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  "body": "\n## Proof\n\nWe show that the complement of the closed ball is open.\n\n1. Let $(X, d)$ be a [[metric space|063W5JMY85]], let $x \\in X$ and let $r \\ge 0$. Let $\\bar{B}(x, r) = \\set{z \\in X : d(x, z) \\le r}$ be the [[closed ball|05DSV40EB9]].\n2. Put $C = X \\setminus \\bar{B}(x, r)$. Let $y \\in C$. Then $d(x, y) \u003e r$.\n3. Put $\\varepsilon = d(x, y) - r$. By step 2, $\\varepsilon \u003e 0$.\n4. Let $z$ be in the [[open ball|0DC3MPR7C9]] $B(y, \\varepsilon)$, so $d(y, z) \u003c \\varepsilon$.\n5. By the [[triangle inequality|01M6CW9VM6]] and the symmetry of $d$, $d(x, y) \\le d(x, z) + d(z, y) = d(x, z) + d(y, z)$.\n6. Hence $d(x, z) \\ge d(x, y) - d(y, z) \u003e d(x, y) - \\varepsilon = r$.\n7. So $z \\notin \\bar{B}(x, r)$, that is, $z \\in C$. Thus $B(y, \\varepsilon) \\subseteq C$.\n8. Each point of $C$ has a ball about it inside $C$. By the [[description of the open sets of a metric space|0A26X79RTB]], $C$ is open.\n9. By the definition of a [[closed set|0G5AV9Q4RR]], $\\bar{B}(x, r)$ is closed in $X$.\n\n## Idea\n\nA point outside the closed ball is farther than $r$ from the centre, by some margin. The triangle inequality shows that a ball of that margin about the point stays outside. So the complement is open.\n",
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