LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-25

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Each rational cut $q^{*}$ is a Dedekind cut

Statement

For every $q \in Q$ the set $q^{*} = {r \in Q : r < q}$ (The real numbers $R$ as Dedekind cuts) is a Dedekind cut (Dedekind cut). In particular $0^{*}$ and $1^{*}$ are Dedekind cuts, hence elements of $R$ , so they are legitimate as the additive and multiplicative identities of $R$ .

Facts & Assumptions

Given: A rational $q$ and the set $q^{*} = {r \in Q : r < q}$ , with the Dedekind-cut axioms (C1) proper and nonempty, (C2) downward closed, (C3) no greatest element (Dedekind cut).

[L1]

$Q$ is a totally ordered field: $<$ is transitive and total, $q - 1 < q < q + 1$ , and whenever $p < q$ the midpoint $\frac{p + q}{2}$ satisfies $p < \frac{p + q}{2} < q$ (The rationals form a totally ordered field).

Proof

technique · direct

1.1givenL1

(C1) $q^{*}$ is nonempty and proper: $q - 1 < q$ gives $q - 1 \in q^{*}$ , while $q ≮ q$ gives $q \notin q^{*}$ , so $q^{*} \neq \emptyset$ and $q^{*} \neq Q$ .

1.2givenL1

(C2) $q^{*}$ is downward closed: if $p \in q^{*}$ , so $p < q$ , and $r < p$ , then $r < q$ by transitivity, hence $r \in q^{*}$ .

1.3givenL1

(C3) $q^{*}$ has no greatest element: if $p \in q^{*}$ then $p < q$ , so the midpoint $m = \frac{p + q}{2}$ satisfies $p < m < q$ , giving $m \in q^{*}$ with $m > p$ .

2.1step 1.1step 1.2step 1.3∎

Satisfying (C1), (C2), (C3), $q^{*}$ is a Dedekind cut; applied at $q = 0$ and $q = 1$ this shows $0^{*}$ and $1^{*}$ are Dedekind cuts and hence elements of $R$ .

Each rational cut $q^{*}$ is a Dedekind cut

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Each rational cut q∗q∗ is a Dedekind cut

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Each rational cut $q^{*}$ is a Dedekind cut