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  "body": "\n## Proof\n\n1. Let $S \\subseteq \\R$ be nonempty and [[bounded below|0B5H2V4JW8]]. Let $m$ be a lower bound of $S$.\n2. In the [[ordered field|0QSPNASE50]] $\\R$, $u \\le v$ implies $-v \\le -u$. Indeed, add $-u - v$ to both sides.\n3. Let $-S = \\set{-s : s \\in S}$. It is nonempty, because $S$ is nonempty.\n4. For each $s \\in S$ we have $m \\le s$, so $-s \\le -m$ by step 2. Thus $-m$ is an upper bound of $-S$.\n5. By the [[completeness axiom|0AXWET80YW]], $-S$ has a [[supremum|0BTMQ04556]] $b$. Let $a = -b$.\n6. The number $a$ is a lower bound of $S$. For $s \\in S$, we have $-s \\in -S$, so $-s \\le b$. By step 2, $a = -b \\le s$.\n7. Let $\\ell$ be any lower bound of $S$. As in step 4, $-\\ell$ is an upper bound of $-S$.\n8. Since $b$ is the least upper bound, $b \\le -\\ell$. By step 2, $\\ell \\le -b = a$.\n9. By steps 6 and 8, $a$ is a lower bound of $S$ that is larger than or equal to each lower bound. So $a$ is an [[infimum|04KFTA34C5]] of $S$.\n\n## Idea\n\nNegation reverses the order, so it turns lower bounds into upper bounds. The supremum of the negated set exists by completeness, and its negative is the infimum.\n",
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