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

1.1

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 $1 < 2$ .

L2L3L4

1.2

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 $2^{1} = 2^{0} \cdot 2 = 1 \cdot 2 = 2$ .

L1L4L6

2.1

$2^{ω} = ω$ : 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.

step 1.1step 1.2L1L4

3.1

$ω$ 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.

step 2.1L5∎

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.

Depends on

Used by

Ordinal α^β and cardinal κ^λ are different operations that share one notation Remark

Dependency tree · two levels

40 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

Ordinal arithmetic (Wikipedia) (standard reference, not scraped)
Cardinal number (Wikipedia) (standard reference, not scraped)
T. Jech, Set Theory, 3rd millennium ed., Ch. 2 (Ordinal numbers) (standard reference, not scraped)

FALSE: the ordinal 2ω2ω is uncountable