CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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$x_{k} = (- 1)^{k}$ , $y_{k} = (- 1)^{k + 1}$ 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 · two levels

42 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on 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​