LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (openai/gpt-5.4)verified 2026-07-26 (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.

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

Supremum of a translate: $\sup (a + S) = a + \sup S$

Statement

Let $S \subseteq R$ be nonempty and bounded above and let $a \in R$ . Write $a + S : = {a + s : s \in S}$ . Then $a + S$ is nonempty and bounded above, and

$\sup (a + S) = a + \sup S .$

Facts & Assumptions

Given: A nonempty $S \subseteq R$ that is bounded above, an element $a \in R$ , and the translate $a + S = {a + s : s \in S}$ .

[L1]

Epsilon characterisation of the supremum: for a nonempty $X \subseteq R$ bounded above and an upper bound $v$ of $X$ , one has $v = \sup X$ if and only if for every $ε > 0$ there is $x \in X$ with $v - ε < x$ (Epsilon characterisation of the supremum).

[L2]

Adding a constant preserves the order: $x < y$ implies $x + c < y + c$ , and hence $x \leq y$ if and only if $x + c \leq y + c$ , since one may add $- c$ to return (Order is preserved by adding a constant and by adding inequalities).

[L3]

Supremum and the least-upper-bound property: every nonempty $X \subseteq R$ bounded above has a least upper bound $\sup X$ , an upper bound that is $\leq$ every upper bound of $X$ (Complete ordered field (least-upper-bound property)).

Proof

technique · direct

1.1givenL3

Since $S$ is nonempty and bounded above, the least-upper-bound property gives $u : = \sup S$ , which is an upper bound of $S$ .

1.2given

The set $a + S$ is nonempty, because $S$ has an element $s$ and then $a + s \in a + S$ .

2.1step 1.1L2

Every $s \in S$ satisfies $s \leq u$ , hence $a + s \leq a + u$ ; as the elements of $a + S$ are exactly these $a + s$ , the number $a + u$ is an upper bound of $a + S$ , so $a + S$ is bounded above.

2.2step 1.1L1L2algebra

Let $ε > 0$ . Applying the epsilon characterisation to $S$ and its supremum $u$ produces $s \in S$ with $u - ε < s$ , and adding $a$ gives $(a + u) - ε = a + (u - ε) < a + s$ , where $a + s \in a + S$ .

3.1step 1.2step 2.1L3

The set $a + S$ is nonempty and bounded above, so $\sup (a + S)$ exists.

4.1step 2.1step 2.2step 3.1L1∎

Now $a + u$ is an upper bound of $a + S$ and for every $ε > 0$ some element of $a + S$ exceeds $(a + u) - ε$ , so the epsilon characterisation applied to $a + S$ gives $\sup (a + S) = a + u = a + \sup S$ .

Supremum of a translate: $\sup (a + S) = a + \sup S$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Supremum of a translate: sup⁡(a+S)=a+sup⁡Ssup(a+S)=a+supS

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Supremum of a translate: $\sup (a + S) = a + \sup S$