Min yields Infimum of Parameters

From ProofWiki
Jump to navigation Jump to search

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 $S$.


Also see