X$. Let $\\mathcal{F}$ be the family of all [[closed sets|0G5AV9Q4RR]] $F$ of $X$ with $S \\subseteq F$.\n2. By the definition of the [[closure|0DRKT1QJR7]], $\\overline{S}$ is the intersection of the members of $\\mathcal{F}$.\n3. For $\\varepsilon \u003e 0$, let $B(x, \\varepsilon)$ be the [[open ball|0DC3MPR7C9]]. A point $s$ lies in $B(x, \\varepsilon)$ exactly when $d(x, s) \u003c \\varepsilon$.\n4. First let $x \\in \\overline{S}$. Suppose, for a contradiction, that some $\\varepsilon \u003e 0$ has no $s \\in S$ with $d(x, s) \u003c \\varepsilon$. Then $S \\cap B(x, \\varepsilon) = \\varnothing$.\n5. Put $F = X \\setminus B(x, \\varepsilon)$. Its complement $B(x, \\varepsilon)$ is open, because [[open balls are open|0B2305GNAZ]]. So $F$ is closed.\n6. By step 4, $S \\subseteq F$. Hence $F \\in \\mathcal{F}$, and $\\overline{S} \\subseteq F$ by step 2.\n7. But $d(x, x) = 0 \u003c \\varepsilon$, so $x \\in B(x, \\varepsilon)$ and $x \\notin F$. This contradicts $x \\in \\overline{S} \\subseteq F$.\n8. So for each $\\varepsilon \u003e 0$ there is $s \\in S$ with $d(x, s) \u003c \\varepsilon$.\n9. Now suppose that for each $\\varepsilon \u003e 0$ there is $s \\in S$ with $d(x, s) \u003c \\varepsilon$. Let $F \\in \\mathcal{F}$.\n10. Suppose, for a contradiction, that $x \\notin F$. Then $x$ lies in $X \\setminus F$, which is open because $F$ is closed.\n11. By the [[description of the open sets of a metric space|0A26X79RTB]], there is $\\varepsilon \u003e 0$ with $B(x, \\varepsilon) \\subseteq X \\setminus F$.\n12. By hypothesis, there is $s \\in S$ with $d(x, s) \u003c \\varepsilon$. Then $s \\in B(x, \\varepsilon) \\subseteq X \\setminus F$, so $s \\notin F$.\n13. This contradicts $s \\in S \\subseteq F$. Hence $x \\in F$.\n14. So $x$ lies in each member of $\\mathcal{F}$, and $x \\in \\overline{S}$ by step 2.\n\n## Idea\n\nIf a ball about $x$ misses $S$, its complement is a closed set that holds $S$ but not $x$. So $x$ is not in the closure. If every ball about $x$ meets $S$, no closed superset of $S$ can leave $x$ out, since its open complement would hold such a ball.\n",
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