We will prove the result by induction on the number of operands $n$. For all $n \in \Z_{>0}$, let $\map P n$ be the proposition: :$\max \set {x_1, x_2, \dotsc, x_n} = \sup \set {x_1, x_2, \dotsc, x_n}$ === Basis for the Induction === $\map P 1$ is the case: :$\max \set {x_1} = \sup \set {x_1}$ By definition of the max operation: :$\max \set {x_1} = x_1$ From Supremum of Singleton: :$\sup \set {x_1} = x_1$ So: :$\max \set {x_1} = \sup \set {x_1} = x_1$ Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\max \set {x_1, x_2, \dots ,x_k} = \sup \set {x_1, x_2, \dots ,x_k}$ from which it is to be shown that: :$\max \set {x_1, x_2, \dots ,x_k, x_{k+1}} = \sup \set {x_1, x_2, \dots ,x_k, x_{k+1}}$ === Induction Step === This is the induction step. Now: {{begin-eqn}} {{eqn | l = \max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} }       | r = \max \set {\max \set {x_1, x_2, \dotsc, x_k}, x_{k + 1} }       | c = {{Defof|Max Operation (General Case)}} }} {{eqn | r = \max \set {\sup \set {x_1, x_2, \dotsc, x_k}, x_{k + 1} }       | c = Induction hypothesis }} {{end-eqn}} As $\struct {S, \preceq}$ is a totally ordered set, all elements of $S$ are comparable by $\preceq$. Therefore there are two cases to consider: ==== Case 1: $x_{k + 1} \preceq \sup \set {x_1, x_2, \dotsc, x_k}$ ==== Let $x_{k + 1} \preceq \sup \set {x_1, x_2, \dotsc, x_k}$. By definition of the max operation: :$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \sup \set {x_1, x_2, \dotsc, x_k}$ By definition, $\sup \set {x_1, x_2, \dotsc, x_k}$ is an upper bound of $\set {x_1, x_2, \dotsc, x_k}$. That is: :$\forall 1 \le i \le k : x_i \preceq \sup \set {x_1, x_2, \dotsc, x_k}$ Thus: :$\forall 1 \le i \le k + 1 : x_i \preceq \sup \set {x_1, x_2, \dotsc, x_k}$ So $\sup \set {x_1, x_2, \dots ,x_k}$ is an upper bound of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$. Let $y$ be any other upper bound of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$. Then: :$y$ is an upper bound of $\set {x_1, x_2, \dotsc, x_k}$. By definition of the supremum: :$\sup \set {x_1, x_2, \dotsc, x_k} \preceq y$ It has been shown that the supremum of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$ is: :$\sup \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \sup \set {x_1, x_2, \dotsc, x_k}$ Thus it follows: :$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \sup \set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$ ==== Case 2: $\sup \set {x_1, x_2, \dotsc, x_k} \preceq x_{k + 1}$ ==== Let $\sup \set {x_1, x_2, \dotsc, x_k} \preceq x_{k + 1}$. By definition of the max operation: :$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = x_{k + 1}$ By definition, $\sup \set {x_1, x_2, \dotsc, x_k}$ is an upper bound of $\set {x_1, x_2, \dotsc, x_k}$. That is: :$\forall 1 \le i \le k : x_i \preceq \sup \set {x_1, x_2, \dotsc, x_k}$ By definition of an ordering: :$\preceq$ is transitive. Thus: :$\forall 1 \le i \le k : x_i \preceq x_{k + 1}$ By definition of an ordering: :$\preceq$ is reflexive. Thus: :$x_{k + 1} \preceq x_{k + 1}$ It follows that $x_{k + 1}$ is an upper bound of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$. Let $y$ be any other upper bound of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$. Then: :$x_{k + 1} \preceq y$. It has been shown that :$\sup \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = x_{k + 1}$. Thus it follows: :$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1}} = \sup \set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$ In either case, the result holds. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\max \set {x_1, x_2, \dotsc, x_n} = \sup \set {x_1, x_2, \dotsc, x_n}$ {{qed}} Category:Max and Min Operations ldrb4v46u61m7u5znizcgonxjhj334k