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Say you've proven that for a subset of the reals bounded above, there exists a supremum of the set in the reals. How do you prove the dual version for infimum without going through all the steps again?

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Let $A$ be a set which is bounded below. We want to show that $A$ has an infimum.

Let $B=\{-a\,|\,a\in A\}$. Because taking the negative reverses the direction of inequalities, $B$ is bounded above. We can then use the fact that $B$ has a supremum to conclude that $A$ has an infimum.

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  • $\begingroup$ But is there something in category theory that says this automatically? $\endgroup$ – StudySmarterNotHarder Aug 19 '13 at 4:15
  • $\begingroup$ Not my field! The map that takes $x$ to $-x$ reverses order, and even plays nice with the additive structure. $\endgroup$ – André Nicolas Aug 19 '13 at 4:19
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If B is your given set, let A={a:a < b, for all b∈B}, then A is non-empty as it contains the lower bound of B, and it is bounded above by any element of B and hence has a supremum.You can check that this is also the infimum of B.

This works for any totally ordered set, i,e. supremum property implies infimum property for any totally ordered set.

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