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  "body": "\n## Proof\n\n1. Let $X$ be a [[topological space|0CNQXVBHCX]] and let $S \\subseteq X$.\n2. Let $\\mathcal{U}$ be the family of all [[open sets|01ZQQDCERP]] $U$ of $X$ with $U \\subseteq S$. By the definition of the [[interior|0C38S0KMJK]], $\\operatorname{int} S$ is the [[union|05GK32H72H]] of the members of $\\mathcal{U}$.\n3. Let $x \\in \\operatorname{int} S$. Then $x \\in U$ for some $U \\in \\mathcal{U}$. Since $U \\subseteq S$, we get $x \\in S$. So $\\operatorname{int} S \\subseteq S$.\n4. By [[interior is open|0PNWCM559P]], the set $\\operatorname{int} S$ is open.\n5. Let $U$ be an open set of $X$ with $U \\subseteq S$. Then $U \\in \\mathcal{U}$.\n6. Each member of a family is contained in the union of the family. So $U \\subseteq \\operatorname{int} S$.\n\n## Idea\n\nThe interior is the union of all open sets that fit inside $S$. Each of them lies in the union, and the union still fits inside $S$. So the interior is the biggest open set that fits.\n",
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