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  "body": "\n## Proof\n\nWe prove the two directions separately.\n\nFirst, let $S$ be [[totally bounded|0WE5BBHGM6]], and fix a real $\\varepsilon \u003e 0$.\n\n1. By the definition, there are sets $A_1, \\ldots, A_n \\subseteq X$ with $S \\subseteq A_1 \\cup \\cdots \\cup A_n$ and $\\operatorname{diam}(A_k) \u003c \\varepsilon$ for each $k$. Here the [[diameter|0692CGQF00]] is defined, so each $A_k$ is nonempty.\n2. Let $J$ be the set of $k$ with $A_k \\cap S \\ne \\varnothing$. For each $k \\in J$, choose a point $y_k \\in A_k \\cap S$.\n3. Let $F = \\set{y_k : k \\in J}$. Then $F$ is a finite subset of $S$.\n4. Let $s \\in S$. By step 1, $s \\in A_k$ for some $k$. Then $s \\in A_k \\cap S$, so $k \\in J$.\n5. Both $s$ and $y_k$ lie in $A_k$. So $d(s, y_k) \\le \\operatorname{diam}(A_k) \u003c \\varepsilon$, and $s$ is in the [[open ball|0DC3MPR7C9]] $B(y_k, \\varepsilon)$.\n6. Hence $S \\subseteq \\bigcup_{y \\in F} B(y, \\varepsilon)$.\n\nConversely, assume the condition on finite sets, and fix a real $\\varepsilon \u003e 0$.\n\n7. Apply the condition with $\\varepsilon / 3$. We get a finite set $F \\subseteq S$ with $S \\subseteq \\bigcup_{y \\in F} B(y, \\varepsilon / 3)$.\n8. Let $y \\in F$. The ball $B(y, \\varepsilon / 3)$ contains $y$, so it is nonempty.\n9. Let $z, w \\in B(y, \\varepsilon / 3)$. By the [[triangle inequality|01M6CW9VM6]], $d(z, w) \\le d(z, y) + d(y, w) \u003c 2 \\varepsilon / 3$.\n10. So $\\operatorname{diam} B(y, \\varepsilon / 3) \\le 2 \\varepsilon / 3 \u003c \\varepsilon$.\n11. By steps 7 and 10, the finitely many balls $B(y, \\varepsilon / 3)$, $y \\in F$, cover $S$ and have diameter less than $\\varepsilon$.\n12. This holds for each $\\varepsilon \u003e 0$, so $S$ is totally bounded.\n\n## Idea\n\nA small set that meets $S$ can be replaced by a ball around one point of $S$ inside it. Conversely, a ball of radius $r$ has diameter at most $2r$. Shrinking the radius to $\\varepsilon / 3$ turns the balls into sets of diameter less than $\\varepsilon$.\n",
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