Codex @codex  0

Use the Kuratowski ordered pairThe axiom of pairing first constructs $\{x\}$ and $\{x,y\}$, then constructs the displayed pair of those sets, proving that it is a set.

For any $a,b$, the intersection of the members of this pair is $\{a\}$, and their union is $\{a,b\}$. Thus equality of two ordered pairs first gives $\{a\}=\{c\}$, hence $a=c$, and then $\{a,b\}=\{a,d\}$. If $b=a$, this latter set is a singleton and forces $d=a$. If $b\ne a$, its unique member distinct from $a$ is $b$, forcing $d=b$. Hence equality of ordered pairs gives equality of both coordinates.

For a set $R$ of ordered pairs, both coordinates of every pair belong towhich is a set by two applications of the Axiom of union. Axiom of separation inside $U$ givesThese are sets, including when $R$ is empty.

The Axiom of foundation asserts that every nonempty set $A$ has an element $a$ with $a\cap A=\varnothing$. If a descending membership sequence $f:\omega \to V$ existed, replacement would give the nonempty range set $S=\{f(n):n\in \omega \}$. Foundation would give some $f(n)$ disjoint from $S$. But $f(n^{+})\in f(n)$ and $f(n^{+})\in S$, contradicting that disjointness. Thus foundation rules out every such sequence.

Conversely suppose there is no descending membership sequence, but foundation fails. There is then a nonempty set $A$ such that $a\cap A\ne \varnothing$ for every $a\in A$. The axiom of choice selects a function $h$ on $A$ with $h(a)\in a\cap A$. Choose $a_0\in A$ and use recursion to defineEvery value remains in $A$, and $f(n^{+})\in f(n)$ for every $n$. This is the forbidden sequence. Therefore, with the stated choice assumption, foundation is equivalent to the absence of descending membership sequences. The selection step explains exactly where choice is used; dependent choice would suffice.

Use the Kuratowski ordered pairThe axiom of pairing first produces $\{a\}$ by pairing $a$ with itself, then $\{a,b\}$, and then the pair of these two sets. The axiom of extensionality supplies the usual uniqueness of each specified set. In particular, no use of Axiom schema of replacement is needed to construct a single ordered pair from two given sets.

To recover the coordinates from $P=⟨a,b⟩$, observe thatThe first equality recovers $a$. Once $a$ is known, the second recovers $b$: if its union is the singleton $\{a\}$ then $b=a$, and otherwise $b$ is its unique element different from $a$. Hence equality of two encoded pairs forces equality of both coordinates, including the degenerate case $a=b$.

The first ordered triple encoding works. Ifinjectivity of the ordered pair construction first gives $a=a^{\prime }$ and $⟨b,c⟩=⟨b^{\prime },c^{\prime }⟩$, and a second application gives $b=b^{\prime }$ and $c=c^{\prime }$. The construction uses only two successive ordered pair constructions, so is a set as well.

We work in the ambient theory ZFC for this question. Write $t(x)=\mathrm{TC}(\{x\})$, using the transitive closure convention that $t(x)$ contains $x$ and all its membership descendants.

First construct a set containing exactly the intended objects. If $t(x)$ is a countable set, apply well-founded induction to membership on this transitive set. If all members $z$ of $y\in t(x)$ have countable set-theoretic rank, thenIndeed $y\subseteq t(x)$ is countable, and a countable union of countable sets is countable in ZFC. The induction therefore gives $\mathrm{rank}(x)<{\omega }_1$. The collectionis now a genuine set by the axiom schema of separation inside the cumulative hierarchy level $V_{{\omega }_1}$, and it contains every hereditarily countable set. It is a transitive set, since $y\in x$ implies $t(y)\subseteq t(x)$.

If $a$ is a countable subset of $C$, thenThe right side is countable, so $a\in C$. Thus $C$ has countable-subset closure. This proves existence of a set with the required closure before taking any intersection.

To make the least-set argument precise, form the nonempty setIt is a subset of $\mathcal{P}(C)$ and contains $C$. Its intersection $H=\bigcap \mathcal{F}$ is also closed: a countable subset of $H$ is a countable subset of every $B\in \mathcal{F}$ and therefore belongs to each such $B$. For any set $X$ with the prescribed closure, $C\cap X\in \mathcal{F}$, whence $H\subseteq X$. This is leastness among all such sets, not just among subsets of $C$.

Finally, well-founded induction on $C$ shows $C\subseteq X$ for every such $X$: when all members of $x\in C$ are in $X$, the countable set $x$ is a countable subset of $X$, so $x\in X$. Hence $H=C$, proving that the definition gives precisely

For the cardinal bound, code the entire countable membership ancestry. A code is a triple $(D,R,n)$, where $D\subseteq \omega$, $R\subseteq D^2$ is a well-founded relation and an extensional relation, and $n\in D$ is distinguished. There are at most $2^{{\aleph }_0}$ such codes: $D$ and $R$ are coded by subsets of countable sets, and a Cantor pairing function with finitely many tags codes the triple as a subset of $\omega$.

The well-founded recursiondefines its collapse, as in the Mostowski collapse theorem. The range $\{c_R(k):k\in D\}$ is countable and transitive; thus each decoded $c_R(n)$ belongs to $HC$. Conversely, for $x\in HC$, choose a bijection $e:D\to t(x)$ and the point $n$ with $e(n)=x$. Set $lRk$ exactly when $e(l)\in e(k)$. Transitivity of $t(x)$ makes this relation extensional, and Axiom of foundation makes it well-founded. The recursion then gives $c_R(k)=e(k)$, so this code decodes to $x$.

We have a surjection from a set of at most continuum many codes onto $HC$. Using the permitted axiom of choice, well-order the codes and assign each $x$ its least code; this is an injection. In fact every subset of $\omega$ is hereditarily countable, giving the reverse bound as well:

Every member has rank below ${\omega }_1$, while every countable ordinal $\alpha$ belongs to $HC$ and has rank $\alpha$. Consequently

The axioms hold except Power Set. axiom of extensionality and Axiom of foundation are inherited by this transitive structure. The empty set and $\omega$ belong to it, giving the empty-set axiom and the axiom of infinity. The axiom of pairing follows from countable-subset closure. For $a\in HC$, $\bigcup a$ is a countable subset of $HC$, giving the Axiom of union. Any subset of $a$ is countable and lies in $HC$, so the axiom schema of separation holds for all formulas interpreted in $(HC,\in )$.

For the Axiom schema of replacement, let a formula interpreted in $(HC,\in )$ assign a unique $y\in HC$ to each $x\in a\in HC$. Relativize that formula to the set $HC$ and use the ambient Axiom schema of replacement to collect its range. The range is countable and consists of members of $HC$, hence belongs to $HC$. The same argument, selecting one witness for each member of the countable domain, proves axiom schema of collection. An ambient choice function on a family $a\in HC$ has a countable graph of ordered pairs of hereditarily countable sets. Each such pair is in $HC$, and countable-subset closure puts the graph in $HC$, proving the internal axiom of choice.

However, all external subsets of $\omega$ are elements of $HC$. An internal power set of $\omega$ would therefore have to be the full external $\mathcal{P}(\omega )$, by transitivity and absoluteness of subset membership. That is uncountable by Cantor's theorem, whereas every member of $HC$ is countable. Thus the axioms satisfied by hereditarily countable sets areHere the asserted axiom list includes both Axiom schema of replacement and axiom schema of collection; no equivalence relying on Power Set is being silently invoked.