CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-06 (claude-sonnet-5)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The sequence $1, 1, 2, 1, 3, 1, 4, \dots$ is unbounded and has a convergent subsequence

Statement refuted

Refuted claim: a sequence of reals with a convergent subsequence is bounded, which is the converse of Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence (FALSE: a sequence with a convergent subsequence is bounded (the converse of Bolzano-Weierstrass), Sequences of reals: bounded, eventually, frequently, tails, subsequences).

The witness is the interleaving

$1,\; 1,\; 2,\; 1,\; 3,\; 1,\; 4,\; \dots$

whose terms at even indices are $1, 2, 3, \dots$ and whose terms at odd indices are all $1$ . It is unbounded, and its odd-indexed subsequence is constant, hence convergent. The refutation is carried out in full in FALSE: a sequence with a convergent subsequence is bounded (the converse of Bolzano-Weierstrass); this item records the witness and adds the computation of its subsequential limit set.

Facts & Assumptions

Given: The strictly increasing index maps $e, o$ of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , which satisfies $|s_k| = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv -1$ , whose ranges partition $\mathbb{N}$ , and the sequence $(y_n)$ of reals with $y_n := (j+1)\cdot 1_{\mathbb{R}}$ when $n = e_j$ and $y_n := 1$ when $n = o_j$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

The witness: $(y_n)$ is well defined, the subsequence $(y_{o_j})$ is constantly $1$ and converges to $1$ , and $(y_n)$ is unbounded (FALSE: a sequence with a convergent subsequence is bounded (the converse of Bolzano-Weierstrass)).

[L2]

The index maps are strictly increasing and their ranges partition $\mathbb{N}$ (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , which satisfies $|s_k| = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv -1$ ).

[L3]

Canonical naturals: positive for $m \ge 1$ and strictly increasing in the index (Canonical naturals are positive and strictly increasing); the Archimedean property (Every complete ordered field is Archimedean).

[L4]

Boundedness of a sequence and of a subset of $\mathbb{R}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Lower bound, bounded below, bounded set); a constant sequence converges to its value (Limits and Cauchy sequences of reals); and every convergent sequence of reals is bounded (Every convergent sequence is bounded).

[L5]

Subsequential limits (Subsequential limit of a real sequence, and the subsequential limit set).

[L6]

Absolute value: $|t| = t$ for $t \ge 0$ (Basic properties of the absolute value); trichotomy of the order (Complete ordered field (least-upper-bound property), Ordered field).

[L7]

Bolzano-Weierstrass: every bounded sequence of reals has a convergent subsequence (Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence).

[L8]

The refuted claim: a sequence of reals with a convergent subsequence is bounded.

Counterexample

technique · direct

1.1

$(y_n)$ has a convergent subsequence, namely the constant subsequence along $o$ , which converges to $1$ ; so it satisfies the hypothesis of the claim.

givenL1L2L4L5L8

1.2

$(y_n)$ is unbounded: no real $M$ satisfies $|y_n| \le M$ at every index.

givenL1L3L4L6

2.1

The claim is therefore refuted: a convergent subsequence does not force boundedness, and the converse of Bolzano-Weierstrass fails.

step 1.1step 1.2L7L8

3.1

The subsequential limit set of $(y_n)$ is exactly $\{1\}$ . It contains $1$ by step 1.1. Conversely, let $(y_{n_i})$ converge; a convergent sequence is bounded, and along the even indices the values $y_{e_j} = j+1$ exceed every bound, so only finitely many $n_i$ can lie in the range of $e$ ; all later $n_i$ lie in the range of $o$ , where the value is $1$ , so the subsequence is eventually constantly $1$ and its limit is $1$ .

step 1.1step 1.2L2L3L4L5∎

Remarks

A one-point subsequential limit set does not imply convergence. By step 3.1 the set is $\{1\}$ , yet the sequence is unbounded and hence divergent. What is true is the converse implication: a convergent sequence has exactly one subsequential limit (Subsequential limit of a real sequence, and the subsequential limit set). So the subsequential limit set being a single point is necessary and not sufficient for convergence, and the missing hypothesis is boundedness.
The interleaving is what makes the example work, and it needs the partition. Defining $(y_n)$ by cases on whether $n$ is even or odd is legitimate only because every natural number is exactly one of the two (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , which satisfies $|s_k| = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv -1$ ). That is the same fact that makes the two subsequences between them exhaust the sequence.
Compare the bounded case. With the unbounded branch replaced by a second constant $c \ne 1$ , the same interleaving is bounded and divergent, its two subsequential limits being $1$ and $c$ . Bolzano-Weierstrass applies there and produces one of the two; here it does not apply at all, and the conclusion nevertheless happens to hold, which is exactly why the converse is not a theorem. A different bounded sequence with two subsequential limits is worked out in full at The sequence $(-1)^k(1 + 1/k)$ is bounded with subsequential limit set exactly $\{-1, 1\}$ , which reaches its two limits by an alternating sign carrying a null perturbation rather than by interleaving constants, so that neither limit is a value of the sequence.

Depends on

Used by

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Sources

Bolzano-Weierstrass theorem (Wikipedia) (standard reference, not scraped)
Subsequence (Wikipedia) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §6.6 (standard reference, not scraped)

The sequence 1,1,2,1,3,1,4,…1, 1, 2, 1, 3, 1, 4, \dots1,1,2,1,3,1,4,… is unbounded and has a convergent subsequence