Minimally Inductive Set Exists

Theorem
There exists a minimal infinite successor set $\omega$ that is a subset of every other infinite successor set.

Proof
From the Axiom of Infinity, there is a set $S$ such that:


 * $\O \in S$
 * $x \in S \implies x^+ \in S$

where $\O$ denotes the empty set and $x^+$ is the successor set of $x$.

That is, there exists an infinite successor set.

Next, by the Axiom of Specification, the minimal infinite successor set $\omega$:


 * $\omega := \displaystyle \bigcap \left\{{S' \subseteq S: \text{$S'$ is an infinite successor set}}\right\}$

exists.

It remains to be shown that if $T$ is an infinite successor set, then $\omega \subseteq T$.

By Intersection of Infinite Successor Sets, $S \cap T$ is an infinite successor set.

Moreover, by Intersection is Subset, $S \cap T$, so that:


 * $S \cap T \in \left\{{S' \subseteq S: \text{$S'$ is an infinite successor set}}\right\}$

from which we conclude by definition of $\omega$ and Intersection is Subset: General Result that:


 * $\omega \subseteq S \cap T \subseteq T$

as desired.