Min Operation on Toset forms Semigroup

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Theorem

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

Let $\map \min {x, y}$ denote the min operation on $x, y \in S$.


Then $\struct {S, \min}$ is a semigroup.


Proof

By the definition of the min operation, either:

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

or

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


So $\min$ is closed on $S$.


From Min Operation is Associative:

$\forall x, y, z \in S: \map \min {x, \map \min {y, z} } = \map \min {\map \min {x, y}, z}$

Hence the result, by definition of semigroup.

$\blacksquare$


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