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Let X be a partially ordered set so that every nonempty subset of X with an upper bound in X has an supremum in X. Prove that every nonempty sebset of X with a lower bound in X has an infimum in X. I would try to explain what I've tried but I'm stuck. This might be a duplicate but I can't find it on this site.

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For a given set $A$, let $b(A)=\{x\in X\mid\forall a\in A:x\leq a\}$. Namely, the set of all lower bounds of $A$. Then by assumption $b(A)$ is non-empty, and by definition all the members of $A$ are upper bounds of $b(A)$. Therefore $b(A)$ has a supremum. I will leave it for you to verify that $\sup b(A)=\inf A$.

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