CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The irrationals are uncountable

Statement

Let $R$ be a complete ordered field (Complete ordered field (least-upper-bound property)) and let $ι : Q \to R$ be the canonical embedding (The unique embedding of ℚ into an ordered field); write $Q_{R} = ι [Q]$ for the copy of the rationals inside $R$ , the set usually written $Q$ once the identification is made. Then the set of irrationals

$R ∖ Q_{R}$

is uncountable (Finite, countably infinite, countable, uncountable).

Only the union of two sets is used, and that needs no choice whatsoever. If the irrationals were at most countable, then $R$ would be the union of the two at most countable sets $Q_{R}$ and $R ∖ Q_{R}$ , and countability of a two-set union is proved by interleaving two given enumerations. The countable union theorem, which does spend ${A C}_{ω}$ , is not invoked here and is not needed; see the remarks below.

Facts & Assumptions

Given: A complete ordered field $R$ , the canonical embedding $ι : Q \to R$ , the subset $Q_{R} = ι [Q]$ and its complement $X = R ∖ Q_{R}$ , so that $R = Q_{R} \cup X$ .

[L1]

$ι$ is injective (The unique embedding of ℚ into an ordered field), hence a bijection of $Q$ onto $Q_{R}$ ; $\approx$ is transitive (Equinumerous sets, $A \approx B$ and $A ⪯ B$ , Injection, surjection, bijection).

[L2]

$Q \approx N$ , so $Q$ is at most countable ( $Q$ is countably infinite).

[L3]

A nonempty set is at most countable if and only if some surjection $N \to$ it exists (A nonempty set is at most countable iff it is a surjective image of $N$ ); uncountable means not at most countable (Finite, countably infinite, countable, uncountable).

[L4]

There is a bijection $β : N \to N \times N$ ( $N \times N \approx N$ ).

[L5]

$R$ is uncountable ( $R$ is uncountable (Cantor's nested intervals, 1874)).

Proof

technique · contradiction

1.1assume-contra

Suppose, for contradiction, that $X = R ∖ Q_{R}$ is at most countable.

1.2L1L2

$Q_{R} \approx Q \approx N$ by [L1] and [L2], so $Q_{R}$ is at most countable, and it is nonempty since $0 \in Q_{R}$ .

1.3L4

Fix the bijection $β : N \to N \times N$ of [L4].

2.1step 1.2given

If $X = \emptyset$ then $R = Q_{R}$ , which is at most countable by step 1.2.

2.2step 1.1step 1.2L3

Otherwise $X \neq \emptyset$ , and since $X$ is at most countable by assumption and $Q_{R}$ is nonempty and at most countable by step 1.2, [L3] provides surjections $f : N \to Q_{R}$ and $g : N \to X$ .

3.1step 2.2given

Define $u : N \times N \to R$ by $u (0, k) = f (k)$ and $u (n, k) = g (k)$ for $n \neq 0$ . Every element of $R$ lies in $Q_{R}$ or in $X$ , hence is $f (k)$ or $g (k)$ for some $k$ , so $u$ is surjective onto $R$ . The two surjections were obtained one after the other, not selected simultaneously from an infinite family, so no choice principle is used.

4.1step 1.3step 3.1L3

Hence $u \circ β : N \to R$ is a surjection and $R \neq \emptyset$ , so $R$ is at most countable by [L3].

5.1step 2.1step 4.1L3L5discharge-contradiction∎

In either case $R$ is at most countable, by step 2.1 in the first and step 4.1 in the second; this contradicts [L5]. Therefore $X = R ∖ Q_{R}$ is uncountable.

Remarks

The same argument shows that removing any at most countable set from $R$ leaves an uncountable set. In particular the algebraic numbers, once they are available, can be removed to show transcendental numbers exist, which is how Cantor's 1874 paper presented the result: an existence proof for transcendentals with no example constructed.
The corollary is a statement about the set of irrationals only. It says nothing about any individual irrational, and it does not exhibit one; the library exhibits $\sqrt{2}$ separately ( $\sqrt{2}$ exists in every complete ordered field, and is irrational, FALSE: some rational number squares to 2).
Keeping the two-set union separate from the countable union is not pedantry. The supplied countable-union theorem explicitly assumes ${A C}_{ω}$ (Countable unions of at most countable sets, assuming ${A C}_{ω}$ ), whereas this corollary, like $R$ is uncountable (Cantor's nested intervals, 1874) itself, is proved in ZF.

The irrationals are uncountable

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources