TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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$ω_{1}$ is uncountable, every ordinal below it is at most countable, it is a cardinal and a limit ordinal, and its existence is a theorem of ZF

Statement

Let $ω_{1} = ℵ (ω)$ (The first uncountable ordinal $ω_{1} : = ℵ (ω)$ ). Then:

(a) The bridge. An ordinal $α$ (Ordinal (von Neumann)) injects into $N$ if and only if $α$ is at most countable (Finite, countably infinite, countable, uncountable).

(b) $ω_{1}$ is uncountable.

(c) Every ordinal $α \in ω_{1}$ is at most countable; so $ω_{1}$ is the least uncountable ordinal.

(d) $ω_{1}$ is a cardinal, that is an initial ordinal (Cardinal (initial ordinal) and cardinality): no $α \in ω_{1}$ is equinumerous with $ω_{1}$ .

(e) $ω_{1}$ is a limit ordinal (Successor and limit ordinals).

All of this is a theorem of ZF and uses no choice principle. That matters here and is stated deliberately: Hartogs: an ordinal that does not inject into a given set is choice free, Every subset of an at most countable set is at most countable and A nonempty set is at most countable iff it is a surjective image of $N$ are choice free, so $ω_{1}$ and every property listed above exist in ZF alone. The cost begins two items later on this page, at the boundedness theorem for at most countable subsets of $ω_{1}$ , which genuinely needs countable choice.

Facts & Assumptions

Given: $ω_{1} = ℵ (ω)$ , the least ordinal admitting no injection into $N = ω$ (The first uncountable ordinal $ω_{1} : = ℵ (ω)$ , Hartogs: an ordinal that does not inject into a given set).

[L1]

$ℵ (A)$ is the least ordinal that does not inject into $A$ ; in particular every ordinal strictly below $ℵ (A)$ does inject into $A$ , and $ℵ (A)$ does not. The construction is choice free (Hartogs: an ordinal that does not inject into a given set).

[L2]

$A$ is finite when $A \approx n$ for some $n \in N$ , countably infinite when $A \approx N$ , at most countable when one of the two holds, and uncountable when neither does (Finite, countably infinite, countable, uncountable, Equinumerous sets, $A \approx B$ and $A ⪯ B$ ).

[L3]

Every subset of an at most countable set is at most countable, and no choice principle is used (Every subset of an at most countable set is at most countable).

[L4]

A nonempty set $A$ is at most countable if and only if there is a surjection $N \to A$ , and no choice principle is used (A nonempty set is at most countable iff it is a surjective image of $N$ , Injection, surjection, bijection).

[L5]

An injection $f : X \to Y$ is a bijection of $X$ onto $f [X] \subseteq Y$ , and $\approx$ is symmetric and transitive (Injection, surjection, bijection, Equinumerous sets, $A \approx B$ and $A ⪯ B$ ).

[L6]

An ordinal $κ$ is a cardinal when no $α \in κ$ satisfies $α \approx κ$ (Cardinal (initial ordinal) and cardinality).

[L7]

Every ordinal is exactly one of $0$ , a successor, or a limit (Successor and limit ordinals); $μ^{+} = μ \cup {μ}$ is an ordinal, $μ \subseteq ν$ iff $μ \in ν$ or $μ = ν$ , and $μ \notin μ$ (Basic closure properties of ordinals); trichotomy holds (Trichotomy and well-ordering of the ordinals).

[L8]

Every natural number is an ordinal, $ω$ is an ordinal and a limit ordinal, and $n \subseteq ω$ for $n \in ω$ ( $ω$ is the least limit ordinal, The natural numbers $N$ (von Neumann)).

Proof

technique · direct

1.1

Claim (a), forwards: if $f : α \to N$ is injective then $α \approx f [α] \subseteq N$ by [L5], and $f [α]$ is at most countable by [L3], so $α$ is at most countable by [L2] and transitivity of $\approx$ .

L2L3L5

1.2

Claim (a), backwards: if $α$ is at most countable then $α \approx n$ for some $n \in N$ or $α \approx N$ ; a bijection $α \to n$ followed by the inclusion $n \subseteq N$ is an injection $α \to N$ by [L8], and a bijection $α \to N$ is one outright.

L2L5L8

1.3

$ω \in ω_{1}$ : the identity is an injection $ω \to N$ , so $ω \neq ω_{1}$ by [L1]; and $ω_{1} \in ω$ or $ω_{1} = ω$ would give $ω_{1} \subseteq ω$ by [L7] and hence an injection $ω_{1} \to N$ by inclusion, which [L1] forbids; so $ω \in ω_{1}$ by trichotomy.

L1L5L7L8

2.1

Claim (b): $ω_{1}$ does not inject into $N$ by [L1], so it is not at most countable by step 1.2, that is, it is uncountable.

step 1.2L1L2

2.2

Claim (c): every $α \in ω_{1}$ injects into $N$ by [L1], hence is at most countable by step 1.1; and by [L7] any uncountable ordinal $γ$ satisfies $ω_{1} \leq γ$ , since $γ \in ω_{1}$ would make $γ$ at most countable.

step 1.1L1L7

3.1

Claim (d): suppose $α \in ω_{1}$ satisfies $α \approx ω_{1}$ ; then $α$ is at most countable by step 2.2, so $ω_{1}$ is at most countable by [L2] and symmetry of $\approx$ , contradicting step 2.1; hence $ω_{1}$ is a cardinal in the sense of [L6].

step 2.2step 2.1L2L5L6

3.2

Claim (e): $ω_{1} \neq 0$ by step 1.3, since $ω \in ω_{1}$ ; and $ω_{1}$ is not a successor, for if $ω_{1} = δ^{+}$ then $ω \in δ^{+}$ gives $ω \subseteq δ$ by [L7], so $δ$ is a nonempty ordinal in $ω_{1}$ and is therefore at most countable by step 2.2, so [L4] supplies a surjection $s : N \to δ$ , and the function $t : N \to δ^{+}$ with $t (0) = δ$ and $t (σ (n)) = s (n)$ is a surjection onto $δ^{+} = δ \cup {δ}$ , making $ω_{1}$ at most countable by [L4] and contradicting step 2.1; so $ω_{1}$ is a limit ordinal by [L7].

step 1.3step 2.2step 2.1L4L7L8

4.1

Claims (a) to (e) are established, and every step used only Hartogs: an ordinal that does not inject into a given set, Every subset of an at most countable set is at most countable and A nonempty set is at most countable iff it is a surjective image of $N$ , all of which are choice free, so the whole statement is a theorem of ZF.

step 3.1step 3.2step 2.1step 2.2step 1.1step 1.2L1L3L4∎

Remarks

The bridge is the whole trick. Hartogs: an ordinal that does not inject into a given set produces the least ordinal that does not inject into $N$ . What is wanted is the least uncountable ordinal. Claim (a) is what identifies the two notions on ordinals, and it is two lines in each direction; without it, quoting Hartogs for uncountability would be citing a theorem for a claim it does not make.

No choice, and why it is worth saying. A reader who has met $ω_{1}$ through cardinal arithmetic often expects the well-ordering theorem to be somewhere in the background. It is not. Hartogs' construction collects the order types of well-ordered subsets of $N$ , and the well-ordering comes with each subset as part of the datum, so nothing is selected (Hartogs: an ordinal that does not inject into a given set, remarks). The first genuine choice principle on this page appears at Assuming countable choice: every at most countable subset of $ω_{1}$ is bounded below $ω_{1}$ , so no at most countable subset of $ω_{1}$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable, and Choice ledger for this page: $ω_{1}$ exists in ZF, and the boundedness theorem does not keeps the ledger.

" $ω_{1}$ is a cardinal" is a property of an ordinal, not an assignment of a size. Cardinal (initial ordinal) and cardinality separates the two: being an initial ordinal is choice free, whereas attaching a cardinality $∣ X ∣$ to an arbitrary set $X$ needs the Axiom of Choice. Claim (d) is the first, and only the first.

What is deliberately absent. Nothing here says $ω_{1}$ is regular, or computes its cofinality, or compares it with the size of $P (N)$ . Regularity of $ω_{1}$ is the boundedness theorem two items later and costs countable choice; the comparison with $P (N)$ is the continuum hypothesis (The continuum hypothesis, and what this page does not prove) and is independent of ZFC.

Depends on

Used by

Cited to discharge well-definedness by The first uncountable ordinal ω₁ := ℵ(ω).

Dependency tree · two levels

43 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

First uncountable ordinal (Wikipedia) (standard reference, not scraped)
Hartogs number (Wikipedia) (standard reference, not scraped)
T. Jech, Set Theory, 3rd millennium ed., Ch. 3 (Cardinal numbers) (standard reference, not scraped)

ω1ω1​ is uncountable, every ordinal below it is at most countable, it is a cardinal and a limit ordinal, and its existence is a theorem of ZF