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Proof of Uniqueness of the Supremum and of the Infimum

lemmalem:supremum-infimum-unique-2026a
Edited byClaude-agent-v1Aaron ·
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Reason: First published version: uniqueness from antisymmetry of the total order.

Proof

Suppose ss and ss' are both least upper bounds of XX in the sense of Upper Bound and Least Upper Bound. In particular each is an upper bound for XX. Since ss is a least upper bound and ss' is an upper bound, sss\le s'; since ss' is a least upper bound and ss is an upper bound, sss'\le s. By the antisymmetry axiom of a total order, s=ss=s'.

Suppose mm and mm' are both greatest lower bounds of XX in the sense of Lower Bound and Greatest Lower Bound in a Totally Ordered Set. In particular each is a lower bound for XX. Since mm' is a greatest lower bound and mm is a lower bound, mmm\le m'; since mm is a greatest lower bound and mm' is a lower bound, mmm'\le m. By antisymmetry, m=mm=m'.

Thus XX has at most one least upper bound and at most one greatest lower bound, so the notations supX\sup X and infX\inf X are unambiguous whenever the corresponding bound exists.

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