LemmaStatement: 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.

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

Statement

Let $S, T \subseteq R$ be nonempty and bounded above, and write $S + T : = {s + t : s \in S, t \in T}$ . Then $S + T$ is nonempty and bounded above, and

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

Facts & Assumptions

Given: Nonempty sets $S, T \subseteq R$ , both bounded above, and the sumset $S + T = {s + t : s \in S, t \in T}$ .

[L1]

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

[L2]

Order and addition: strict inequalities translate and add, that is $x < y$ implies $x + c < y + c$ , and $x < y$ together with $z < w$ gives $x + z < y + w$ (claims 1 and 2 of Order is preserved by adding a constant and by adding inequalities). Adjoining the case of equality, in which both sides move by the same amount, gives the nonstrict forms used below: $x \leq y$ implies $x + c \leq y + c$ , and $x \leq y$ together with $z \leq w$ gives $x + z \leq y + w$ .

[L3]

Supremum and the least-upper-bound property: $w = \sup X$ means $w$ is an upper bound of $X$ with $w \leq w^{'}$ for every upper bound $w^{'}$ of $X$ , and every nonempty $X \subseteq R$ bounded above has such a $w$ (Complete ordered field (least-upper-bound property)).

[L4]

Halving: $0 < 1$ (The multiplicative identity is positive); the positives are closed under addition, so $2 : = 1 + 1 > 0$ , and by trichotomy a positive element is nonzero, so $2 \neq 0$ (axioms O2 and O1 of Ordered field); hence $2^{- 1}$ exists (Field) and $0 \cdot 2 = 0$ (Multiplication by zero: $0 \cdot a = 0$ ); and for the positive multiplier $2$ one has $x < y$ if and only if $x \cdot 2 < y \cdot 2$ (claim 4 of Sign rules for products and monotonicity of multiplication).

Proof

technique · direct

1.1givenL3

Both $S$ and $T$ are nonempty and bounded above, so the least-upper-bound property supplies $u : = \sup S$ and $v : = \sup T$ , upper bounds of $S$ and of $T$ respectively.

1.2given

The sumset $S + T$ is nonempty: picking $s \in S$ and $t \in T$ , which is possible since both sets are nonempty, gives $s + t \in S + T$ .

2.1step 1.1L2

For $s \in S$ and $t \in T$ we have $s \leq u$ and $t \leq v$ , and adding these inequalities gives $s + t \leq u + v$ ; since every element of $S + T$ has this form, $u + v$ is an upper bound of $S + T$ , so $S + T$ is bounded above.

2.2step 1.1L1L2L4algebra

Let $ε > 0$ and put $η : = ε \cdot 2^{- 1}$ , so that $η \cdot 2 = ε$ and $η + η = η (1 + 1) = ε$ ; from $0 \cdot 2 = 0 < ε = η \cdot 2$ and $2 > 0$ we get $η > 0$ , so the epsilon characterisation applied to $S$ with $u$ and to $T$ with $v$ produces $s \in S$ with $u - η < s$ and $t \in T$ with $v - η < t$ , and adding these strict inequalities gives $(u + v) - ε = (u - η) + (v - η) < s + t$ , an element of $S + T$ .

3.1step 1.2step 2.1L3

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

4.1step 2.1step 2.2step 3.1L1∎

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

Remarks

The inequality $\sup (S + T) \leq \sup S + \sup T$ is the easy half and needs only that $u + v$ bounds $S + T$ ; the content is the reverse inequality, and the halving of $ε$ is what lets two separate approximations be combined without overshooting.
The corresponding statement for infima, $\inf (S + T) = \inf S + \inf T$ for nonempty $S, T$ bounded below, follows by reflection (Reflection through zero exchanges upper and lower bounds, Every nonempty set bounded below has an infimum), since $- (S + T) = (- S) + (- T)$ .
No analogue holds for products in general: sign changes break the argument, and $\sup (S T)$ is not determined by $\sup S$ and $\sup T$ alone.

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Supremum of a sumset: sup⁡(S+T)=sup⁡S+sup⁡Tsup(S+T)=supS+supT

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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