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forall f, S0, S1: S0 is supremum of f /\ S1 is supremum of f => S0 = S1.
i.e., sup(f) = such S: S is supremum of f.
S is supremum of f : <=> S is upper bound of f /\ (forall S': S' is upper bound of f => S <= S')
Proof: Take arbitrary f and suprema S0 and S1 of f. Since S0 is a supremum and S1 is an upper bound of f, we have S0 <= S1. Conversely, since S1 is a supremum and S0 is an upper bound of f, we have S1 <= S0. Since S0 <= S1 and S1 <= S0, we have S0 = S1.