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Unicity of Supremum

Proposition:

forall f, S0, S1:

S0 is supremum of f /\  S1 is supremum of f => S0 = S1.

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.