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In ZF, an injection $h:\omega \to X$ yields an injective nonsurjective self-map by shifting $h(n)$ to $h(n+1)$ and fixing the complement. Conversely, if an injection $g:X\to X$ omits $a$, the natural-number recursion theorem defines $h(n)=g^n(a)$. Any repetition would, after cancellation, put $a$ in the image of $g$. Thus $h$ is injective. The argument uses no axiom of choice, even when some infinite sets are Dedekind-finite sets.

Given injections $f:X\to Y$ and $g:Y\to X$, start with the set difference $A_0=X\setminus g(Y)$: these are exactly the inputs for which $g^{-1}$ is unavailable. Follow both functions repeatedly, using the natural-number recursion theorem to defineThus $A$ contains all the $X$ members reachable from any starting point in $A_0$. The notation $f(A_k)$ and $g(f(A_k))$ denotes images of a function. If $A_0$ is the empty set, then every $A_k$ is empty and $g^{-1}$ alone gives the desired bijection.

Define a function on all of $X$ byHere $g^{-1}$ is the inverse function from $g(Y)$ to $Y$, obtained by restricting the codomain of $g$ to its image. It is well defined because $g$ is injective. Every $x\notin A$ belongs to $g(Y)$, since $A_0\subseteq A$, so the second rule is always available where it is used.

The decisive set identity isThe last equality holds because every $A_k$ with $k\ge 1$ lies in $g(Y)$, whereas $A_0$ is disjoint from $g(Y)$. Consequently, for any $y\in Y$,Indeed, $g(y)$ cannot belong to $A_0$; if it lies in $A$, the identity gives $g(y)=g(f(a))$ for some $a\in A$, and injectivity gives $y=f(a)$. The converse follows directly by applying $g$.

Now $f$ restricts to a bijection from $A$ onto $f(A)$. The displayed equivalence says that $g$ restricts to a bijection from $Y\setminus f(A)$ onto $X\setminus A$, so its inverse function bijects the remaining inputs with the remaining outputs. The two input sets partition $X$ and the two output sets partition $Y$. Their combined function $h$ is therefore both injective and surjective.

No choice function is needed: $A$ is fixed by the given injections, and each required $g^{-1}(x)$ is unique. We neither choose a representative from each chain nor enumerate $X$. This is a proof in ordinary classical ZF, even when no well-ordering of the sets exists there.^[ref]

Suppose $h:\omega \hookrightarrow X$ is an injection. Shift its image by definingThe two pieces have disjoint images, so $g$ is injective, but $h(0)$ is omitted. Thus $g$ is not surjective.

Conversely, let $g:X\to X$ be an injection which omits some $a\in X$. The natural-number recursion theorem defines $h(0)=a$, $h(n+1)=g(h(n))$. If $h(m)=h(n)$ with $m<n$, repeated injectivity cancels $m$ iterations and gives $a=g^{n-m}(a)$, putting $a$ in the image of $g$, a contradiction. Hence $h$ is injective. ThereforeChoosing one witness $a$ from a single known nonempty set uses no axiom of choice. This is the Dedekind-infinite characterization.

An approximation is a function $s:n+1\to X$ with $s(0)=x_0$ and $s(k+1)=g(s(k))$ for $k<n$. By induction on $n$, there is such a function: start with $\{(0,x_0)\}$ and append the uniquely specified value $g(s(n))$.

Any two approximations agree on their common domain, by induction from their common initial value and the same transition function $g$. The collection $\mathcal{S}$ of all approximations is a set, being a definable subset of $\mathcal{P}(\omega \times X)$. Therefore $f=\bigcup \mathcal{S}$ is a function. Existence of approximations of every finite length gives $\mathrm{dom}(f)=\omega$. Each defining equation appears in a sufficiently long approximation, soThis proves the natural-number recursion theorem without the axiom of choice; no arbitrary sequence of choices was made.

By the natural-number recursion theorem, let ${\alpha }_0=0$ and ${\alpha }_{n+1}=F({\alpha }_n)$. These ordinals form a nondecreasing sequence by the preceding bound. If ${\alpha }_{n+1}={\alpha }_n$ for some $n$, it is already a fixed point.

Otherwise the sequence is strictly increasing. Its supremum $\lambda ={\bigcup }_{n<\omega }{\alpha }_n$ is a limit ordinal, and the sequence is cofinal in it. Continuity and monotonicity giveFor the middle equality, every $\beta <\lambda$ is below some ${\alpha }_n$, while every ${\alpha }_n<\lambda$. Therefore $F$ has a fixed point. The sequence is a set by Axiom schema of replacement, since $F$ is a definable set-theoretic class function.

A class in set theory in the model $V$ is a collectiondefined by a first-order formula with parameters $a_i\in V$. A set-theoretic class function is a definable class relation $F$ for which every input in its domain has exactly one output. Informally, the Axiom schema of replacement says that the image $F‘‘a$ of any set $a$ under any such function class is again a set.

Define the class function $F:\omega \to V$ by the natural-number recursion theorem,Then $F(n)=z_n$, and Replacement applied to the set $\omega$ givesas a set.

Call a set small when it injects into some $z_n$. Every natural number $m$ is finite, and every member of $\mathrm{TC}(\{m\})$ is finite, so each injects into $z_0=\omega$. Hence $m\in \mathbf{HS}$. The set $\omega$ itself injects into $z_0$, and its hereditary members are natural numbers, so $\omega \in \mathbf{HS}$.

We next prove $z_n\in \mathbf{HS}$ by mathematical induction. The case $z_0=\omega$ was just proved. If $z_n\in \mathbf{HS}$ and $u\in z_{n+1}=\mathcal{P}(z_n)$, then $u\subseteq z_n$, so inclusion injects $u$ into $z_n$ and makes $u$ small. Every set below $u$ in its transitive closure is already below $z_n$ and is small by the induction hypothesis. The set $z_{n+1}$ itself injects into $z_{n+1}$. Thus every member of $\mathrm{TC}(\{z_{n+1}\})$ is small.

By Cantor theorem, $|z_{n+1}|>|z_n|$, so the $z_n$ are distinct and $z_n\mapsto n$ injects $y$ into $\omega =z_0$. Thus $y$ is small. Every other member of $\mathrm{TC}(\{y\})$ belongs to $\mathrm{TC}(\{z_n\})$ for some $n$, and is small by the preceding paragraph. Therefore $y\in \mathbf{HS}$. This proves the finite-power-set hereditary-small construction.

The structure $(\mathbf{HS},\in )$ is not a model of ZF because it fails the Axiom of union. If $\bigcup y$ were small, it would inject into some $z_m$. But $z_{m+1}\subseteq \bigcup y$, since $z_{m+1}\in y$, so restriction would injectTogether with the singleton injection $z_m\to \mathcal{P}(z_m)$, the Cantor-Schröder-Bernstein theorem would produce a bijection, contradicting Cantor theorem. Hence $\bigcup y$ is not small and therefore does not belong to $\mathbf{HS}$. Since $y\in \mathbf{HS}$ has no union inside the class, the Union axiom fails.