Equivalence of Definitions of Minimally Closed Class

Theorem

Let $A$ be a class.

Let $g$ be a mapping on $A$.

The following definitions of the concept of minimally closed class under $g$ are equivalent:

Definition 1

$A$ is minimally closed under $g$ with respect to $b$ if and only if:

\((1)\)	$:$							$A$ is closed under $g$
\((2)\)	$:$							There exists $b \in A$ such that no proper subclass of $A$ containing $b$ is closed under $g$.

Definition 2

$A$ is minimally closed under $g$ with respect to $b$ if and only if:

\((1)\)	$:$							$A$ is closed under $g$
\((2)\)	$:$							There exists $b \in A$ such that every subclass of $A$ containing $b$ which is closed under $g$ contains all the elements of $A$.

Proof

Let it be given that $A$ is closed under $g$.

$(1)$ implies $(2)$

Let $A$ be a minimally closed class under $g$ by definition 1.

Then by definition:

there exists $b \in A$ such that $A$ has no proper subclass $B$ such that:

$b \in B$

$B$ is closed under $g$.

Let $A$ have a subclass $C$ which is closed under $g$ such that $b \in C$.

Then by definition, $C$ is not a proper subclass of $A$.

Thus by definition of proper subclass:

$C = A$

By definition of subclass:

$A \subseteq C$

and:

$C \subseteq A$

That is:

$\forall x \in A: x \in C$

and:

$\forall x \in C: x \in A$

That is: $C$ contains all elements of $A$.

Thus $A$ is a minimally closed class under $g$ by definition 2.

$\Box$

$(2)$ implies $(1)$

Let $A$ be a minimally closed class under $g$ by definition 2.

Then by definition:

Every subclass of $A$ containing $b$ which is closed under $g$ contains all the elements of $A$.

Let $C$ be a subclass of $A$ which is closed under $g$ such that $b \in C$.

By definition of subclass:

$C \subseteq A$

But by hypothesis:

$\forall x \in A: x \in C$

That is:

$A \subseteq C$

By definition of equality of classes it follows that:

$A = C$

and so by definition $C$ cannot be a proper subclass of $A$.

Thus $A$ is a minimally closed class under $g$ by definition 1.

$\blacksquare$