Max of Subfamily of Operands Less or Equal to Max
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Theorem
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $x_1, x_2, \dotsc, x_n \in S$ for some $n \in \N_{>0}$.
Let $\set{k_1, k_2, \dotsc, k_m} \subseteq \set{1, 2, \dotsc, n}$
Then:
- $\max \set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m}} \preceq \max \set {x_1, x_2, \dotsc, x_n}$
where:
- $\max$ denotes the max operation
Proof
From Max Operation Yields Supremum of Operands:
- $\max \set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m}} = \sup \set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m} }$
and
- $\max \set {x_1, x_2, \dotsc, x_n} = \sup \set {x_1, x_2, \dotsc, x_n}$
Since $\set {k_1, k_2, \dotsc, k_m} \subseteq \set {1, 2, \dotsc, n}$ then:
- $\set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m} } \subseteq \set {x_1, x_2, \dotsc, x_n}$
From Supremum of Subset:
- $\sup \set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m} } \preceq \sup \set {x_1, x_2, \dotsc, x_n}$
$\blacksquare$