Min Operation Yields Infimum of Parameters

Theorem

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

Let $x, y \in S$.

Then:

$\map \min {x, y} = \map \inf {\set {x, y} }$

where:

$\min$ denotes the min operation

$\inf$ denotes the infimum.

General Case

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

Then:

$\min \set {x_1, x_2, \dotsc, x_n} = \inf \set {x_1, x_2, \dotsc, x_n}$

Proof

As $\struct {S, \preceq}$ be a totally ordered set, all elements of $S$ are $\preceq$-comparable.

Therefore there are two cases to consider:

Case 1: $x \preceq y$

In this case:

$\map \min {x, y} = x = \map \inf {\set {x, y} }$

Case 2: $y \preceq x$

In this case:

$\map \min {x, y} = y = \map \inf {\set {x, y} }$

In either case, the result holds.

$\blacksquare$

Warning

Note that it is considered abuse of notation to write

$\min = \inf$

This is because

$\min: S \times S \to S$

while

$\inf: \powerset S \to S$

where $\powerset S$ is the power set of $S$.

Also see

• Max Operation Yields Supremum of Parameters