Modification to axiom of infinity for the sake of asserting uniqueness?

Question

I noticed that if one writes $$\exists X:(\varnothing\in X\space\wedge\space(n\in X\rightarrow n\cup \{n\}\in X)) $$ As the axiom of infinity, one must define The Natural numbers as the smallest set that satisfies the above condition. But by replacing the conditional with an iff, ie: $$\exists\mathbb{N}!:(\varnothing\in\mathbb{N}\space\wedge\space(n\in\mathbb{N}\leftrightarrow n\cup\{n\}\in\mathbb{N}))$$ One could simply call the set proclaimed to exist and be unique above the natural numbers. Am I correct? Is there any reason that the Axiom of Infinity could not be replaced with this statement?

Answer 1

Your proposed condition does not define $\mathbb N$ uniquely.

For example, it is also satisfied by $V_\omega$, the set of all hereditarily finite sets.

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We could take as an axiom that there exists an inductive set that has no inductive proper subset, and that would be unique. But that seems just to be added complexity that doesn't really buy us anything, given that we have the axiom of separation available anyway.

Alternatively we could take: $$\exists X: X=\{\varnothing\}\cup \{n\cup\{n\}\mid n\in X\}$$ which in the presence of the Axion of Regularity would pinpoint $\omega$ uniquely. (It is clearly inductive, and if $Y$ is any inductive subset of it, then $Y=X$, or else $X\setminus Y$ would violate Regularity).