Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-05 (gpt-5.6-sol-codex-subscription)
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.

Subsequences inherit the limit

Statement

Let (xk) be a sequence of reals converging to x∈R (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals), and let n:N→N be strictly increasing. Then the subsequence (xnk) converges to x.

Divergence test. Consequently, if two subsequences of (xk) converge to different limits, then (xk) does not converge.

Facts & Assumptions

Given: A sequence (xk) of reals converging to a real x, and a strictly increasing n:N→N, so that (xnk) is a subsequence of (xk) (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

[L1]

(xk) converges to x when for every rational ε>0 there is K∈N with ∣xk−x∣<ε^ for all k≥K (Limits and Cauchy sequences of reals).

[L2]

Growth of the index map: nk≥k for every k (A strictly increasing index map satisfies nk≥k).

[L3]

The order on N is transitive, so nk≥k≥K gives nk≥K (≤ is a linear order on N, Order on the natural numbers).

[L4]

A sequence has at most one limit (A sequence has at most one limit).

Proof

technique · direct
1.1

Let ε>0 be rational. By convergence of (xk) there is K∈N with ∣xj−x∣<ε for every j≥K.

L1
2.1

For every k≥K we have nk≥k≥K, so the estimate of step 1.1 applies at the index j=nk and gives ∣xnk−x∣<ε.

step 1.1L2L3
3.1

Since the rational ε>0 was arbitrary, and the same threshold K served, (xnk) converges to x.

step 2.1L1
4.1

Divergence test: suppose subsequences (xnk) and (xmk) converge to a and b with a≠b. If (xk) converged, to x say, then by step 3.1 both subsequences would converge to x, and uniqueness of limits would force a=x and b=x, hence a=b, contrary to hypothesis; so (xk) does not converge.

step 3.1L4∎

Remarks

  • A sequence is a subsequence of itself, via the identity index map nk=k, which is strictly increasing. So "every subsequence converges to x" and "the sequence converges to x" are equivalent, and the lemma is the non-trivial half of that equivalence.

  • The converse of the first claim is false: one convergent subsequence says nothing about the sequence (FALSE: a convergent subsequence forces the sequence to converge). It is the divergence test, not the convergence of a single subsequence, that is usable.

  • The divergence test is the standard way to show a concrete sequence diverges, and it is how FALSE: every bounded sequence converges is refuted. The opposite direction, extracting a convergent subsequence from a bounded sequence, is Bolzano-Weierstrass; it is not available at this point in the reading order, being the subject of the next page of this track, Monotone Sequences, Bolzano-Weierstrass, and Cauchy Completeness, where it is proved.

Depends on

Used by

Dependency tree · two levels

30 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