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  "body": "\n## Proof\n\nWe argue by contradiction.\n\n1. Let $S$ be a nonempty set of reals that is bounded above, let $b = \\sup S$ be its [[supremum|0BTMQ04556]], and let $c \u003c b$.\n2. Suppose that no $s \\in S$ satisfies $c \u003c s \\le b$.\n3. Let $s \\in S$. Since $b$ is an [[upper bound|0B5H2V4JW8]] of $S$, $s \\le b$. By step 2, $c \u003c s$ fails.\n4. The order of $\\R$ is total, by the [[ordered field property|0QSPNASE50]]. So $s \\le c$.\n5. Thus $c$ is an upper bound of $S$. Since $b$ is the least upper bound, $b \\le c$.\n6. This contradicts $c \u003c b$. So some $s \\in S$ satisfies $c \u003c s \\le b$.\n\n## Idea\n\nA number below the supremum cannot be an upper bound, because the supremum is the least one. So some element of the set lies above it.\n",
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