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*Proposition:*

where

forallf,S_{0},S_{1}:S_{0}is supremum off/\S_{1}is supremum off=>S_{0}=S_{1}.

i.e., sup(

Sis supremum off: <=>Sis upper bound off/\( forallS':S' is upper bound off=>S<=S')

*Proof:*
Take arbitrary `f` and suprema `S`_{0} and `S`_{1} of `f`.
Since `S`_{0} is a supremum and `S`_{1} is an upper bound of
`f`, we have `S`_{0} <= `S`_{1}. Conversely,
since `S`_{1} is a supremum and `S`_{0} is an upper bound of
`f`, we have `S`_{1} <= `S`_{0}. Since
`S`_{0} <= `S`_{1} and `S`_{1} <= `S`_{0},
we have `S`_{0} = `S`_{1}.

Author: Wolfgang Schreiner

Last Modification: December 14, 1999