CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck 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.
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A sequence with $lim sup = + \infty$ : the greatest subsequential limit exists only in $\overline{R}$

Statement refuted

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

The witness below has a nonempty $SL (x)$ with a greatest element, so the failure is not that the real set is empty: it is that the greatest element of $SL (x)$ is $0$ while ${lim sup}_{k} x_{k} = + \infty$ . The dominant behaviour of the sequence is invisible to $SL (x)$ and is recorded only by $\overset{}{SL} (x)$ (Convergence in $\overline{R}$ and the extended subsequential limit set: $L \in \overline{R}$ 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 $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ; the canonical naturals $ι (k) = k \cdot 1_{R}$ with $ι (0) = 0$ ; and the sequence $x_{k} : = ι (k)$ 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 $e_{j} \geq j$ and $o_{j} \geq j$ (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ , A strictly increasing index map satisfies $n_{k} \geq k$ ).

[L2]

Limit superior in $\overline{R}$ : existence for every sequence, the tail supremum being the least upper bound of the tail range and $lim sup$ the greatest lower bound of the family of tail suprema (Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ , The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence, Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ , 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 $\overline{R}$ is total, $+ \infty$ is greatest, every real is $< + \infty$ and $> - \infty$ , and the order restricts on $R$ to the order of $R$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

[L4]

Extended subsequential limits and convergence in $\overline{R}$ ; divergence to $+ \infty$ means that for every real $M$ one has $x_{k} > M$ eventually (Convergence in $\overline{R}$ and the extended subsequential limit set: $L \in \overline{R}$ 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: $ι$ is strictly increasing with $ι (k) \geq 0$ , and for every real $M$ there is a natural $p \geq 1$ with $M < ι (p)$ (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 $N$ 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, $\leq$ is a linear order on $N$ , 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 $x_{k} \geq 0$ for every $k$ ; moreover $x_{e_{j}} = ι (e_{j})$ and $x_{o_{j}} = 0$ for every $j$ .

givenL1L5L7

1.2

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

givenL1L4L6

2.1

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

step 1.1L1L2L3L5L7

3.1

Every real subsequential limit of $(x_{k})$ equals $0$ . Let $n$ be strictly increasing with $x_{n_{i}} \to L \in R$ ; the subsequence is then bounded, say $∣ x_{n_{i}} ∣ \leq B$ for every $i$ . Suppose $s_{n_{i}} = 1$ for arbitrarily large $i$ : taking a natural $p \geq 1$ with $B < ι (p)$ and such an index $i \geq p$ , we get $x_{n_{i}} = ι (n_{i}) \geq ι (i) \geq ι (p) > B$ , contradicting the bound. So there is $I$ with $s_{n_{i}} = - 1$ , hence $x_{n_{i}} = 0$ , for every $i \geq I$ ; 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 ${lim sup}_{k} x_{k}$ is the greatest lower bound of the family ${+ \infty}$ , namely $+ \infty$ , while $SL (x) = {0}$ by steps 1.2 and 3.1, whose greatest element is the real number $0$ . Since $0 \neq + \infty$ , 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 $\overline{R}$ and is the greatest one the element ${lim sup}_{k} x_{k} = + \infty$ lies in $\overset{}{SL} (x)$ and is its greatest element, so $\overset{}{SL} (x) = {0, + \infty}$ : the value $- \infty$ is excluded because $x_{k} \geq 0$ for every $k$ , so no subsequence can be eventually below a negative real. The real set $SL (x) = {0}$ is exactly the finite part of it, as Convergence in $\overline{R}$ and the extended subsequential limit set: $L \in \overline{R}$ 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 $lim sup$ is real and the two statements agree; the point of stating The limit superior is itself a subsequential limit in $\overline{R}$ and is the greatest one in $\overline{R}$ 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 $x_{k} = ι (k)$ has $SL (x) = \emptyset$ , so the refuted claim fails there only because an empty set has no greatest element. Interleaving with $0$ makes $SL (x)$ nonempty with a greatest element, so the claim fails for the substantive reason: the greatest real subsequential limit is not the limit superior.

Depends on

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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=+∞: the greatest subsequential limit exists only in R‾R