False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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FALSE: $lim sup (x_{k} + y_{k}) = lim sup x_{k} + lim sup y_{k}$

Statement

False claim: for all sequences $(x_{k})$ , $(y_{k})$ of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) whose limit superiors have a defined sum in $\overline{R}$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined),

${lim sup}_{k} (x_{k} + y_{k}) = {lim sup}_{k} x_{k} + {lim sup}_{k} y_{k} .$

The corresponding statement with $=$ replaced by $\leq$ is true and is $lim sup (x_{k} + y_{k}) \leq lim sup x_{k} + lim sup y_{k}$ whenever the right-hand side is defined in $\overline{R}$ , and dually for $lim inf$ . The claim above is what one gets by strengthening that inequality to an equality, and it fails: the two sides can differ by as much as the whole oscillation of the sequences, because the two limit superiors may be attained along different sets of indices while the sum of the sequences never sees either of them.

The witness is $x_{k} = (- 1)^{k}$ and $y_{k} = - (- 1)^{k}$ , refuted below; it is recorded separately as a named counterexample on the companion page.

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 sequences $x_{k} : = s_{k}$ and $y_{k} : = - s_{k}$ ; and the tail ranges and extended tail bounds of 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 $\overline{R}$ .

[L1]

The alternating sequence: $∣ s_{k} ∣ = 1$ for every $k$ , $s_{e_{j}} = 1$ and $s_{o_{j}} = - 1$ for every $j$ , and $e$ , $o$ are strictly increasing (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 $s \circ o \equiv - 1$ ).

[L2]

A strictly increasing index map satisfies $n_{j} \geq j$ (A strictly increasing index map satisfies $n_{k} \geq k$ ).

[L3]

Limit superior: ${lim sup}_{k} z_{k} = \inf {\sup T_{n} (z)}$ with $T_{n} (z) = {z_{k} : k \geq n}$ ; all these bounds exist in $\overline{R}$ , $\sup$ being the least upper bound and $\inf$ the greatest lower bound (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).

[L4]

The order on $\overline{R}$ is total and restricts on $R$ to the order of $R$ ; $\pm \infty$ are not real (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

[L5]

Absolute value: $∣ t ∣ = 1$ forces $t = 1$ or $t = - 1$ (Basic properties of the absolute value, Absolute value in an ordered field).

[L6]

Order arithmetic: $0 < 1$ , so $- 1 < 1$ and $0 < 1 + 1 = 2$ ; in particular $0 \neq 2$ (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)).

[L7]

Subadditivity: ${lim sup}_{k} (z_{k} + w_{k}) \leq {lim sup}_{k} z_{k} + {lim sup}_{k} w_{k}$ 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 $lim inf$ ).

[L8]

The refuted claim: for all sequences whose limit superiors have a defined sum, ${lim sup}_{k} (z_{k} + w_{k}) = {lim sup}_{k} z_{k} + {lim sup}_{k} w_{k}$ .

Refutation

technique · direct

1.1

The sequences $x_{k} = s_{k}$ and $y_{k} = - s_{k}$ are sequences of reals, and $x_{k} + y_{k} = s_{k} + (- s_{k}) = 0$ for every $k$ .

givenL1

1.2

Every value $s_{k}$ is $1$ or $- 1$ , since $∣ s_{k} ∣ = 1$ ; and for every $n$ both values occur at an index $\geq n$ , since $s_{e_{n}} = 1$ with $e_{n} \geq n$ and $s_{o_{n}} = - 1$ with $o_{n} \geq n$ .

givenL1L2L5

2.1

Hence $T_{n} (x) = {1, - 1}$ for every $n$ . Its least upper bound in $\overline{R}$ is $1$ : the element $1$ bounds both $1$ and $- 1$ from above because $- 1 < 1$ , and any upper bound $u$ satisfies $1 \leq u$ because $1 \in T_{n} (x)$ . So $\sup T_{n} (x) = 1$ for every $n$ , and ${lim sup}_{k} x_{k}$ is the greatest lower bound of the one-element family ${1}$ , namely $1$ .

step 1.2L3L4L6

2.2

The sequence $y_{k} = - s_{k}$ takes the value $1$ at every $o_{n}$ and the value $- 1$ at every $e_{n}$ , and takes no other value, so $T_{n} (y) = {1, - 1}$ for every $n$ as well, and the same computation gives ${lim sup}_{k} y_{k} = 1$ .

step 1.2L1L2L3L4L5L6

2.3

The sum sequence is constantly $0$ , so $T_{n} (x + y) = {0}$ , whose least upper bound is $0$ , and ${lim sup}_{k} (x_{k} + y_{k}) = 0$ .

step 1.1L3L4

3.1

Both limit superiors are the real number $1$ , so their sum is defined and equals $1 + 1 = 2$ , and the claim asserts $0 = 2$ for this pair. But $0 < 2$ , so $0 \neq 2$ and the claim fails.

step 2.1step 2.2step 2.3L6L8

4.1

The claim is therefore false. What survives is the inequality of [L7], which for this pair reads $0 \leq 2$ and is strict.

step 3.1L7L8∎

Remarks

Which half of the equality fails. Only $\geq$ ; the inequality $\leq$ is a theorem ( $lim sup (x_{k} + y_{k}) \leq lim sup x_{k} + lim sup y_{k}$ whenever the right-hand side is defined in $\overline{R}$ , and dually for $lim inf$ ). So the claim is not wrong by accident of the witness: the reverse inequality has no proof, and this pair shows it has no proof because it is false.
The mechanism is a mismatch of index sets. ${lim sup}_{k} x_{k} = 1$ is achieved along the even indices and ${lim sup}_{k} y_{k} = 1$ along the odd ones. The sum can only see both at once if some index is large in both sequences simultaneously, and here no index is: where $x_{k} = 1$ one has $y_{k} = - 1$ . Equality does hold when one of the two sequences converges, because then its limit superior is achieved along every subsequence.
The witness is named on the companion page as $x_{k} = (- 1)^{k}$ , $y_{k} = (- 1)^{k + 1}$ give $lim sup (x_{k} + y_{k}) = 0 < 2 = lim sup x_{k} + lim sup y_{k}$ ↗, which quotes the computation made here.
The defined-sum hypothesis is inherited from $lim sup (x_{k} + y_{k}) \leq lim sup x_{k} + lim sup y_{k}$ whenever the right-hand side is defined in $\overline{R}$ , and dually for $lim inf$ and is not what fails here. Both limit superiors in the witness are real, so the right-hand side is a perfectly good real number; the equality is false anyway.

Depends on

Used by

xₖ = (-1)ᵏ, yₖ = (-1)ᵏ⁺¹ give limsup(xₖ + yₖ) = 0 < 2 = limsup xₖ + limsup yₖ Counterexample

Dependency tree · two levels

46 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)

FALSE: lim sup⁡(xk+yk)=lim sup⁡xk+lim sup⁡yklimsup(xk​+yk​)=limsupxk​+limsupyk​