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:

$\sup \set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m} } \preceq \sup \set {x_1, x_2, \dotsc, x_n}$

$\blacksquare$