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

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$x_{k} = (- 1)^{k}$ , $y_{k} = (- 1)^{k + 1}$ give $lim sup (x_{k} + y_{k}) = 0 < 2 = lim sup x_{k} + lim sup y_{k}$

Statement refuted

That the subadditivity of $lim sup (x_{k} + y_{k}) \leq lim sup x_{k} + lim sup y_{k}$ whenever the right-hand side is defined in $\overline{R}$ , and dually for $lim inf$ can be improved to an equality: that for all sequences $(x_{k})$ , $(y_{k})$ of reals whose limit superiors have a defined sum, ${lim sup}_{k} (x_{k} + y_{k}) = {lim sup}_{k} x_{k} + {lim sup}_{k} y_{k}$ . The claim is recorded and refuted as FALSE: $lim sup (x_{k} + y_{k}) = lim sup x_{k} + lim sup y_{k}$ ; this item is the named witness.

Facts & Assumptions

Given: The alternating sequence $(s_{k})$ 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 sequences $x_{k} : = s_{k}$ and $y_{k} : = - s_{k}$ , which are the families usually written $(- 1)^{k}$ and $(- 1)^{k + 1}$ .

[L1]

For this pair of sequences: $x_{k} + y_{k} = 0$ for every $k$ , ${lim sup}_{k} x_{k} = 1$ , ${lim sup}_{k} y_{k} = 1$ and ${lim sup}_{k} (x_{k} + y_{k}) = 0$ (FALSE: $lim sup (x_{k} + y_{k}) = lim sup x_{k} + lim sup y_{k}$ ).

[L2]

Subadditivity: ${lim sup}_{k} (x_{k} + y_{k}) \leq {lim sup}_{k} x_{k} + {lim sup}_{k} y_{k}$ whenever the right-hand side is defined ( $lim sup (x_{k} + y_{k}) \leq lim sup x_{k} + lim sup y_{k}$ whenever the right-hand side is defined in $\overline{R}$ , and dually for $lim inf$ ).

[L3]

Limit superior in $\overline{R}$ , and the fact that a sum of two real numbers is the field sum (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 extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L4]

Order arithmetic: $0 < 1$ , hence $0 < 1 + 1 = 2$ and $0 \neq 2$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Ordered field, Complete ordered field (least-upper-bound property)).

Counterexample

technique · direct

1.1

The two sequences $x_{k} = s_{k}$ and $y_{k} = - s_{k}$ are sequences of reals whose termwise sum is constantly $0$ , and their limit superiors are both the real number $1$ .

givenL1L3

1.2

Both limit superiors being real, their sum is defined in $\overline{R}$ and equals the field sum $1 + 1 = 2$ .

givenL1L3L4

2.1

The limit superior of the sum sequence is $0$ , while the sum of the limit superiors is $2$ , and $0 < 2$ ; so the inequality of [L2] holds here and is strict, and the equality asserted by the refuted claim fails.

step 1.1step 1.2L1L2L4∎

Remarks

The gap is as large as the oscillation. Each sequence oscillates between $- 1$ and $1$ , and the two oscillations are exactly out of phase, so the sum never sees either peak. Replacing $y$ by a sequence in phase with $x$ , for instance $y = x$ , restores equality; the failure is a statement about the relative phase, not about the sizes involved.
Both hypotheses of $lim sup (x_{k} + y_{k}) \leq lim sup x_{k} + lim sup y_{k}$ whenever the right-hand side is defined in $\overline{R}$ , and dually for $lim inf$ hold here. The two limit superiors are real, so their sum is defined, and no boundedness is needed for that theorem at all. The counterexample therefore refutes the equality on the theorem's own terms, not by exploiting a degenerate case.
The dual statement fails in the dual way. For the same pair, ${lim inf}_{k} x_{k} = {lim inf}_{k} y_{k} = - 1$ while ${lim inf}_{k} (x_{k} + y_{k}) = 0$ , so the $lim inf$ inequality ${lim inf}_{k} (x_{k} + y_{k}) \geq {lim inf}_{k} x_{k} + {lim inf}_{k} y_{k}$ of $lim sup (x_{k} + y_{k}) \leq lim sup x_{k} + lim sup y_{k}$ whenever the right-hand side is defined in $\overline{R}$ , and dually for $lim inf$ is also strict here, reading $0 \geq - 2$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 71 results over 22 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

Limit superior and limit inferior (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)
N. Donaldson, Math 140A: Real Analysis notes (standard reference, not scraped)

xk=(−1)kxk​=(−1)k, yk=(−1)k+1yk​=(−1)k+1 give lim sup⁡(xk+yk)=0<2=lim sup⁡xk+lim sup⁡yklimsup(xk​+yk​)=0<2=limsupxk​+limsupyk​