False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck 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.

FALSE: a convergent subsequence forces the sequence to converge

Statement

False claim: if some subsequence of a sequence $(x_k)$ of reals converges, then $(x_k)$ itself converges (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

The true statement in this direction runs the other way: Subsequences inherit the limit says that if the sequence converges then every subsequence converges, to the same limit. Reversing it needs "every", not "some".

Facts & Assumptions

Given: The alternating sequence $(s_k)$ of reals and the index map $n : \mathbb{N} \to \mathbb{N}$ constructed in FALSE: every bounded sequence converges, namely the unique sequences with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , and $n_0 = 0$ , $n_{\sigma(j)} = \sigma(\sigma(n_j))$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

Established in FALSE: every bounded sequence converges: the map $n$ is strictly increasing; $s_{n_j} = 1$ for every $j$ ; and $(s_k)$ does not converge.

[L2]

A constant sequence converges to its value (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L3]

Subsequences are the composites along strictly increasing index maps (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L4]

Every subsequence of a convergent sequence converges to the same limit (Subsequences inherit the limit), and a sequence is a subsequence of itself along the identity index map, which is strictly increasing (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

Refutation

technique · direct

1.1

The map $n$ is strictly increasing, so $(s_{n_j})_j$ is a subsequence of $(s_k)$ , and $s_{n_j} = 1$ for every $j$ , so this subsequence is the constant sequence with value $1$ .

L1L3

2.1

A constant sequence converges to its value, so the subsequence $(s_{n_j})_j$ converges to $1$ .

step 1.1L2

3.1

The sequence $(s_k)$ therefore has a convergent subsequence, while $(s_k)$ itself does not converge; the claim is false.

step 2.1L1

4.1

The corrected statement puts "every" where the false claim put "some": a sequence $(x_k)$ of reals converges to $x$ if and only if every subsequence of $(x_k)$ converges to $x$ . The forward direction is [L4]; the backward direction is immediate, because $(x_k)$ is a subsequence of itself along the identity index map, and applying the hypothesis to that subsequence is already the conclusion.

step 3.1L4∎

Remarks

The witness is the same alternating sequence that refutes FALSE: every bounded sequence converges. Its subsequence along the index map $n$ , the even indices, is constant $1$ , and its subsequence along the index map $m$ , the odd indices, is constant $-1$ ; either one alone converges, and it is the disagreement between them that kills convergence of the whole sequence, by the divergence test in Subsequences inherit the limit.
A second repair exists and is not proved on this page: if $(x_k)$ is Cauchy, in the sense of Limits and Cauchy sequences of reals whose other direction is Every convergent sequence is Cauchy, and some subsequence converges to $x$ , then $(x_k)$ converges to $x$ . That is the standard bridge from Cauchy to convergence, and it belongs with the completeness material on the next page of this track, which is not available at this point in the reading order. It is named here only to make clear which extra hypothesis repairs the false claim; nothing above uses it. It is worth adding, so that the reader is not left thinking the repair is unavailable, that for the $\mathbb{R}$ of this library the conclusion is already in hand by a shorter route: The reals are complete gives that a Cauchy sequence of reals converges outright, with no subsequence hypothesis at all, and uniqueness of limits (A sequence has at most one limit) with Subsequences inherit the limit then identifies its limit as $x$ . What the next page supplies is that same conclusion proved from the least-upper-bound property rather than from a construction.
A useful way to remember the asymmetry: a subsequence sees only part of the sequence, so it can only ever certify what happens along the indices it keeps. Convergence is a statement about all indices, and no single subsequence carries that information.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 75 results over 25 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.

Sources

J. K. Hunter, An Introduction to Real Analysis, Ch. 3 (standard reference, not scraped)
CMU 21-269 notes, Compactness — subsequences (standard reference, not scraped)
Subsequence (Wikipedia) (standard reference, not scraped)
Limit of a sequence (Wikipedia) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §6.6 (standard reference, not scraped)