Max of Subfamily of Operands Less or Equal to Max

Theorem
Let $\struct {S, \preceq}$ be a totally ordered set.

Let $x_1, x_2, \dots ,x_n \in S$ for some $n \in \N_{>0}$.

Let $\set{k_1, k_2, \dots, k_m} \subseteq \set{1, 2, \dots, n}$

Then:
 * $\max \set {x_{k_1}, x_{k_2}, \dots ,x_{k_m}} \preceq \max \set {x_1, x_2, \dots ,x_n}$

where:
 * $\max$ denotes the max operation

Proof
From Max yields Supremum of Operands:
 * $\max \set {x_{k_1}, x_{k_2}, \dots ,x_{k_m}} = \sup \set {x_{k_1}, x_{k_2}, \dots ,x_{k_m}}$

and
 * $\max \set {x_1, x_2, \dots ,x_n} = \sup \set {x_1, x_2, \dots ,x_n}$

Since $\set{k_1, k_2, \dots, k_m} \subseteq \set{1, 2, \dots, n}$ then:
 * $\set {x_{k_1}, x_{k_2}, \dots ,x_{k_m}} \subseteq \set {x_1, x_2, \dots ,x_n}$

From Supremum of Subset:
 * $\sup \set {x_{k_1}, x_{k_2}, \dots ,x_{k_m}} \preceq \sup \set {x_1, x_2, \dots ,x_n}$