False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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FALSE: the ordinal $2^{ω}$ is uncountable

Statement

FALSE. The ordinal $2^{ω}$ (Ordinal exponentiation $α^{β}$ , with the conventions $α^{0} = 1$ and $0^{0} = 1$ ) is uncountable (Finite, countably infinite, countable, uncountable).

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

Facts & Assumptions

Given: The ordinals with the operations of Ordinal exponentiation $α^{β}$ , with the conventions $α^{0} = 1$ and $0^{0} = 1$ , and $ω = N$ the least limit ordinal ( $ω$ is the least limit ordinal, Successor and limit ordinals, The natural numbers $N$ (von Neumann)).

[L1]

$α^{0} = 1$ , $α^{δ^{+}} = α^{δ} \cdot α$ , and $α^{λ} = ⋃ {α^{β} : 0 < β < λ}$ for limit $λ$ (Ordinal exponentiation $α^{β}$ , with the conventions $α^{0} = 1$ and $0^{0} = 1$ ).

[L2]

For $α > 1$ one has $β \leq α^{β}$ for every ordinal $β$ (claim (d) of $α^{β + γ} = α^{β} \cdot α^{γ}$ and $(α^{β})^{γ} = α^{β \cdot γ}$ ; and for $α > 1$ exponentiation is strictly increasing with $β \leq α^{β}$ ).

[L3]

For $m, n \in ω$ the ordinal $m^{n}$ lies in $ω$ (claim (a) of On $ω$ the ordinal $+$ and $\cdot$ are the Peano operations: $ω$ is closed under ordinal $+$ , $\cdot$ and exponentiation, and for naturals $m, n$ the ordinal $m + n$ and $m \cdot n$ are the natural-number sum and product).

[L4]

$ω$ is a limit ordinal, so $⋃ ω = ω$ and $ξ \in ω$ implies $ξ^{+} \in ω$ ( $ω$ is the least limit ordinal, Successor and limit ordinals); every ordinal is transitive, $μ \subseteq ν$ iff $μ \in ν$ or $μ = ν$ , and $μ \notin μ$ (Ordinal (von Neumann), Basic closure properties of ordinals, Trichotomy and well-ordering of the ordinals); $μ < ν$ iff $μ^{+} \leq ν$ ; and $1 \in 2$ , so $1 < 2$ .

[L5]

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

[L6]

$μ \cdot 1 = μ$ (claim (a) of Monotonicity of ordinal $+$ and $\cdot$ : strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0 + β = β$ and $1 \cdot β = β$ ).

Refutation

technique · direct

1.1L2L3L4

For every $n \in ω$ the ordinal $2^{n}$ lies in $ω$ by [L3], hence $2^{n} \subseteq ω$ by [L4]; and $n \leq 2^{n}$ by [L2], since $1 < 2$ .

1.2L1L4L6

The set united in the limit clause at $λ = ω$ is ${2^{n} : n \in ω and n \neq 0}$ , and it is nonempty, since $1 \in ω$ and $1 \neq 0$ , with $2^{1} = 2^{0} \cdot 2 = 1 \cdot 2 = 2$ .

2.1step 1.1step 1.2L1L4

$2^{ω} = ω$ : the union is contained in $ω$ because each $2^{n} \subseteq ω$ by step 1.1; and it contains $ω$ , because a given $m \in ω$ has $m^{+} \in ω$ with $m^{+} \neq 0$ by [L4], and $m \in m^{+} \leq 2^{m^{+}}$ by step 1.1, so $m \in 2^{m^{+}}$ , one of the sets united.

3.1step 2.1L5∎

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

Remarks

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

Order type against cardinality. $2^{ω} = ω$ is a statement about order type. It says nothing about the size of $P (N)$ , which is uncountable by Cantor's theorem: $A ≺ P (A)$ . The two operations that both get written $2^{ω}$ are compared in Ordinal $α^{β}$ and cardinal $κ^{λ}$ are different operations that share one notation, which is where the clash of notation is set out.

A weaker true statement. Every ordinal below $ω_{1}$ is at most countable ( $ω_{1}$ 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), and $2^{ω} = ω < ω_{1}$ , so countability of $2^{ω}$ also follows from that theorem. The computation above is preferred because it identifies the ordinal exactly.

FALSE: the ordinal $2^{ω}$ is uncountable

Statement

Facts & Assumptions

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Dependency tree · 2 levels

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FALSE: the ordinal 2ω2ω is uncountable

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Dependency tree · 2 levels

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FALSE: the ordinal $2^{ω}$ is uncountable