Min Operation Yields Infimum of Parameters

Theorem
Let $\left({S, \preceq}\right)$ be a totally ordered set.

Let $x, y \in S$.

Then:
 * $\min \left({x, y}\right) = \inf \left({\left\{{x, y}\right\}}\right)$

where:
 * $\min$ denotes the min operation
 * $\inf$ denotes the infimum.

Proof
As $\left({S, \preceq}\right)$ be a totally ordered set, all elements of $S$ are comparable by $\preceq$.

Therefore there are two cases to consider:

Case 1: $x \preceq y$
In this case:
 * $\min \left({x, y}\right) = x = \inf \left({\left\{{x, y}\right\}}\right)$

Case 2: $y \preceq x$
In this case:
 * $\min \left({x, y}\right) = y = \inf \left({\left\{{x, y}\right\}}\right)$

In either case, the result holds.

Comment
Note that it is considered abuse of notation to write
 * $\min = \inf$

This is because
 * $\min: S \times S \to S$

while
 * $\inf: \mathcal P \left({S}\right) \to S$

where $\mathcal P \left({S}\right)$ is the power set of $S$.

Also see

 * Max Operation Yields Supremum of Parameters