False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-05 (gpt-5.6-sol-codex-subscription)

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FALSE: every bounded sequence converges

Statement

False claim: every bounded sequence of reals converges (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

The implication in the opposite direction is true and is Every convergent sequence is bounded: every convergent sequence is bounded. The claim above asserts the converse. It is refuted by the alternating sequence $(s_k)$ and the index map defined by $n_0 = 0$ and $n_{\sigma(j)} = \sigma(\sigma(n_j))$ : the refutation proves that $n$ is strictly increasing, $s_{n_j} = 1$ for every $j$ , and $(s_k)$ does not converge.

The sequence usually written $x_k = (-1)^k$ is introduced here by recursion (The recursion theorem), as the unique $(s_k)$ with $s_0 = 1$ and $s_{\sigma(k)} = -s_k$ . That is the same sequence under a different presentation, and it is chosen because the three inductions of the refutation read straight off those two recursion equations, whereas the power notation would first have to be unwound into them.

Facts & Assumptions

Given: By the recursion theorem (The recursion theorem) applied to the set $\mathbb{R}$ , the element $1$ and the function $u \mapsto -u$ , there is a unique sequence $(s_k)$ of reals with $s_0 = 1$ and $s_{\sigma(k)} = -s_k$ for every $k$ . Applied to the set $\mathbb{N}$ , the element $0$ and the function $i \mapsto \sigma(\sigma(i))$ , it gives a unique $n : \mathbb{N} \to \mathbb{N}$ with $n_0 = 0$ and $n_{\sigma(j)} = \sigma(\sigma(n_j))$ ; applied to $\mathbb{N}$ , the element $\sigma(0) = 1$ and the same function, it gives a unique $m : \mathbb{N} \to \mathbb{N}$ with $m_0 = \sigma(0)$ and $m_{\sigma(j)} = \sigma(\sigma(m_j))$ (The natural numbers $\mathbb{N}$ (von Neumann), Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

For any set $A$ , any $a \in A$ and any $f : A \to A$ there is a unique $g : \mathbb{N} \to A$ with $g(0) = a$ and $g(\sigma(n)) = f(g(n))$ for every $n \in \mathbb{N}$ (The recursion theorem); and if $P(0)$ holds and $P(n)$ implies $P(\sigma(n))$ for every $n$ , then $P(n)$ holds for every $n \in \mathbb{N}$ (The principle of mathematical induction).

[L2]

Absolute value and field arithmetic: $|-u| = |u|$ (Basic properties of the absolute value); $|1| = 1$ , because $1 > 0$ (The multiplicative identity is positive) and $|v| = v$ whenever $v \ge 0$ by the definition of the absolute value (Order on the reals, Absolute value in an ordered field); and $-(-u) = u$ (Field).

[L3]

Order in $\mathbb{R}$ : $0 < 1$ , sums of positives are positive, and adding a constant preserves the order, so $1 - (-1) = 1 + 1 > 0$ and hence $-1 < 1$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Complete ordered field (least-upper-bound property), Ordered field).

[L4]

Order on $\mathbb{N}$ : $i < \sigma(i)$ for every $i$ , because $\sigma(i) = i + 1$ and $\sigma(i) \ne i$ ; and the order is transitive (Order on the natural numbers, Addition of natural numbers, No natural number equals its own successor, $\le$ is a linear order on $\mathbb{N}$ ).

[L5]

If $n : \mathbb{N} \to \mathbb{N}$ satisfies $n_i < n_{\sigma(i)}$ for every $i \in \mathbb{N}$ , then $n$ is strictly increasing (claim 1 of A strictly increasing index map satisfies $n_k \ge k$ ).

[L6]

Convergence, and the fact that a constant sequence converges to its value (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L7]

Divergence test: a sequence with two subsequences converging to different limits does not converge (Subsequences inherit the limit).

[L8]

$(x_k)$ is bounded if there is $M \in \mathbb{R}$ with $|x_k| \le M$ for every $k \in \mathbb{N}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

Refutation

technique · direct

1.1

By induction, $|s_k| = 1$ for every $k$ : the base case is $|s_0| = |1| = 1$ , and if $|s_k| = 1$ then $|s_{\sigma(k)}| = |-s_k| = |s_k| = 1$ . Hence $|s_k| \le 1$ for every $k$ , so $(s_k)$ is bounded.

L1L2L8

1.2

By induction, $s_{n_j} = 1$ for every $j$ : the base case is $s_{n_0} = s_0 = 1$ , and if $s_{n_j} = 1$ then $s_{n_{\sigma(j)}} = s_{\sigma(\sigma(n_j))} = -s_{\sigma(n_j)} = -(-s_{n_j}) = s_{n_j} = 1$ .

L1L2

1.3

By induction, $s_{m_j} = -1$ for every $j$ : the base case is $s_{m_0} = s_{\sigma(0)} = -s_0 = -1$ , and if $s_{m_j} = -1$ then $s_{m_{\sigma(j)}} = s_{\sigma(\sigma(m_j))} = -(-s_{m_j}) = s_{m_j} = -1$ .

L1L2

1.4

Both $n$ and $m$ are strictly increasing: for every $j$ we have $n_j < \sigma(n_j) < \sigma(\sigma(n_j)) = n_{\sigma(j)}$ by [L4], and likewise $m_j < m_{\sigma(j)}$ , so [L5] applies to each.

L1L4L5

1.5

The reals $1$ and $-1$ are distinct, since $1 - (-1) = 1 + 1 > 0$ .

2.1

By step 1.4 the maps $n$ and $m$ define subsequences $(s_{n_j})_j$ and $(s_{m_j})_j$ of $(s_k)$ ; by steps 1.2 and 1.3 these are the constant sequences $1$ and $-1$ , so they converge to $1$ and to $-1$ respectively, and by step 1.5 those two limits differ.

step 1.2step 1.3step 1.4step 1.5L6L8

3.1

By the divergence test, $(s_k)$ does not converge.

step 2.1L7

4.1

So $(s_k)$ is a bounded sequence of reals that does not converge, by steps 1.1 and 3.1; the claim that every bounded sequence converges is therefore false.

step 1.1step 3.1∎

Remarks

The refutation is self-contained: the witness is constructed by recursion, its boundedness and its two subsequential limits are each proved by induction, and the failure of convergence comes from the divergence test of Subsequences inherit the limit together with uniqueness of limits (A sequence has at most one limit).
What is true in this direction is the Bolzano-Weierstrass theorem: every bounded sequence of reals has a convergent subsequence. That is a genuine theorem and it needs the least-upper-bound property. It is not available at this point in the reading order: it is the subject of the next page of this track, Monotone Sequences, Bolzano-Weierstrass, and Cauchy Completeness, where it is proved. It is named here only to say what the correct statement is; nothing above uses it, and no item available here may be cited for it. The false claim above is what one gets by deleting the word "subsequence" from it.
The error is tempting because boundedness feels like "no room to escape". It is not: boundedness forbids running away, but it does not forbid oscillating forever, and oscillation is exactly what $(s_k)$ does.
The same witness refutes FALSE: a convergent subsequence forces the sequence to converge, and it is the sequence referred to in the remarks of The absolute value is compatible with limits.

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 67 results over 26 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)
Limit of a sequence (Wikipedia) (standard reference, not scraped)
Subsequence (Wikipedia) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §6.4 (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)