CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-08 (gpt-5.6-terra-codex-subscription)

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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.
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A sequence with $\limsup = +\infty$ : the greatest subsequential limit exists only in $\overline{\mathbb{R}}$

Statement refuted

That The limit superior is itself a subsequential limit in $\overline{\mathbb{R}}$ and is the greatest one can be stated inside $\mathbb{R}$ : that for every sequence $(x_k)$ of reals the set $\operatorname{SL}(x)$ of real subsequential limits (Subsequential limit of a real sequence, and the subsequential limit set) has a greatest element and that element is $\limsup_k x_k$ .

The witness below has a nonempty $\operatorname{SL}(x)$ with a greatest element, so the failure is not that the real set is empty: it is that the greatest element of $\operatorname{SL}(x)$ is $0$ while $\limsup_k x_k = +\infty$ . The dominant behaviour of the sequence is invisible to $\operatorname{SL}(x)$ and is recorded only by $\overline{\operatorname{SL}}(x)$ (Convergence in $\overline{\mathbb{R}}$ and the extended subsequential limit set: $L \in \overline{\mathbb{R}}$ is an extended subsequential limit when some subsequence converges to $L$ , or diverges to $L = \pm\infty$ ).

Facts & Assumptions

Given: The alternating sequence $(s_k)$ and the 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 $s \circ o \equiv -1$ ; the canonical naturals $\iota(k) = k \cdot 1_{\mathbb{R}}$ with $\iota(0) = 0$ ; and the sequence $x_k := \iota(k)$ when $s_k = 1$ and $x_k := 0$ when $s_k = -1$ .

[L1]

The alternating sequence: $|s_k| = 1$ , $s_{e_j} = 1$ , $s_{o_j} = -1$ , and $e$ , $o$ are strictly increasing, so $e_j \ge j$ and $o_j \ge j$ (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 $s \circ o \equiv -1$ , A strictly increasing index map satisfies $n_k \ge k$ ).

[L2]

Limit superior in $\overline{\mathbb{R}}$ : existence for every sequence, the tail supremum being the least upper bound of the tail range and $\limsup$ the greatest lower bound of the family of tail suprema (Limit superior and limit inferior of a real sequence as $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$ , The tail suprema of any real sequence are nonincreasing in $\overline{\mathbb{R}}$ , so the limit superior exists for every sequence, Every subset of $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\overline{\mathbb{R}}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$ , Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L3]

The order on $\overline{\mathbb{R}}$ is total, $+\infty$ is greatest, every real is $< +\infty$ and $> -\infty$ , and the order restricts on $\mathbb{R}$ to the order of $\mathbb{R}$ (The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$ , its order, and the arithmetic that is left undefined).

[L4]

Extended subsequential limits and convergence in $\overline{\mathbb{R}}$ ; divergence to $+\infty$ means that for every real $M$ one has $x_k > M$ eventually (Convergence in $\overline{\mathbb{R}}$ and the extended subsequential limit set: $L \in \overline{\mathbb{R}}$ is an extended subsequential limit when some subsequence converges to $L$ , or diverges to $L = \pm\infty$ , Divergence to $+\infty$ and to $-\infty$ , Subsequential limit of a real sequence, and the subsequential limit set, Limits and Cauchy sequences of reals).

[L5]

Canonical naturals: $\iota$ is strictly increasing with $\iota(k) \ge 0$ , and for every real $M$ there is a natural $p \ge 1$ with $M < \iota(p)$ (Canonical naturals are positive and strictly increasing, Every complete ordered field is Archimedean).

[L6]

A convergent sequence of reals is bounded, a limit is unique, and a sequence agreeing with a constant from some index on converges to that constant (Every convergent sequence is bounded, A sequence has at most one limit, Convergence depends only on the tail, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L7]

Absolute value and order: $|t| = 1$ forces $t = 1$ or $t = -1$ ; $0 < 1$ ; the order on $\mathbb{N}$ is total (Basic properties of the absolute value, Absolute value in an ordered field, The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Order on the natural numbers, $\le$ is a linear order on $\mathbb{N}$ , Ordered field, Complete ordered field (least-upper-bound property)).

Counterexample

technique · direct

1.1

Each $s_k$ is $1$ or $-1$ , so $(x_k)$ is a well-defined sequence of reals with $x_k \ge 0$ for every $k$ ; moreover $x_{e_j} = \iota(e_j)$ and $x_{o_j} = 0$ for every $j$ .

givenL1L5L7

1.2

The subsequence along $o$ is constantly $0$ , and $o$ is strictly increasing, so $0 \in \operatorname{SL}(x)$ .

givenL1L4L6

2.1

For every $n \in \mathbb{N}$ the tail supremum $\sup T_n(x)$ is $+\infty$ . Given a real $M$ , take a natural $p \ge 1$ with $M < \iota(p)$ and an index $j$ at least as large as both $n$ and $p$ ; then $e_j \ge j \ge n$ , so $x_{e_j} = \iota(e_j) \in T_n(x)$ , and $e_j \ge j \ge p$ gives $\iota(e_j) \ge \iota(p) > M$ . So no real number bounds $T_n(x)$ above, and the least upper bound in $\overline{\mathbb{R}}$ must be $+\infty$ .

step 1.1L1L2L3L5L7

3.1

Every real subsequential limit of $(x_k)$ equals $0$ . Let $n$ be strictly increasing with $x_{n_i} \to L \in \mathbb{R}$ ; the subsequence is then bounded, say $|x_{n_i}| \le B$ for every $i$ . Suppose $s_{n_i} = 1$ for arbitrarily large $i$ : taking a natural $p \ge 1$ with $B < \iota(p)$ and such an index $i \ge p$ , we get $x_{n_i} = \iota(n_i) \ge \iota(i) \ge \iota(p) > B$ , contradicting the bound. So there is $I$ with $s_{n_i} = -1$ , hence $x_{n_i} = 0$ , for every $i \ge I$ ; a sequence equal to $0$ from an index on converges to $0$ , so $L = 0$ by uniqueness of limits.

step 1.1step 2.1L1L4L5L6L7

4.1

Consequently $\limsup_k x_k$ is the greatest lower bound of the family $\{+\infty\}$ , namely $+\infty$ , while $\operatorname{SL}(x) = \{0\}$ by steps 1.2 and 3.1, whose greatest element is the real number $0$ . Since $0 \ne +\infty$ , the refuted claim fails for this sequence.

step 2.1step 1.2step 3.1L2L3∎

Remarks

What the extended set records. By The limit superior is itself a subsequential limit in $\overline{\mathbb{R}}$ and is the greatest one the element $\limsup_k x_k = +\infty$ lies in $\overline{\operatorname{SL}}(x)$ and is its greatest element, so $\overline{\operatorname{SL}}(x) = \{0, +\infty\}$ : the value $-\infty$ is excluded because $x_k \ge 0$ for every $k$ , so no subsequence can be eventually below a negative real. The real set $\operatorname{SL}(x) = \{0\}$ is exactly the finite part of it, as Convergence in $\overline{\mathbb{R}}$ and the extended subsequential limit set: $L \in \overline{\mathbb{R}}$ is an extended subsequential limit when some subsequence converges to $L$ , or diverges to $L = \pm\infty$ says it must be.
Why the theorem cannot simply be restricted to bounded sequences. For a bounded sequence $\limsup$ is real and the two statements agree; the point of stating The limit superior is itself a subsequential limit in $\overline{\mathbb{R}}$ and is the greatest one in $\overline{\mathbb{R}}$ is that it then holds for every sequence, with no hypothesis to check, and this witness shows the hypothesis-free version is strictly stronger.
A simpler witness would prove less. The sequence $x_k = \iota(k)$ has $\operatorname{SL}(x) = \emptyset$ , so the refuted claim fails there only because an empty set has no greatest element. Interleaving with $0$ makes $\operatorname{SL}(x)$ nonempty with a greatest element, so the claim fails for the substantive reason: the greatest real subsequential limit is not the limit superior.

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Sources

Limit superior and limit inferior (Wikipedia) (standard reference, not scraped)
Subsequential limit (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)

A sequence with lim sup⁡=+∞\limsup = +\inftylimsup=+∞: the greatest subsequential limit exists only in R‾\overline{\mathbb{R}}R