Zero is Identity in Naturally Ordered Semigroup

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Theorem

Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.

Let $0$ be the zero of $\struct {S, \circ, \preceq}$.


Then $0$ is the identity for $\circ$.

That is:

$\forall n \in S: n \circ 0 = n = 0 \circ n$


Proof

By definition of an ordering:

$0 \preceq 0$


Thus from axiom $(\text {NO} 3)$:

$\exists p \in S: 0 \circ p = 0$


By the definition of zero:

$0 \preceq 0 \circ 0$ and $0 \preceq p$


Thus since $\preceq$ is compatible with $\circ$:

$0 \circ 0 \preceq 0 \circ p = 0$


Thus:

$0 \circ 0 \preceq 0$ and $0 \preceq 0 \circ 0$

Hence, as $\preceq$ is antisymmetric, it follows that:

$0 \circ 0 = 0$


Because $\struct {S, \circ, \preceq}$ is a semigroup, $\circ$ is associative.


So:

$\forall n \in S: \paren {n \circ 0} \circ 0 = n \circ \paren {0 \circ 0} = n \circ 0$

Thus from axiom $(\text {NO} 2)$:

$\forall n \in S: n \circ 0 = n$


Similarly:

$\forall n \in S: 0 \circ \paren {0 \circ n} = \paren {0 \circ 0} \circ n = 0 \circ n$

meaning:

$\forall n \in S: 0 \circ n = n$


Thus:

$\forall n \in S: n \circ 0 = n = 0 \circ n$

and so $0$ is the identity for $\circ$.

$\blacksquare$


Sources