Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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.

The limit at c depends only on the restriction of f to a punctured neighbourhood of c, and passes to any subset of the domain having c as a limit point

Statement

Let A⊆R and let c be a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R).

  1. Locality. Let f,g:A→R and L∈R, and suppose there is a real η>0 with f(x)=g(x) for every x∈A satisfying 0<∣x−c∣<η. Then lim⁡x→cf(x)=L  ⟺  lim⁡x→cg(x)=L (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A).

  2. Restriction. Let B⊆A with c a limit point of B, let f:A→R and suppose lim⁡x→cf(x)=L. Then c is a limit point of A as well, and lim⁡x→cf∣B(x)=L, where f∣B:B→R is the restriction of f.

So the limit at c sees only the values of f on an arbitrarily small punctured neighbourhood of c, and it survives shrinking the domain, provided the smaller domain still accumulates at c. Together with At a limit point of the domain a function has at most one limit this is what makes the phrase the limit at c a local notion.

The converse of claim 2 is false in general: a restriction may have a limit where the function has none, as the one-sided limits of the sign function on the companion page show.

Facts & Assumptions

Given: A set A⊆R and a limit point c of A; for claim 1 functions f,g:A→R, a real L and a real η>0 with f(x)=g(x) for every x∈A satisfying 0<∣x−c∣<η; for claim 2 a subset B⊆A having c as a limit point, a function f:A→R and a real L with lim⁡x→cf(x)=L (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A, Limit point, isolated point, adherent point, derived set, and dense subset of R).

[L1]

The limit condition: lim⁡x→ch(x)=L means that for every real ε>0 there is a real δ>0 such that every x in the domain of h with 0<∣x−c∣<δ satisfies ∣h(x)−L∣<ε (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A).

[L2]

Limit point: c is a limit point of a set S when for every real δ>0 there is x∈S with 0<∣x−c∣<δ (Limit point, isolated point, adherent point, derived set, and dense subset of R, The ε-neighbourhood and the punctured ε-neighbourhood of a point of R).

[L3]

Order arithmetic: of two positive reals the smaller is positive, the order being total; and u<v≤w gives u<w (Ordered field).

[L4]

Absolute value (Basic properties of the absolute value); and uniqueness of the limit at a limit point (At a limit point of the domain a function has at most one limit), which is what makes the phrase "the limit" in the statement denote.

Proof

technique · direct
1.1

For claim 1, assume lim⁡x→cf(x)=L and let ε>0 be an arbitrary real.

assume-hypL1
1.2

For claim 2, B⊆A and c is a limit point of B; hence c is a limit point of A, since for every real δ>0 a point x∈B with 0<∣x−c∣<δ is also a point of A with 0<∣x−c∣<δ.

L2
1.3

For claim 2, assume lim⁡x→cf(x)=L and let ε>0 be an arbitrary real.

assume-hypL1
2.1

By [L1] fix a real δ0>0 such that every x∈A with 0<∣x−c∣<δ0 satisfies ∣f(x)−L∣<ε, and put δ to be the smaller of δ0 and η, so δ>0.

step 1.1L1L3choose
2.2

By [L1] fix a real δ>0 such that every x∈A with 0<∣x−c∣<δ satisfies ∣f(x)−L∣<ε.

step 1.3L1choose
3.1

Every x∈A with 0<∣x−c∣<δ satisfies both 0<∣x−c∣<δ0 and 0<∣x−c∣<η, so g(x)=f(x) and ∣g(x)−L∣=∣f(x)−L∣<ε; as ε>0 was arbitrary, lim⁡x→cg(x)=L.

step 2.1L1L3L4
3.2

Every x∈B with 0<∣x−c∣<δ lies in A and satisfies 0<∣x−c∣<δ, so f∣B(x)=f(x) and therefore ∣f(x)−L∣<ε; as ε>0 was arbitrary, and c is a limit point of B, lim⁡x→cf∣B(x)=L.

step 2.2L1L4
4.1

The hypothesis of claim 1 is symmetric in f and g, so interchanging their roles in steps 1.1, 2.1 and 3.1 gives the implication in the other direction, and claim 1 is proved; claim 2 is steps 1.2 and 3.2.

step 1.2step 3.1step 3.2∎

Remarks

Depends on

Used by

Dependency tree · two levels

20 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