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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Epsilon characterisation of the infimum

Statement

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

$ℓ = \inf S ⟺ for every ε > 0 there exists s \in S with s < ℓ + ε .$

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 $S \subseteq R$ that is bounded below, a lower bound $ℓ$ of $S$ , and the reflection $T : = - S = {- s : s \in S}$ .

[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 $S \subseteq R$ 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 $X \subseteq R$ 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).

Proof

technique · direct

1.1givenL1

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.

1.2L1algebra

For $s \in R$ and $ε > 0$ , negation turns the inequality $(- ℓ) - ε < - s$ into $s < ℓ + ε$ and back, because $- (- s) = s$ and $- ((- ℓ) - ε) = ℓ + ε$ .

2.1step 1.1L2L1

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 $- ℓ = \sup T$ .

3.1step 1.1step 2.1L3

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 $t \in T$ with $(- ℓ) - ε < t$ .

4.1step 1.2step 3.1L1

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

5.1step 2.1step 3.1step 4.1∎

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

Epsilon characterisation of the infimum

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources