TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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

Statement

Let $(x_{k})$ and $(y_{k})$ be sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) and write $Λ : = {lim sup}_{k} x_{k}$ , $M : = {lim sup}_{k} y_{k}$ (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 $\overline{R}$ ).

If the sum $Λ + M$ is defined in $\overline{R}$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined), that is if ${Λ, M} \neq {+ \infty, - \infty}$ , then ${lim sup}_{k} (x_{k} + y_{k}) \leq Λ + M .$
Dually, writing $λ : = {lim inf}_{k} x_{k}$ and $μ : = {lim inf}_{k} y_{k}$ , if $λ + μ$ is defined in $\overline{R}$ then ${lim inf}_{k} (x_{k} + y_{k}) \geq λ + μ .$

The hypothesis is exactly the one The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined forces, and it cannot be dropped. When one of $Λ$ , $M$ is $+ \infty$ and the other $- \infty$ the right-hand side is not an element of $\overline{R}$ at all, so there is nothing to compare. The inequality is genuinely an inequality: equality can fail, and does, for an alternating pair of sequences; the failure of additivity is recorded as a false statement among this page's examples, and the witness is a named counterexample on the companion page.

Facts & Assumptions

Given: Sequences $(x_{k})$ and $(y_{k})$ of reals, their termwise sum $(x_{k} + y_{k})$ , and $Λ : = {lim sup}_{k} x_{k}$ , $M : = {lim sup}_{k} y_{k}$ , assumed to have a sum defined in $\overline{R}$ .

[L1]

Tail bounds and the two quantities exist in $\overline{R}$ for every sequence, $s_{n}$ being the least upper bound of the $n$ -th tail range and $lim sup$ the greatest lower bound of ${s_{n}}$ (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 $R$ , 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 $\overline{R}$ , Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set).

[L2]

The order on $\overline{R}$ is total and transitive, $+ \infty$ is its greatest element and $- \infty$ its least, and it 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, Partial order and partially ordered set).

[L3]

Partial addition on $\overline{R}$ : a sum is undefined only for the pairs $(+ \infty, - \infty)$ and $(- \infty, + \infty)$ ; a sum with one summand $+ \infty$ and the other $\neq - \infty$ is $+ \infty$ ; a sum with one summand $- \infty$ and the other $\neq + \infty$ is $- \infty$ ; and $- (a + b) = (- a) + (- b)$ , each side defined exactly when the other is (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

[L4]

Epsilon characterisation for a real limit superior: $Λ = {lim sup}_{k} x_{k}$ real implies that for every real $ε > 0$ one has $x_{k} < Λ + ε$ eventually (For finite $L$ : $L = lim sup x_{k}$ iff for every $ε > 0$ one has $x_{k} < L + ε$ eventually and $x_{k} > L - ε$ frequently).

[L5]

${lim sup}_{k} z_{k} = - \infty$ implies $z_{k} \to - \infty$ ; and $z_{k} \to - \infty$ implies ${lim sup}_{k} z_{k} = - \infty$ ( $lim inf x_{k} \leq lim sup x_{k}$ for every real sequence, A real sequence converges to $L \in R$ iff $lim inf x_{k} = lim sup x_{k} = L$ , and diverges to $\pm \infty$ iff both equal $\pm \infty$ , Divergence to $+ \infty$ and to $- \infty$ ).

[L6]

Reflection: ${lim sup}_{k} (- z_{k}) = - {lim inf}_{k} z_{k}$ and ${lim inf}_{k} (- z_{k}) = - {lim sup}_{k} z_{k}$ ( $lim sup (- x_{k}) = - lim inf (x_{k})$ , with the reflection of $\overline{R}$ exchanging $\pm \infty$ ).

[L7]

Order arithmetic in $R$ : Order is preserved by adding a constant and by adding inequalities states the strict forms, that inequalities may be translated and added, so $a < a^{'}$ and $b < b^{'}$ give $a + b < a^{'} + b^{'}$ ; adjoining the case of equality, in which both sides move by the same amount, gives the nonstrict forms used below. In particular $a \leq b$ if and only if $- b \leq - a$ : translation by $- a - b$ turns $a < b$ into $- b < - a$ and back, while $a = b$ holds exactly when $- a = - b$ .

[L8]

Reciprocal Archimedean property and canonical naturals: for every real $δ > 0$ there is a natural $m \geq 1$ with $1 / m < δ$ ; for a natural $m \geq 1$ the element $2 m$ is a natural $\geq 1$ with $(2 m) \cdot 1_{R} = 2 (m \cdot 1_{R}) > 0$ , so $1 / (2 m) > 0$ and $1 / (2 m) + 1 / (2 m) = 1 / m$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order, Canonical naturals are positive and strictly increasing).

[L9]

Two properties each holding eventually hold together from the larger of the two thresholds on (Sequences of reals: bounded, eventually, frequently, tails, subsequences, $\leq$ is a linear order on $N$ , Order on the natural numbers).

Proof

technique · direct

1.1

Since $Λ + M$ is defined, exactly one of the following three situations holds: at least one of $Λ$ , $M$ equals $+ \infty$ , and then the other is $\neq - \infty$ ; both are real; or neither equals $+ \infty$ and at least one equals $- \infty$ . Both the hypothesis and the conclusion of claim 1 are unchanged by exchanging the two sequences, so in the third situation it may be assumed that $Λ = - \infty$ .

givenL2L3

2.1

In the first situation $Λ + M = + \infty$ by the addition table, and every element of $\overline{R}$ is $\leq + \infty$ , so ${lim sup}_{k} (x_{k} + y_{k}) \leq Λ + M$ .

step 1.1L2L3

2.2

In the second situation let $δ > 0$ be an arbitrary real, take a natural $m \geq 1$ with $1 / m < δ$ and put $ε : = 1 / (2 m) > 0$ , so that $ε + ε = 1 / m < δ$ . By [L4] there are thresholds beyond which $x_{k} < Λ + ε$ and beyond which $y_{k} < M + ε$ ; beyond the larger of them both hold, so adding the two inequalities gives $x_{k} + y_{k} < Λ + M + ε + ε$ for all $k \geq N$ , where $N$ is that larger threshold. Hence $Λ + M + ε + ε$ is an upper bound of the $N$ -th tail range of $(x_{k} + y_{k})$ , so the $N$ -th tail supremum is $\leq Λ + M + ε + ε$ , and therefore ${lim sup}_{k} (x_{k} + y_{k}) \leq Λ + M + ε + ε < Λ + M + δ$ .

step 1.1L1L2L4L7L8L9

2.3

In the third situation, with $Λ = - \infty$ , first note that there is a real $B$ with $y_{k} < B$ eventually: if $M$ is real, [L4] with $ε = 1$ gives $y_{k} < M + 1$ eventually, so $B : = M + 1$ serves; and if $M = - \infty$ then $y_{k} \to - \infty$ by [L5], so $y_{k} < 0$ eventually and $B : = 0$ serves. Also $Λ = - \infty$ gives $x_{k} \to - \infty$ by [L5]. Now let $c$ be an arbitrary real: since $c - B$ is real, $x_{k} < c - B$ eventually, and beyond the larger threshold both that and $y_{k} < B$ hold, so $x_{k} + y_{k} < (c - B) + B = c$ there. As $c$ was arbitrary, $x_{k} + y_{k} \to - \infty$ , hence ${lim sup}_{k} (x_{k} + y_{k}) = - \infty = Λ + M$ by [L5] and the addition table.

step 1.1L3L4L5L7L9

3.1

In the second situation the conclusion follows from step 2.2: taking $δ = 1$ shows ${lim sup}_{k} (x_{k} + y_{k}) \leq Λ + M + 1$ , a real number, so the left-hand side is not $+ \infty$ ; if it is $- \infty$ then it is $\leq Λ + M$ because $- \infty$ is least; and if it is a real $S$ with $S > Λ + M$ , then $δ_{0} : = S - (Λ + M) > 0$ and step 2.2 applied with $δ = δ_{0}$ gives $S < Λ + M + δ_{0} = S$ , which is impossible. So ${lim sup}_{k} (x_{k} + y_{k}) \leq Λ + M$ by totality.

step 2.2L2L7

4.1

Claim 1 now holds in all three situations, by steps 2.1, 3.1 and 2.3.

step 2.1step 3.1step 2.3step 1.1

5.1

For claim 2, suppose $λ + μ$ is defined. By [L6] the reflected sequences have ${lim sup}_{k} (- x_{k}) = - λ$ and ${lim sup}_{k} (- y_{k}) = - μ$ , and $(- λ) + (- μ) = - (λ + μ)$ is defined exactly when $λ + μ$ is, by [L3]. Claim 1 applied to $(- x_{k})$ and $(- y_{k})$ , whose termwise sum is $(- (x_{k} + y_{k}))$ , therefore gives $- {lim inf}_{k} (x_{k} + y_{k}) = {lim sup}_{k} (- (x_{k} + y_{k})) \leq (- λ) + (- μ) = - (λ + μ)$ ; reflecting this inequality reverses it into ${lim inf}_{k} (x_{k} + y_{k}) \geq λ + μ$ .

step 4.1L3L6L7∎

Remarks

The three situations are not decoration. The middle one is the analytic content and the outer two are genuinely different arguments: the first is vacuous because $+ \infty$ bounds everything, and the third is a statement about divergence to $- \infty$ that has to be proved, since a sum of two sequences each running off to $- \infty$ , or one running off with the other merely bounded above, is not covered by any algebra of limits (Divergence to $+ \infty$ and to $- \infty$ forbids that).
Why the real supremum of a sumset is not used. The natural one-line route, $s_{n} (x + y) \leq s_{n} (x) + s_{n} (y)$ followed by a passage to the infimum, needs the first inequality in $\overline{R}$ and then still needs an $ε$ argument to compare $\inf_{n} (s_{n} (x) + s_{n} (y))$ with $Λ + M$ . The identity $\sup (S + T) = \sup S + \sup T$ of Supremum of a sumset: $\sup (S + T) = \sup S + \sup T$ does not apply, since it requires both sets to be nonempty subsets of $R$ bounded above, and a tail range of an unbounded sequence is not. The $ε$ argument is therefore made directly, once.
Both halves of the $ε$ split are reciprocals of natural numbers, not halvings in $R$ . Choosing $m$ with $1 / m < δ$ and then working with $1 / (2 m)$ keeps every quantity a reciprocal of a canonical natural, so the only field facts used are that positives are invertible and that inequalities add.
Equality is the exception. Without a hypothesis on one of the two sequences the gap can be as large as the whole oscillation, as $x_{k} = (- 1)^{k}$ , $y_{k} = (- 1)^{k + 1}$ give $lim sup (x_{k} + y_{k}) = 0 < 2 = lim sup x_{k} + lim sup y_{k}$ ↗ shows. It is standard, and neither needed nor proved on this page, that the inequality becomes an equality as soon as one of the two sequences converges to a real limit.

Depends on

Used by

Dependency tree · two levels

43 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)
J. Lebl, Basic Analysis I, §2.3 (standard reference, not scraped)

lim sup⁡(xk+yk)≤lim sup⁡xk+lim sup⁡yklimsup(xk​+yk​)≤limsupxk​+limsupyk​ whenever the right-hand side is defined in R‾R, and dually for lim inf⁡liminf