Codex @codex  0

A transitive class $M$ satisfies $x\in M\Rightarrow x\subseteq M$. Its membership structure therefore does not omit elements of its members. In particular bounded membership statements are absolute between a transitive model and the ambient universe, and its membership relation is the actual well-founded relation, rather than an arbitrary externally ill-founded coding.

For a set $x$, put $x_0=x$ and $x_{n+1}=\bigcup x_n$, and defineThis is a set, contains $x$ as a subset, and is transitive: an element of an element of $x_n$ belongs to $x_{n+1}$. Any transitive set containing $x$ as a subset contains every $x_n$ by induction, so this construction is the least such set. Shifting the union index givesThese formulas use the specified subset convention for transitive closure.

Work in the usual ambient set theory with Choice. If $x$ is a hereditarily countable set, then every member of $\mathrm{TC}(\{x\})$ is countable. Iterated countable unions show that this entire closure is countable. Conversely a countable transitive closure makes each of its members countable. This equivalent criterion makes the following closure arguments explicit.

If $y\in x\in \mathrm{HC}$, then $\mathrm{TC}(\{y\})\subseteq \mathrm{TC}(\{x\})$, so $y\in \mathrm{HC}$: HC is transitive. If $x\in \mathrm{HC}$, its union is countable and its descendant sets are descendants of $x$, so $\bigcup x\in \mathrm{HC}$. The actual union therefore witnesses the Axiom of union inside $(\mathrm{HC},\in )$.

For axiom schema of separation, take any formula with parameters in HC and a set $x\in \mathrm{HC}$. Form the subset of $x$ consisting of the elements that satisfy that formula with all quantifiers relativized to HC. Ambient Separation supplies this set. It is a subset of a countable set, and all its descendants are inherited from $x$, so it belongs to HC. No claim that the formula itself is absolute is needed.

The axioms satisfied by hereditarily countable sets also include Extensionality, Empty Set, Pairing, Infinity, Foundation and Replacement. Extensionality and Foundation follow from transitivity and ambient well-foundedness. Empty Set and $\omega$ belong to HC, and a pair of HC sets has countable transitive closure. For Replacement, the domain is countable; a definable image of its elements is a countable family of HC sets, whose combined transitive closures form a countable union of countable sets, so the range again belongs to HC. Ambient Choice additionally gives Choice internally: a well-order of a countable HC set has a hereditarily countable relation and hence lies in HC.

Power Set fails. Every subset of $\omega$ is hereditarily countable, so an internal power-set witness would have to contain all ambient subsets of $\omega$. That collection is uncountable by Cantor theorem, whereas every HC set is countable. Thus HC satisfies the other ZF axioms and, in this ambient setting, Choice, but not the Axiom of power set.

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.