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  "body": "\n## Proof\n\n1. Let $(X, d)$ be a [[metric space|063W5JMY85]], let $n \\ge 1$, and let $U_1, \\ldots, U_n$ be [[open subsets|01ZQQDCERP]] of $X$.\n2. Put $U = U_1 \\cap \\cdots \\cap U_n$, the [[intersection|0HBRW4E6G1]] of these sets.\n3. Let $x \\in U$. Then $x \\in U_k$ for each $k = 1, \\ldots, n$.\n4. Each $U_k$ is open. By the [[description of the open sets of a metric space|0A26X79RTB]], there is $\\varepsilon_k \u003e 0$ with $B(x, \\varepsilon_k) \\subseteq U_k$.\n5. Here $B(x, \\varepsilon_k)$ is the [[open ball|0DC3MPR7C9]] with centre $x$ and radius $\\varepsilon_k$.\n6. Put $\\varepsilon = \\min(\\varepsilon_1, \\ldots, \\varepsilon_n)$. This is the least of finitely many positive reals, so $\\varepsilon \u003e 0$.\n7. Let $y \\in B(x, \\varepsilon)$. Then $d(x, y) \u003c \\varepsilon \\le \\varepsilon_k$, so $y \\in B(x, \\varepsilon_k) \\subseteq U_k$ for each $k$.\n8. Thus $y \\in U$, and so $B(x, \\varepsilon) \\subseteq U$.\n9. Each point of $U$ has a ball about it inside $U$. By the same description of open sets, $U$ is open in $X$.\n\n## Idea\n\nEach of the sets gives a radius that works for it. The smallest radius works for all of them at once. Finiteness matters: the minimum of finitely many positive numbers is positive, but an infimum of infinitely many can be $0$.\n",
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