Strictly Inductive Semigroup is Inductive Semigroup

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Theorem

Let $\struct {S, \circ}$ be a strictly inductive semigroup.


Then $\struct {S, \circ}$ is an inductive semigroup.


Proof

In accordance with our assertion, let $\struct {S, \circ}$ be a strictly inductive semigroup.

By definition of strictly inductive semigroup, there exists $\beta \in S$ such that the only subset of $S$ containing both $\beta$ and $x \circ \beta$ whenever it contains $x$ is $S$ itself.

That is:

$\exists \beta \in S: \forall A \subseteq S: \paren {\beta \in S \land \paren {\forall x \in A: x \circ \beta \in A} } \implies A = S$


Let $\alpha = \beta$.

Then we may say that there exist $\alpha, \beta \in S$ such that the only subset of $S$ containing both $\alpha$ and $x \circ \beta$ whenever it contains $x$ is $S$ itself.

That is:

$\exists \alpha, \beta \in S: \forall A \subseteq S: \paren {\alpha \in A \land \paren {\forall x \in A: x \circ \beta \in A} } \implies A = S$


Hence, by identifying $\alpha$ with $\beta$, $\struct {S, \circ}$ fulfils the conditions of being an inductive semigroup.

$\blacksquare$


Sources

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