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.

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.

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