TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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Comparability of arbitrary sets, that any two sets admit an injection one way or the other, is equivalent to the Axiom of Choice

Statement

Over ZF the following two statements are equivalent.

(1) The Axiom of Choice (The Axiom of Choice).

(2) Comparability. For any two sets $A$ and $B$ , either $A \preceq B$ or $B \preceq A$ ; that is, there is an injection $A \to B$ or an injection $B \to A$ (Equinumerous sets, $A \approx B$ and $A \preceq B$ , Injection, surjection, bijection).

So comparability is not a triviality about sizes but a choice principle in disguise. Everything on this page that compares two cardinals uses trichotomy of ordinals, which is a theorem of ZF; comparing two arbitrary sets is a different matter, and this theorem says exactly how different.

Facts & Assumptions

Given: ZF. Neither statement is assumed; the theorem asserts their equivalence.

[L1]

For every set $A$ there is a least ordinal $\aleph(A)$ admitting no injection into $A$ , and its construction is choice free (Hartogs: an ordinal that does not inject into a given set).

[L2]

Over ZF the Axiom of Choice is equivalent to the well-ordering theorem, that every set can be well ordered (Choice, Zorn and well-ordering are equivalent, Well-order and well-ordered set).

[L3]

Assuming the Axiom of Choice, every set carries a well-order (The well-ordering theorem).

[L4]

A set is well-orderable exactly when it is equinumerous with an ordinal, and it then has a least such ordinal $\lvert A \rvert$ , with $A \approx \lvert A \rvert$ (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, Cardinal (initial ordinal) and cardinality, Ordinal (von Neumann)).

[L5]

Ordinals satisfy trichotomy and $\alpha \subseteq \beta$ iff $\alpha \in \beta$ or $\alpha = \beta$ (Trichotomy and well-ordering of the ordinals, Basic closure properties of ordinals).

[L6]

A subset inclusion is an injection, and a composition of injections and bijections is an injection (Injection, surjection, bijection).

Proof

technique · direct

1.1

Assume (1). Given sets $A$ and $B$ , both are well-orderable by [L3], so $\lvert A\rvert$ and $\lvert B\rvert$ exist with $A \approx \lvert A\rvert$ and $B \approx \lvert B\rvert$ by [L4]; trichotomy in [L5] gives $\lvert A\rvert \subseteq \lvert B\rvert$ or $\lvert B\rvert \subseteq \lvert A\rvert$ , and composing the corresponding inclusion with the two bijections gives an injection one way or the other by [L6], which is (2).

L3L4L5L6

1.2

Assume (2), and let $A$ be any set. Put $\kappa = \aleph(A)$ , which by [L1] admits no injection into $A$ ; comparability applied to $A$ and $\kappa$ therefore leaves an injection $j : A \to \kappa$ . Defining $a <_A b :\iff j(a) \in j(b)$ transports the well-order of the ordinal $\kappa$ back to $A$ : irreflexivity, transitivity and trichotomy are immediate from injectivity, and a nonempty $S \subseteq A$ has the $<_A$ -least element $j^{-1}$ of the $\in$ -least element of $j[S]$ , which exists by [L5]. So every set can be well ordered, and (1) follows by [L2].

L1L2L5L6

2.1

Both implications are established, so (1) and (2) are equivalent over ZF.

step 1.1step 1.2∎

Remarks

Why Hartogs' theorem is the whole engine of the hard direction. Comparability by itself says nothing about ordinals; what makes it bite is that ZF alone produces, for each set $A$ , an ordinal too long to sit inside $A$ . Comparability then has only one way to resolve the pair $(A, \aleph(A))$ , and that resolution is precisely a well-ordering of $A$ . This is Hartogs' 1915 argument.

What is not being claimed. The theorem does not say that two sets are always comparable, nor that they are not. It says that "always comparable" and "the Axiom of Choice" are the same assumption over ZF. Whether ZF alone refutes comparability is a different question, not addressed here.

Where this sits relative to the rest of the page. Trichotomy for the alephs is free: they are ordinals, and Trichotomy and well-ordering of the ordinals is a theorem of ZF. The step that costs the Axiom of Choice is getting from an arbitrary set to an aleph in the first place, which is 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. This theorem is that observation sharpened into an equivalence.

Depends on

Used by

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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 67 results over 23 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Encyclopedia of Mathematics, Comparability of cardinals (standard reference, not scraped)
Hartogs number (Wikipedia) (standard reference, not scraped)
Axiom of choice — equivalents (Wikipedia) (standard reference, not scraped)