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  "body": "\n## Proof\n\nWe argue by contradiction, with the complements of the sets $C_n$ as an open cover.\n\n1. Let $X$ be a [[topological space|0CNQXVBHCX]]. Let $C_1 \\supseteq C_2 \\supseteq \\cdots$ be nonempty [[compact|00R0X1WGPX]] [[closed|0G5AV9Q4RR]] subsets of $X$.\n2. Assume that $\\bigcap_{n \\ge 1} C_n = \\emptyset$. For $n \\ge 1$, let $U_n = X \\setminus C_n$. Each $U_n$ is open, because $C_n$ is closed.\n3. Let $x \\in C_1$. By step 2, $x$ is not in some $C_n$, so $x \\in U_n$. Hence $C_1 \\subseteq \\bigcup_{n \\ge 1} U_n$.\n4. The set $C_1$ is compact. By [[compactness via open covers|0P6PMJB4MA]], there are finitely many indices $n_1, \\ldots, n_p$ with $C_1 \\subseteq U_{n_1} \\cup \\cdots \\cup U_{n_p}$.\n5. Since $C_1$ is nonempty, $p \\ge 1$. Let $m = \\max(n_1, \\ldots, n_p)$.\n6. For $n \\le m$ we have $C_m \\subseteq C_n$, so $U_n \\subseteq U_m$. Hence $C_1 \\subseteq U_m = X \\setminus C_m$.\n7. But $C_m \\subseteq C_1$, so $C_m \\subseteq X \\setminus C_m$. This forces $C_m = \\emptyset$.\n8. This contradicts the hypothesis that $C_m$ is nonempty. So $\\bigcap_{n \\ge 1} C_n \\ne \\emptyset$.\n\n## Idea\n\nIf the intersection were empty, the complements would form an open cover of the compact set $C_1$.\nBecause the sets are nested, a finite subcover reduces to a single complement $X \\setminus C_m$, and that would make $C_m$ empty.\n",
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