Minimally Inductive Set Exists

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

Proof
Assuming the Axiom of Infinity, Axiom of Power Set, and the Axiom of Specification,

Let $\mathcal{C} :=\{Z\in\mathcal{P}(I):\varnothing\in Z \wedge \forall x\in Z (x^+\in Z)\}$

$\mathcal{C}$ consists of all the infinite successor sets that we can reach with the above axioms.

Then define $\omega := \{x\in I:\forall Z\in\mathcal{C}(x\in Z)\} = \bigcap\mathcal{C}$

From Intersection of Infinite Successor Sets, $\omega$ is also an infinite successor set.

Then from Intersection Subset, $\forall Z\in\mathcal{C}\,(\omega\subseteq Z)$.

Hence, $\omega$ is 'minimal' in that it is contained in every infinite successor set that can be constructed from the above axioms.

Furthermore, if we have the Axiom of Extensionality in our axioms, $\omega$ is immediately unique.