# Min yields Infimum of Parameters

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## 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.

$\blacksquare$

## 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 $S$.