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  "body": "\n## Proof\n\nWe show that each stage $C_n$ is closed, and then intersect.\n\n1. In the construction of the [[Cantor set|0BXNDZFQME]], $C_0 = [0, 1]$ is one closed interval. Removing the open middle third of a closed interval $[a, b]$ leaves the two closed intervals $[a, a + \\tfrac{b-a}{3}]$ and $[b - \\tfrac{b-a}{3}, b]$.\n2. So, by induction on $n$, each $C_n$ is a union of $2^n$ closed intervals.\n3. A closed interval $[a, b]$ is [[closed|0G5AV9Q4RR]] in $\\mathbb{R}$. Its complement is $(-\\infty, a) \\cup (b, \\infty)$. Each point $x$ of it has the ball of radius $\\min(|x - a|, |x - b|)$ inside the complement, so the complement is open by [[the ball criterion|0A26X79RTB]].\n4. A finite union $F_1 \\cup \\cdots \\cup F_k$ of closed sets is closed. Its complement is $(\\mathbb{R} \\setminus F_1) \\cap \\cdots \\cap (\\mathbb{R} \\setminus F_k)$, which is open by [[finite intersection of open sets is open|09NDE9V7X1]].\n5. By steps 2 to 4, each $C_n$ is closed in $\\mathbb{R}$.\n6. By [[intersection of closed sets is closed|09YMV8H4X5]], with the nonempty index set $\\{0, 1, 2, \\ldots\\}$, the set $C = \\bigcap_{n \\ge 0} C_n$ is closed in $\\mathbb{R}$.\n\n## Idea\n\nEach stage of the construction is a finite union of closed intervals, so it is closed. Closed sets stay closed under arbitrary intersections, and the Cantor set is the intersection of all stages.\n",
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