The clauses at $0$, at a successor and at a limit determine exactly one operation $\alpha \mapsto {\mathrm{\aleph }}_{\alpha }$, in ZF, and — assuming the Axiom of Choice — exactly one operation $\alpha \mapsto {\mathrm{\beth }}_{\alpha }$; each value is an infinite cardinal, each is strictly increasing and continuous at limits, and $\alpha \le {\mathrm{\aleph }}_{\alpha }$

(a) The alephs, in ZF. There is exactly one class operation $\alpha \mapsto {\mathrm{\aleph }}_{\alpha }$, defined at every ordinal (Ordinal (von Neumann)) and given by a formula, satisfying

${\mathrm{\aleph }}_0=\omega ,{\mathrm{\aleph }}_{\alpha +1}=\mathrm{\aleph }({\mathrm{\aleph }}_{\alpha }),{\mathrm{\aleph }}_{\lambda }=\bigcup \{{\mathrm{\aleph }}_{\alpha }:\alpha \in \lambda \}(\lambda \text{ a limit ordinal}),$

where $\mathrm{\aleph }(\cdot )$ is the Hartogs number (For every set $A$ the Hartogs number $\mathrm{\aleph }(A)$ is a cardinal, and for every cardinal $\kappa$ it is the least cardinal strictly above $\kappa$; this is a theorem of ZF) and $\alpha +1=\alpha \cup \{\alpha \}$.

(b) Every ${\mathrm{\aleph }}_{\alpha }$ is an infinite cardinal (Cardinal (initial ordinal) and cardinality); the operation is strictly increasing, $\alpha \in \beta \Rightarrow {\mathrm{\aleph }}_{\alpha }\in {\mathrm{\aleph }}_{\beta }$; and it is continuous at limits, which is the third clause read as a supremum.

(c) $\alpha \le {\mathrm{\aleph }}_{\alpha }$ for every ordinal $\alpha$.

(d) The beths, assuming the Axiom of Choice (The Axiom of Choice). There is exactly one class operation $\alpha \mapsto {\mathrm{\beth }}_{\alpha }$, defined at every ordinal, with

${\mathrm{\beth }}_0=\omega ,{\mathrm{\beth }}_{\alpha +1}=2^{{\mathrm{\beth }}_{\alpha }},{\mathrm{\beth }}_{\lambda }=\bigcup \{{\mathrm{\beth }}_{\alpha }:\alpha \in \lambda \}(\lambda \text{ a limit ordinal}),$

and it too takes infinite cardinal values, is strictly increasing, and is continuous at limits.

Like Transfinite recursion along the ordinals: a class rule determines exactly one operation defined at every ordinal these are theorem schemas: an instance for each of the defining formulas. The aleph half uses Replacement and no choice; the beth half needs the Axiom of Choice, and needs it only because $2^{\kappa }$ does (Assuming the Axiom of Choice, $2^{\kappa }=\mid \mathcal{P}(\kappa )\mid$, and Cantor's theorem in cardinal form: $\kappa <2^{\kappa }$).

Why the published recursion theorem is not enough on its own. Transfinite recursion is stated for a well-order, that is for a set, and $\alpha \mapsto {\mathrm{\aleph }}_{\alpha }$ has to be defined at every ordinal. The bridge is Transfinite recursion along the ordinals: a class rule determines exactly one operation defined at every ordinal, and it is used here exactly as Ordinal addition $\alpha +\beta$ uses it.

Where the two hierarchies part company. The successor clause for the alephs is the Hartogs number, built in ZF from well-ordered subsets; the successor clause for the beths is the power set, which ZF cannot well-order. So the alephs exist without choice and the beths do not, and the question of how the two hierarchies line up is not settled by anything on this page.

Continuity is a clause, not a theorem. The third displayed clause defines the value at a limit to be the supremum, so continuity holds by construction. It is worth naming because it is what makes the cofinality computations of this page work: it is exactly why ${\mathrm{\aleph }}_{\omega }$ is reachable from below by an $\omega$-indexed family.