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False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29
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FALSE: the ordinal 2ω2^{\omega} is uncountable

Statement

FALSE. The ordinal 2ω2^{\omega} (Ordinal exponentiation αβ\alpha^{\beta}, with the conventions α0=1\alpha^{0} = 1 and 00=10^{0} = 1) is uncountable (Finite, countably infinite, countable, uncountable).

The claim comes from importing an expectation about cardinal exponentiation, where the power of 22 by the size of N\mathbb{N} is the size of P(N)\mathcal{P}(\mathbb{N}) and really is uncountable. Ordinal exponentiation is a different operation that happens to share the notation, and here 2ω=ω2^{\omega} = \omega, which is countably infinite.

Facts & Assumptions

[L1]

α0=1\alpha^{0} = 1, αδ+=αδα\alpha^{\delta^{+}} = \alpha^{\delta} \cdot \alpha, and αλ={αβ:0<β<λ}\alpha^{\lambda} = \bigcup\{\alpha^{\beta} : 0 < \beta < \lambda\} for limit λ\lambda (Ordinal exponentiation αβ\alpha^{\beta}, with the conventions α0=1\alpha^{0} = 1 and 00=10^{0} = 1).

[L4]

ω\omega is a limit ordinal, so ω=ω\bigcup \omega = \omega and ξω\xi \in \omega implies ξ+ω\xi^{+} \in \omega (ω\omega is the least limit ordinal, Successor and limit ordinals); every ordinal is transitive, μν\mu \subseteq \nu iff μν\mu \in \nu or μ=ν\mu = \nu, and μμ\mu \notin \mu (Ordinal (von Neumann), Basic closure properties of ordinals, Trichotomy and well-ordering of the ordinals); μ<ν\mu < \nu iff μ+ν\mu^{+} \le \nu; and 121 \in 2, so 1<21 < 2.

[L5]

A set is at most countable when it is finite or equinumerous with N\mathbb{N}, and uncountable when it is neither; ω=N\omega = \mathbb{N} is equinumerous with N\mathbb{N} by the identity (Finite, countably infinite, countable, uncountable, Equinumerous sets, ABA \approx B and ABA \preceq B, The natural numbers N\mathbb{N} (von Neumann)).

Refutation

technique · direct
1.1

For every nωn \in \omega the ordinal 2n2^{n} lies in ω\omega by [L3], hence 2nω2^{n} \subseteq \omega by [L4]; and n2nn \le 2^{n} by [L2], since 1<21 < 2.

L2L3L4
1.2

The set united in the limit clause at λ=ω\lambda = \omega is {2n:nω and n0}\{2^{n} : n \in \omega \text{ and } n \ne 0\}, and it is nonempty, since 1ω1 \in \omega and 101 \ne 0, with 21=202=12=22^{1} = 2^{0} \cdot 2 = 1 \cdot 2 = 2.

L1L4L6
2.1

2ω=ω2^{\omega} = \omega: the union is contained in ω\omega because each 2nω2^{n} \subseteq \omega by step 1.1; and it contains ω\omega, because a given mωm \in \omega has m+ωm^{+} \in \omega with m+0m^{+} \ne 0 by [L4], and mm+2m+m \in m^{+} \le 2^{m^{+}} by step 1.1, so m2m+m \in 2^{m^{+}}, one of the sets united.

step 1.1step 1.2L1L4
3.1

ω\omega is equinumerous with N\mathbb{N} by [L5], so 2ω=ω2^{\omega} = \omega is countably infinite and in particular at most countable, hence not uncountable; the claim is false.

step 2.1L5

Remarks

The general pattern. The same computation gives kω=ωk^{\omega} = \omega for every finite k2k \ge 2. What makes a finite base collapse is that knk^{n} is again a natural number, by On ω\omega the ordinal ++ and \cdot are the Peano operations: ω\omega is closed under ordinal ++, \cdot and exponentiation, and for naturals m,nm, n the ordinal m+nm + n and mnm \cdot n are the natural-number sum and product, so the whole tower stays inside ω\omega and its supremum is ω\omega. An infinite base does not collapse: ωω\omega^{\omega} is computed on the companion examples page and is far above ω\omega.

Order type against cardinality. 2ω=ω2^{\omega} = \omega is a statement about order type. It says nothing about the size of P(N)\mathcal{P}(\mathbb{N}), which is uncountable by Cantor's theorem: AP(A)A \prec \mathcal{P}(A). The two operations that both get written 2ω2^{\omega} are compared in Ordinal αβ\alpha^{\beta} and cardinal κλ\kappa^{\lambda} are different operations that share one notation, which is where the clash of notation is set out.

A weaker true statement. Every ordinal below ω1\omega_1 is at most countable (ω1\omega_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), and 2ω=ω<ω12^{\omega} = \omega < \omega_1, so countability of 2ω2^{\omega} also follows from that theorem. The computation above is preferred because it identifies the ordinal exactly.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 62 results over 27 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