LemmaStatement: AI-adaptedProof: AI-adaptedprecheck 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.

Monotonicity of the supremum under inclusion

Statement

Let $S$ and $T$ be subsets of $R$ with $\emptyset \neq S \subseteq T$ , and suppose $T$ is bounded above. Then $S$ is nonempty and bounded above, both $\sup S$ and $\sup T$ exist, and

$\sup S \leq \sup T .$

Facts & Assumptions

Given: Sets $S, T \subseteq R$ with $S \neq \emptyset$ , $S \subseteq T$ , and $T$ bounded above.

[L1]

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

[L2]

A least upper bound is unique, so $\sup X$ denotes a single real number (Suprema and infima are unique).

Proof

technique · direct

1.1givenchoose

Since $T$ is bounded above, fix an upper bound $u$ of $T$ , so $t \leq u$ for every $t \in T$ .

1.2given

By hypothesis $S$ is nonempty and $S \subseteq T$ , so $T$ is nonempty as well.

2.1step 1.1step 1.2L1

Every $s \in S$ lies in $T$ and therefore satisfies $s \leq u$ ; hence $u$ is an upper bound of $S$ and $S$ is bounded above.

3.1step 1.2step 2.1L1L2

Both $S$ and $T$ are nonempty and bounded above, so by the least-upper-bound property $\sup S$ and $\sup T$ exist, each uniquely.

4.1step 3.1step 1.2L1

As $\sup T$ is an upper bound of $T$ , every $t \in T$ satisfies $t \leq \sup T$ ; since $S \subseteq T$ , every $s \in S$ satisfies $s \leq \sup T$ , so $\sup T$ is an upper bound of $S$ .

5.1step 4.1step 3.1L1∎

The number $\sup S$ is the least of the upper bounds of $S$ , and $\sup T$ is one of them, hence $\sup S \leq \sup T$ .

Remarks

The hypothesis that the larger set is bounded above cannot be weakened to the smaller one being bounded above: $\sup T$ has to exist for the inequality to mean anything.
The dual statement, $\inf T \leq \inf S$ for $\emptyset \neq S \subseteq T$ with $T$ bounded below, follows by applying this lemma to $\emptyset \neq - S \subseteq - T$ , which gives $\sup (- S) \leq \sup (- T)$ , and then negating and using $\inf X = - \sup (- X)$ (Reflection through zero exchanges upper and lower bounds, Every nonempty set bounded below has an infimum).

Monotonicity of the supremum under inclusion

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources