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  "body": "\n## Proof\n\nWe prove the two directions separately.\n\n1. Let $A$ be [[open|01ZQQDCERP]] and let $x \\in A$. Take $U = A$. Then $U$ is open and $x \\in U \\subseteq A$. So $A$ is a neighbourhood of $x$.\n2. Conversely, let $A$ be a neighbourhood of each of its points. For each $x \\in A$, choose an open set $U_x$ with $x \\in U_x \\subseteq A$.\n3. Let $W = \\bigcup_{x \\in A} U_x$, the [[union|05GK32H72H]] of this family. Each $U_x \\subseteq A$, so $W \\subseteq A$.\n4. Each $x \\in A$ lies in $U_x$, so $A \\subseteq W$. Hence $A = W$.\n5. By the axioms of a [[topology|0CNQXVBHCX]], a union of members of $\\tau$ is in $\\tau$. This includes the empty family, whose union is $\\varnothing$. So $A = W$ is open.\n\n## Idea\n\nAn open set is its own open neighbourhood at each point. Conversely, if each point has an open set around it inside $A$, then $A$ is the union of these open sets, and unions of open sets are open.\n",
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