False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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FALSE: a sequence with a convergent subsequence is bounded (the converse of Bolzano-Weierstrass)

Statement

False claim: if a sequence $(y_n)$ of reals has a convergent subsequence, then $(y_n)$ is bounded (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Subsequential limit of a real sequence, and the subsequential limit set).

This is the converse of Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence, which says that boundedness implies the existence of a convergent subsequence. The implication does not reverse, and it fails as badly as it can: a sequence can be unbounded and still have a constant subsequence.

The witness is the interleaving $1, 1, 2, 1, 3, 1, 4, \dots$ , in which the terms at even indices run through $1, 2, 3, \dots$ and every odd-indexed term is $1$ . It is recorded separately as the named counterexample of the companion page. The even and odd index maps are supplied by 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 $s \circ o \equiv -1$ , which also supplies what makes the definition legitimate: every natural number is an even index or an odd index, and never both.

Facts & Assumptions

Given: The strictly increasing index maps $e, o : \mathbb{N} \to \mathbb{N}$ 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 defined by cases on that partition: $y_n := (j+1) \cdot 1_{\mathbb{R}}$ when $n = e_j$ , and $y_n := 1$ when $n = o_j$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

The index maps: $e$ and $o$ are strictly increasing, and every natural number is $e_j$ for exactly one $j$ , or $o_j$ for exactly one $j$ , and never both (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 $s \circ o \equiv -1$ ).

[L2]

Canonical naturals: $m \cdot 1_{\mathbb{R}} > 0$ for $m \ge 1$ , and $m \mapsto m \cdot 1_{\mathbb{R}}$ is strictly increasing (Canonical naturals are positive and strictly increasing).

[L3]

Archimedean property: for every real $x$ there is a natural $m \ge 1$ with $x < m \cdot 1_{\mathbb{R}}$ (Every complete ordered field is Archimedean).

[L4]

Absolute value: $|t| \ge t$ always, and $|t| = t$ when $t \ge 0$ (Basic properties of the absolute value).

[L5]

A constant sequence converges to its value, and a sequence is bounded when some real $M$ satisfies $|y_n| \le M$ at every index (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

[L6]

Subsequences and subsequential limits: for strictly increasing $n$ , $(y_{n_j})$ is a subsequence, and its limit is a subsequential limit of $(y_n)$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Subsequential limit of a real sequence, and the subsequential limit set).

[L7]

Trichotomy of the order on $\mathbb{R}$ (Complete ordered field (least-upper-bound property), Ordered field).

[L8]

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

Refutation

technique · direct

1.1

The sequence $(y_n)$ is well defined: by [L1] each $n \in \mathbb{N}$ falls under exactly one of the two clauses, and the index $j$ realising it is unique, so exactly one value is assigned to each $n$ .

givenL1

2.1

The subsequence along $o$ is the constant sequence with value $1$ : for every $j$ , $y_{o_j} = 1$ by the second clause. Since $o$ is strictly increasing, this is a subsequence of $(y_n)$ .

step 1.1L1L6

2.2

The subsequence along $e$ takes the value $y_{e_j} = (j+1)\cdot 1_{\mathbb{R}}$ for every $j$ .

step 1.1L1

3.1

The constant subsequence $(y_{o_j})$ converges, to $1$ , so $(y_n)$ has a convergent subsequence and $1$ is a subsequential limit of it: $(y_n)$ satisfies the hypothesis of the claim.

step 2.1L5L6L8

3.2

$(y_n)$ is not bounded. Let $M \in \mathbb{R}$ be arbitrary. By [L3] fix a natural $m \ge 1$ with $|M| < m \cdot 1_{\mathbb{R}}$ , and take $j := m - 1 \in \mathbb{N}$ , which is legitimate since $m \ge 1$ . Then $y_{e_j} = m \cdot 1_{\mathbb{R}} > |M| \ge M$ , and $y_{e_j} > 0$ gives $|y_{e_j}| = y_{e_j} > M$ . So no real $M$ satisfies $|y_n| \le M$ at every index.

step 2.2L2L3L4L5L7

4.1

The sequence $(y_n)$ therefore has a convergent subsequence and is unbounded: the claim is false.

step 3.1step 3.2L8∎

Remarks

What survives is exactly Bolzano-Weierstrass in the stated direction. Boundedness gives a convergent subsequence (Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence); a convergent subsequence gives nothing about the sequence. The correct strengthening on the other side is not boundedness at all but a Cauchy hypothesis: a Cauchy sequence with a convergent subsequence does converge, and is bounded, by A Cauchy sequence with a convergent subsequence converges, to that subsequence’s limit and Every Cauchy sequence of reals is bounded.
One subsequence is never evidence about a sequence. The same point in a different form is FALSE: a convergent subsequence forces the sequence to converge on the previous page: a convergent subsequence does not force convergence. Here it does not even force boundedness, which is weaker, so this is the sharper failure of the two.
The witness is as extreme as possible in one direction and as tame as possible in the other. Its subsequential limit set is exactly $\{1\}$ , a single point, while the sequence itself is unbounded; so having a one-point subsequential limit set does not imply convergence either, and Subsequential limit of a real sequence, and the subsequential limit set records that consequence.
The witness is recorded as the named counterexample The sequence $1, 1, 2, 1, 3, 1, 4, \dots$ is unbounded and has a convergent subsequence ↗, which also computes its subsequential limit set.

Depends on

Used by

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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 58 results over 15 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

Bolzano-Weierstrass theorem (Wikipedia) (standard reference, not scraped)
Subsequence (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §6.6 (standard reference, not scraped)