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Assume the supremum property holds and prove that the infimum property must also hold. Formally, assume that for any $A\subset\mathbb R$ that is non-empty and bounded above, $\sup A$ exists and $\sup A\in\mathbb R$. Show that for any $B \subset \mathbb R$ that is non-empty and bounded below, $\inf B \in \mathbb R.$

Proof: Define $B'=-B$. Then $B'$ is bounded above, so by the supremum property $\sup B'\in\mathbb R$. Since $\inf B =-\sup B'$, $\inf B\in \mathbb R$. $\square$

Is that rigorous enough? Please feel free to critique.

Anne Bauval
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Hank
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    If you already know that $\inf(B)=\sup(B')$ (which is actually the hardest part here) then yes, this is a proof. I usually prefer a different proof: the infimum of $B$ is the supremum of the set of lower bounds of $B$. – Mark Apr 21 '24 at 17:29
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    It is fine (up to an important but now corrected typo: $-$sign missing in $\inf B =-\sup B'$), since $x\mapsto-x$ is an order-reversing bijection. However, I would neither say "$\sup A$ exists and $\sup A\in\mathbb R$", nor just "$\sup A\in\mathbb R$", but: "in the ordered set $\Bbb R$, $\sup A$ exists". And similarly for the other sups and infs. – Anne Bauval Apr 21 '24 at 17:44
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    Does this answer your question? How to show $-\sup(-A)=\inf(A)$? – Anne Bauval Apr 21 '24 at 17:55
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    Your proof works in $\mathbb R$. The proof suggested by @Mark is more general and works in every partial order. – drhab Apr 21 '24 at 18:22

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