Smallest Element of Minimally Closed Class under Progressing Mapping

Theorem

Let $N$ be a class which is closed under a progressing mapping $g$.

Let $b$ be an element of $N$ such that $N$ is minimally closed under $g$ with respect to $b$.

$b$ is the smallest element of $N$.

Proof

Aiming for a contradiction, suppose $b$ is not the smallest element of $N$.

Then there exists $m \in N$ such that $b \nsubseteq m$.

In particular:

$m \ne b$

Let $B$ be the subclass of $A$ defined as:

$B = \set {x \in A: \paren {x = b} \lor \paren {\exists y \in B: x = \map g y} }$

This is a subclass of $A$ containing $b$ which is closed under $g$.

Let $z \in B$.

We have that:

$b \nsubseteq m$

Suppose that:

$z \nsubseteq m$

As $g$ is a Progressing Mapping, we have that:

$z \subseteq \map g z$

It follows that:

$\map g z \nsubseteq m$

By the Principle of General Induction for Minimally Closed Class it follows that:

$\forall z \in B: m \nsubseteq \map g z$

and in particular:

$\forall z \in B: m \ne \map g z$

In summary:

$m \ne b$

and:

$\not \exists z \in B: m = \map g z$

It follows that:

$m \notin B$

Thus we have created a proper subclass $B$ of $A$ containing $b$ which is closed under $g$.

This contradicts the assumption that $A$ is a minimally closed class under $g$ with respect to $b$

Hence, by Proof by Contradiction, our supposition that $b$ is not the smallest element of $N$ was false.

The result follows.

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Theorem $4.17 \ (5)$