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  "body": "\n## Proof\n\nWe cover $X$ by balls whose doubles lie in members of the cover, and take the smallest radius of a finite subcover.\n\n1. If $X$ is empty, there is no nonempty $A \\subseteq X$, and the [[diameter|0692CGQF00]] is only defined for nonempty sets. So $\\delta = 1$ works. From now on let $X$ be nonempty.\n2. Let $p \\in X$. Since the $U_i$ cover $X$, choose $i(p) \\in I$ with $p \\in U_{i(p)}$.\n3. The set $U_{i(p)}$ is open in the [[metric topology|0A6YRNZB2R]]. So there is $r_p \u003e 0$ with $B(p, 2 r_p) \\subseteq U_{i(p)}$.\n4. The [[open balls|0DC3MPR7C9]] $B(p, r_p)$, for $p \\in X$, are [[open|0B2305GNAZ]] and each contains its centre. So they form an [[open cover|0NVPNC12JY]] of $X$.\n5. Since $X$ is [[compact|0QPHW703PV]], finitely many of these balls cover $X$. Call their centres $p_1, \\ldots, p_n$, with $n \\ge 1$, and write $r_k = r_{p_k}$.\n6. Let $\\delta = \\min(r_1, \\ldots, r_n)$. Then $\\delta \u003e 0$.\n7. Let $A \\subseteq X$ have $\\operatorname{diam}(A) \u003c \\delta$. The diameter is only defined for nonempty sets, so pick $a \\in A$.\n8. By step 5, $a \\in B(p_k, r_k)$ for some $k$, so $d(a, p_k) \u003c r_k$.\n9. Let $y \\in A$. Then $d(y, a) \\le \\operatorname{diam}(A) \u003c \\delta \\le r_k$.\n10. By the [[triangle inequality|01M6CW9VM6]], $d(y, p_k) \\le d(y, a) + d(a, p_k) \u003c r_k + r_k = 2 r_k$.\n11. So $A \\subseteq B(p_k, 2 r_k)$. By step 3, $A \\subseteq U_{i(p_k)}$. Thus $\\delta$ is a Lebesgue number of the cover.\n\n## Idea\n\nAround each point choose a ball such that the ball of twice the radius fits inside one set of the cover. Compactness leaves finitely many such balls, so the smallest radius is positive. A set of smaller diameter sits in one small ball around a point of it, and hence in the double ball.\n",
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