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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.

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

Epsilon characterisation of the infimum

Statement

Let S⊆R be nonempty and bounded below, and let ℓ be a lower bound of S (Lower bound, bounded below, bounded set). Then

ℓ=inf⁡S⟺for every ε>0 there exists s∈S with s<ℓ+ε.

In words: among the lower bounds of S, the infimum is exactly the one that cannot be raised by any positive amount and still bound S from below.

Facts & Assumptions

Given: A nonempty S⊆R that is bounded below, a lower bound ℓ of S, and the reflection T:=−S={−s:s∈S}.

[L1]

Reflection, at the level of sets: S is nonempty exactly when T is; ℓ is a lower bound of S exactly when −ℓ is an upper bound of T; and S is bounded below exactly when T is bounded above (Reflection through zero exchanges upper and lower bounds). Elementwise, negation reverses the order: −(−a)=a, because (−a)+a=0 and additive inverses are unique (Field, Identities and inverses in a field are unique); and a<b exactly when −b<−a, because translation invariance applied with the constant −a−b turns a<b into −b<−a and, applied with the constant a+b, turns it back (Order is preserved by adding a constant and by adding inequalities).

[L2]

Every nonempty S⊆R bounded below has an infimum, and inf⁡S=−sup⁡(−S)=−sup⁡T (Every nonempty set bounded below has an infimum).

[L3]

Epsilon characterisation of the supremum: for a nonempty X⊆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∈X with v−ε<x (Epsilon characterisation of the supremum).

Proof

technique · direct
1.1

Since S is nonempty and bounded below and ℓ is a lower bound of S, the set T is nonempty and −ℓ is an upper bound of T, so T is nonempty and bounded above.

givenL1
1.2

For s∈R and ε>0, negation turns the inequality (−ℓ)−ε<−s into s<ℓ+ε and back, because −(−s)=s and −((−ℓ)−ε)=ℓ+ε.

L1algebra
2.1

By [L2] the infimum of S exists and equals −sup⁡T; hence ℓ=inf⁡S holds if and only if ℓ=−sup⁡T, which by negating both sides holds if and only if −ℓ=sup⁡T.

step 1.1L2L1
3.1

Applying [L3] to the nonempty bounded-above set T and its upper bound −ℓ: −ℓ=sup⁡T if and only if for every ε>0 there is t∈T with (−ℓ)−ε<t.

step 1.1step 2.1L3
4.1

The elements of T are exactly the −s with s∈S, so by 1.2 the condition "there is t∈T with (−ℓ)−ε<t" is equivalent to "there is s∈S with s<ℓ+ε".

step 1.2step 3.1L1
5.1

Chaining the equivalences, ℓ=inf⁡S if and only if for every ε>0 there is s∈S with s<ℓ+ε.

step 2.1step 3.1step 4.1∎

Depends on

Used by

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources