The customary formulation of the Axiom of Infinity within Zermelo-Fraenkel set theory asserts the existence of an inductive set: a set $ I$ with $\varnothing\in I$ such that $x\in I$ implies $x\cup\{x\}\in I$. Since the intersection of any nonempty set of inductive sets is itself inductive, an instance of the Axiom Schema of Separation implies the existence of a smallest inductive set, namely the set of *von Neumann naturals* $$\mathbb{N}\_{\bf vN} = \{\varnothing, \{\varnothing\},\{\varnothing,\{\varnothing\}\},\ldots\}.$$


Any inductive set is infinite (in fact, Dedekind infinite) but this formulation of the axiom asserts more, namely the existence of a specific countably infinite set. Given one such set, the existence of others, for example the set of *Zermelo naturals* $$\mathbb{N}\_{\bf Zer}=\{\varnothing,\{\varnothing\},\{\{\varnothing\}\},\ldots\}$$ follows from appropriate instances of the Axiom Schema of Replacement. 


Consider the subsystem of Zermelo-Fraenkel set theory with axioms Extensionality, Separation Schema, Union, Power Set, Pair. Augment this Basic System with an Axiom of Infinity which asserts the existence of an infinite set, but not any particular one. Such a formulation requires that the notion of 'finite' be defined prior to that of 'natural number', following Kuratowski for example. Any infinite set $ I$ determines a Dedekind-infinite set of *local naturals* $$\mathbb{N}\_I=\{\mbox{equinumerosity classes of finite subsets of } I\}$$ which (duly equipped with initial element and successorship) yields a Lawvere natural number object, as in the Recursion Theorem. The existence of $\mathbb{N}\_{\bf vN}$ and $\mathbb{N}\_{\bf Zer}$ then follow from appropriate instances of Replacement.


One might wonder if there is some clever way to specify an infinite set without recourse to Replacement. That is, does there exist (in the language of set theory) a formula $\boldsymbol \phi$
with one free variable $x$ such that  

$$ \mbox{Basic+Infinity+Foundation } \vdash\; \exists y ( \forall x (x\in y \leftrightarrow \boldsymbol \phi)\,\wedge \, y \mbox{ is infinite})\>?$$


I'm inclined to guess no, on the following circumstantial grounds:


* For $\mathbb{N}\_{\bf vN}$ and $\mathbb{N}\_{\bf Zer}$ the use of Replacement is essential: Mathias has shown (Theorem 5.6 of *Slim Models of Zermelo Set Theory* that there exist transitive models ${\mathfrak M}\_{\bf vN}$ and ${\mathfrak M}\_{\bf Zer}$ of Basic+Infinity+Foundation with ${\mathbb N}\_{\bf vN}\in {\mathfrak M}\_{\bf vN}$ and ${\mathbb N}\_{\bf Zer}\in{ \mathfrak M}\_{\bf Zer}$, but such that every element of ${\mathfrak M}\_{\bf vN}\cap {\mathfrak M}\_{\bf Zer}$ is hereditarily finite.
* The usual definitions of $\mathbb{N}\_{\bf vN}$ and $\mathbb{N}\_{\bf Zer}$ involve unstratified formulas. Coret has shown (Corollary 9 of *Sur les cas stratifiés du schéma du replacement*) that this is unavoidable:
$$ \mbox{Basic+Infinity } \vdash\; \forall y ( \forall x (x\in y \leftrightarrow \boldsymbol \phi)\,\rightarrow \, y \mbox{ is hereditarily finite})$$
for any stratified $\boldsymbol \phi$. Using the same technique he has shown (Corollary 10) that Basic+Infinity proves every stratified instance of Replacement.



Can one exhibit an explicit Kuratowski infinite set without invoking Replacement?
The answer is no. Let ZC' be ZFC without replacement and infinity and with the assertion there is a Kuratowski infinite set. We will construct a model $M$ of ZC' such that only hereditarily finite elements of $M$ are fixed under all automorphisms of $M.$ The idea is generate a model from a $\mathbb{Z}^2$-array of objects, each of whose only element is the object below it in the array.


Define $M=\bigcup\_{m<\omega} \bigcup\_{n \in \mathbb{Z}} \mathcal{P}^m(\{\omega\} \times \mathbb{Z} \times [n, \infty)),$ where we adjust the $\mathcal{P}$ operator to replace each singleton of the form $\{(\omega, n\_1, n\_2)\}$ with $(\omega, n\_1, n\_2+1).$ (We use $\omega$ to differentiate the $\mathbb{Z}^2$ objects from the hereditarily finite sets).


Define a relation $E$ on $M$ by $E \restriction \mathbb{Z}^2=\{((\omega, n\_1, n\_2),(\omega, n\_1, n\_2+1)): n\_1, n\_2 \in \mathbb{Z}\}.$ Extending $E$ to the iterated power sets is done in the natural way. It is easy to see $(M, E)$ satisfies ZC'.


Notice that the map $(\omega, n\_1, n\_2) \mapsto (\omega, n\_1, n\_2+1)$ extends to a unique automorphism on $M,$ which only fixes hereditarily finite elements. Thus, none of the infinite sets in the model are first-order definable.


Update: The author of this question was curious whether there is such a model which also satisfies transitive containment (TC), the assertion that every set has a transitive closure. The answer is yes; here's a sketch of such a model.


Assume ZFC + "there are non-OD reals" in $V$ (or work in a ctm satisfying a sufficiently large finite fragment of this theory). We'll build an OD model $M$ of ZC'+TC such that, from every infinite element $x,$ we can define (in $V$) a non-OD real $r.$ Then $x$ is not definable in $M,$ or else we would have an ordinal definition for $r.$


For $r \in \mathbb{R} = \mathcal{P}(\omega),$ we recursively define $n\_r$ by $0\_r=\emptyset$ and $(n+1)\_r =\{n\_r\}$ if $n \not \in r,$ and $(n+1)\_r = \{n\_r, \emptyset\}$ if $n \in r.$ Basically, we're defining a new system of natural numbers for each real.


Let $M = \bigcup\_{m=0}^{\infty} \bigcup\_{S \in [\mathbb{R} \setminus OD]^{<\omega}} \mathcal{P}^m ( \{n\_r: n<\omega, r \in S\} ).$


For each infinite $x,$ there are finitely many (and at least one) non-OD reals $r$ such that the transitive closure of $x$ contains every $n\_r.$ The least such real is as desired. And it's routine to verify $M \models ZC'+TC.$


