LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-02 (claude-opus-5)

How statement and proof provenance work

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.

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The Dedekind reals are Archimedean

Statement

$\mathbb{R}$ (Dedekind cuts) is Archimedean: for every cut $A$ there is a natural number $n$ with $A < n^{*}$ . Equivalently, the rational cuts $(n^{*})_{n \in \mathbb{N}}$ are cofinal in $\mathbb{R}$ : no single cut is an upper bound for all of them.

Facts & Assumptions

Given: A cut $A$ .

[L1]

A cut is a proper subset of $\mathbb{Q}$ ( $A \ne \mathbb{Q}$ ), and $a \in A$ , $b \notin A \Rightarrow a < b$ (Dedekind cut); the elements of $\mathbb{R}$ are exactly these cuts (The real numbers $\mathbb{R}$ as Dedekind cuts).

[L2]

Rational Archimedean property: for every $q \in \mathbb{Q}$ there is a natural number $n$ with $n > q$ (The rationals are Archimedean).

[L3]

The embedding preserves order: $p < q \Rightarrow p^{*} \subsetneq q^{*}$ , i.e. $p^{*} < q^{*}$ (The rational cuts embed densely in $\mathbb{R}$ , preserving sums, products, $0$ , $1$ and the order).

[L4]

Inclusion order, and transitivity of $<$ in the totally ordered field (Order on the Dedekind reals, The Dedekind reals form a totally ordered field).

Proof

technique · direct

1.1

Since $A \ne \mathbb{Q}$ , choose a rational $q \notin A$ .

L1choose

1.2

By the rational Archimedean property, choose a natural number $n$ with $n > q$ .

L2choose

2.1

$A \subseteq q^{*}$ : for $a \in A$ , the separation property gives $a < q$ (as $q \notin A$ ), so $a \in q^{*}$ .

step 1.1L1

2.2

$q^{*} < n^{*}$ : from $q < n$ and order preservation, $q^{*} \subsetneq n^{*}$ .

step 1.2L3

3.1

Hence $A \subseteq q^{*} \subsetneq n^{*}$ , so $A < n^{*}$ : the rational cuts are cofinal and $\mathbb{R}$ is Archimedean.

step 2.1step 2.2L4∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 39 results over 18 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

W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (Appendix: construction of ℝ) (standard reference, not scraped)
Archimedean property (Wikipedia) (standard reference, not scraped)