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LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)
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Epsilon characterisation of the infimum

Statement

Let S⊆RS \subseteq \mathbb{R} be nonempty and bounded below, and let ℓ\ell be a lower bound of SS (Lower bound, bounded below, bounded set). Then

ℓ=inf⁡S⟺for every ε>0 there exists s∈S with s<ℓ+ε.\ell = \inf S \quad \Longleftrightarrow \quad \text{for every } \varepsilon > 0 \text{ there exists } s \in S \text{ with } s < \ell + \varepsilon.

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

Facts & Assumptions

Given: A nonempty S⊆RS \subseteq \mathbb{R} that is bounded below, a lower bound ℓ\ell of SS, and the reflection T:=−S={−s:s∈S}T := -S = \{-s : s \in S\}.

[L1]

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

[L2]

Every nonempty S⊆RS \subseteq \mathbb{R} bounded below has an infimum, and inf⁡S=−sup⁡(−S)=−sup⁡T\inf S = -\sup(-S) = -\sup T (Every nonempty set bounded below has an infimum).

[L3]

Epsilon characterisation of the supremum: for a nonempty X⊆RX \subseteq \mathbb{R} bounded above and an upper bound vv of XX, one has v=sup⁡Xv = \sup X if and only if for every ε>0\varepsilon > 0 there is x∈Xx \in X with v−ε<xv - \varepsilon < x (Epsilon characterisation of the supremum).

Proof

technique · direct
1.1

Since SS is nonempty and bounded below and ℓ\ell is a lower bound of SS, the set TT is nonempty and −ℓ-\ell is an upper bound of TT, so TT is nonempty and bounded above.

givenL1
1.2

For s∈Rs \in \mathbb{R} and ε>0\varepsilon > 0, negation turns the inequality (−ℓ)−ε<−s(-\ell) - \varepsilon < -s into s<ℓ+εs < \ell + \varepsilon and back, because −(−s)=s-(-s) = s and −((−ℓ)−ε)=ℓ+ε-\big((-\ell) - \varepsilon\big) = \ell + \varepsilon.

L1algebra
2.1

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

step 1.1L2L1
3.1

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

step 1.1step 2.1L3
4.1

The elements of TT are exactly the −s-s with s∈Ss \in S, so by 1.2 the condition "there is t∈Tt \in T with (−ℓ)−ε<t(-\ell) - \varepsilon < t" is equivalent to "there is s∈Ss \in S with s<ℓ+εs < \ell + \varepsilon".

step 1.2step 3.1L1
5.1

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

step 2.1step 3.1step 4.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 11 results over 7 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

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