What each result on this page costs in choice, and where the continuum escapes what ZFC can decide

This item is bookkeeping, in the manner of The choice ledger: what costs the Axiom of Choice and what does not: it records what each result stated here actually costs, so that anything quoting a result from this page knows whether it is quoting a theorem of ZF or a consequence of the Axiom of Choice (The Axiom of Choice).

Theorems of ZF, using no choice principle at all.

• A set equinumerous with some ordinal has a least such ordinal, that ordinal is a cardinal, and equinumerous sets get the same one; no choice principle is used: a well-orderable set has a least equinumerous ordinal, that ordinal is a cardinal, and equinumerous sets receive the same one. The choice hypothesis of Cardinal (initial ordinal) and cardinality is needed only to make every set well-orderable.
• Disjoint union, cartesian product, function space and power set respect equinumerosity, and for ordinals $\alpha, \beta$ the sets $\alpha \sqcup \beta$ and $\alpha \times \beta$ carry explicit well-orders, so their cardinalities exist in ZF, and with it $\oplus$ and $\otimes$ of Cardinal sum $\kappa \oplus \lambda$, product $\kappa \otimes \lambda$ and exponentiation $\kappa^{\lambda}$, and why they are written apart from the ordinal operations. The disjoint union and the product of two ordinals are well-ordered by orders written down from the ordinal order.
• Every natural number and $\omega$ are cardinals, every infinite cardinal is a limit ordinal, and on the natural numbers the cardinal operations are the published finite counting operations, with $\lvert A \rvert$ in the finite sense equal to $\lvert A \rvert$ in the cardinal sense, the dictionary with finite counting.
• Hessenberg: $\kappa \otimes \kappa = \kappa$ for every infinite cardinal $\kappa$, proved in ZF from the canonical well-order of $\kappa \times \kappa$ and its consequence Absorption: for cardinals $\kappa, \lambda$ with $\kappa$ infinite and $\lambda \le \kappa$, $\kappa \oplus \lambda = \kappa$, and $\kappa \otimes \lambda = \kappa$ when $\lambda \ne 0$.
• For every set $A$ the Hartogs number $\aleph(A)$ is a cardinal, and for every cardinal $\kappa$ it is the least cardinal strictly above $\kappa$; this is a theorem of ZF, and with it the whole aleph hierarchy: The clauses at $0$, at a successor and at a limit determine exactly one operation $\alpha \mapsto \aleph_\alpha$, in ZF, and — assuming the Axiom of Choice — exactly one operation $\alpha \mapsto \beth_\alpha$; each value is an infinite cardinal, each is strictly increasing and continuous at limits, and $\alpha \le \aleph_\alpha$ and The successor cardinal $\kappa^{+}$, the alephs $\aleph_\alpha$, the beths $\beth_\alpha$, successor and limit cardinals, and the identifications $\aleph_0 = \omega$ and $\aleph_1 = \omega_1$ in their aleph clauses, and clause (a) of Every infinite cardinal is $\aleph_\alpha$ for exactly one ordinal $\alpha$, in ZF; and, assuming the Axiom of Choice, every infinite set is equinumerous with exactly one aleph.
• Cofinality $\operatorname{cf}(\alpha)$, and regular and singular cardinals and $\operatorname{cf}(\alpha) \le \alpha$; $\operatorname{cf}(0) = 0$ and $\operatorname{cf}(\alpha + 1) = 1$; for a limit ordinal $\lambda$ the value $\operatorname{cf}(\lambda)$ is an infinite cardinal with $\operatorname{cf}(\operatorname{cf}(\lambda)) = \operatorname{cf}(\lambda)$, so it is regular; and every cofinal subset of $\lambda$ has cardinality at least $\operatorname{cf}(\lambda)$, a value that is attained; and clauses (a) and (c) of $\aleph_0$ is regular in ZF; assuming the Axiom of Choice every successor aleph $\aleph_{\alpha+1}$ is regular; $\operatorname{cf}(\aleph_\omega) = \aleph_0$, so $\aleph_\omega$ is singular, and under choice it is the least singular infinite cardinal, that $\aleph_0$ is regular and $\operatorname{cf}(\aleph_\omega) = \aleph_0$.

Costing the Axiom of Choice, and named as such in their own statements.

• Cardinal exponentiation $\kappa^{\lambda}$ itself, because ZF does not well-order a set of functions. Everything reached through it inherits the hypothesis: Assuming the Axiom of Choice, $2^{\kappa} = \lvert \mathcal{P}(\kappa) \rvert$, and Cantor's theorem in cardinal form: $\kappa < 2^{\kappa}$, the beth clauses of The successor cardinal $\kappa^{+}$, the alephs $\aleph_\alpha$, the beths $\beth_\alpha$, successor and limit cardinals, and the identifications $\aleph_0 = \omega$ and $\aleph_1 = \omega_1$, The sum $\sum_{i \in I} \kappa_i$ and the product $\prod_{i \in I} \kappa_i$ of an indexed family of cardinals, defined under the Axiom of Choice, König's theorem: assuming the Axiom of Choice, if $\kappa_i < \lambda_i$ for every $i \in I$ then $\sum_{i \in I} \kappa_i < \prod_{i \in I} \lambda_i$ and Assuming the Axiom of Choice: $\kappa < \kappa^{\operatorname{cf}(\kappa)}$ for every infinite cardinal $\kappa$, and $\operatorname{cf}(2^{\kappa}) > \kappa$; in particular $\operatorname{cf}(2^{\aleph_0}) > \aleph_0$.
• Clause (b) of Every infinite cardinal is $\aleph_\alpha$ for exactly one ordinal $\alpha$, in ZF; and, assuming the Axiom of Choice, every infinite set is equinumerous with exactly one aleph, that an arbitrary infinite set is equinumerous with an aleph.
• Clauses (b) and (d) of $\aleph_0$ is regular in ZF; assuming the Axiom of Choice every successor aleph $\aleph_{\alpha+1}$ is regular; $\operatorname{cf}(\aleph_\omega) = \aleph_0$, so $\aleph_\omega$ is singular, and under choice it is the least singular infinite cardinal, that successor alephs are regular and that $\aleph_\omega$ is the least singular infinite cardinal.

Equivalent to the Axiom of Choice over ZF, so neither weaker nor stronger: comparability of arbitrary sets (Comparability of arbitrary sets, that any two sets admit an injection one way or the other, is equivalent to the Axiom of Choice) and Tarski's square law (Tarski: the Axiom of Choice is equivalent to the statement that $A \times A \approx A$ for every infinite set $A$, so extending Hessenberg's theorem from the alephs to arbitrary sets is exactly as strong as choice). The first is recorded in The choice ledger: what costs the Axiom of Choice and what does not as Hartogs' result, quoted there and proved here.

Countable choice (The Axiom of Countable Choice ($\mathrm{AC}_\omega$)) is not used anywhere on this page. Where an argument might have needed it, the ordinal structure supplied a canonical least element instead.

How far regularity can fail without choice. That clause (b) above cannot be proved in ZF is recorded rather than proved here, and it is conditional: Gitik 1980: consistently, every uncountable cardinal is singular ‡ states that, relative to a large-cardinal consistency hypothesis, there is a model of ZF in which every uncountable cardinal is singular. Granting that hypothesis, "successor cardinals are regular" is a consequence of choice and not a structural fact about cardinals, which is why clause (b) is stated with its hypothesis.

Where the continuum escapes ZFC. The constraint proved here is $\operatorname{cf}(2^{\aleph_0}) > \aleph_0$ (Assuming the Axiom of Choice: $\kappa < \kappa^{\operatorname{cf}(\kappa)}$ for every infinite cardinal $\kappa$, and $\operatorname{cf}(2^{\kappa}) > \kappa$; in particular $\operatorname{cf}(2^{\aleph_0}) > \aleph_0$), which excludes some candidate values for $2^{\aleph_0}$ and selects none. Whether $2^{\aleph_0} = \aleph_1$ is the continuum hypothesis, stated in The continuum hypothesis, and what this page does not prove; that ZFC proves neither it nor its negation, granted the consistency of ZFC, is recorded in The continuum hypothesis and its generalisation are independent of ZFC ‡ and is proved neither on this page nor on any page this one rests on. The generalised form is stronger than it looks: over ZF it implies the Axiom of Choice, a result of Sierpiński recorded in Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice ‡, and proved neither here nor on any page this one rests on.

What this page therefore does and does not settle about $2^{\aleph_0}$. It settles that $2^{\aleph_0}$ is an aleph, granted choice; that it is strictly above $\aleph_0$; and that its cofinality is uncountable. It settles nothing about which aleph it is, and no statement on this page or its companion asserts a value.