TheoremStatement: Literature-sourcedProof: AI-generatedSession-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.

$\limsup(x_k + y_k) \le \limsup x_k + \limsup y_k$ whenever the right-hand side is defined in $\overline{\mathbb{R}}$ , and dually for $\liminf$

Statement

Let $(x_k)$ and $(y_k)$ be sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) and write $\Lambda := \limsup_{k} x_k$ , $M := \limsup_{k} y_k$ (Limit superior and limit inferior of a real sequence as $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$ ).

If the sum $\Lambda + M$ is defined in $\overline{\mathbb{R}}$ (The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$ , its order, and the arithmetic that is left undefined), that is if $\{\Lambda, M\} \ne \{+\infty, -\infty\}$ , then $\limsup_{k}(x_k + y_k) \;\le\; \Lambda + M .$
Dually, writing $\lambda := \liminf_k x_k$ and $\mu := \liminf_k y_k$ , if $\lambda + \mu$ is defined in $\overline{\mathbb{R}}$ then $\liminf_{k}(x_k + y_k) \;\ge\; \lambda + \mu .$

The hypothesis is exactly the one The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$ , its order, and the arithmetic that is left undefined forces, and it cannot be dropped. When one of $\Lambda$ , $M$ is $+\infty$ and the other $-\infty$ the right-hand side is not an element of $\overline{\mathbb{R}}$ 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 $\Lambda := \limsup_k x_k$ , $M := \limsup_k y_k$ , assumed to have a sum defined in $\overline{\mathbb{R}}$ .

[L1]

Tail bounds and the two quantities exist in $\overline{\mathbb{R}}$ for every sequence, $s_n$ being the least upper bound of the $n$ -th tail range and $\limsup$ the greatest lower bound of $\{s_n\}$ (The tail suprema of any real sequence are nonincreasing in $\overline{\mathbb{R}}$ , so the limit superior exists for every sequence, Every subset of $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\overline{\mathbb{R}}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$ , Limit superior and limit inferior of a real sequence as $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$ , Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set).

[L2]

The order on $\overline{\mathbb{R}}$ is total and transitive, $+\infty$ is its greatest element and $-\infty$ its least, and it restricts on $\mathbb{R}$ to the order of $\mathbb{R}$ (The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$ , its order, and the arithmetic that is left undefined, Partial order and partially ordered set).

[L3]

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

[L4]

Epsilon characterisation for a real limit superior: $\Lambda = \limsup_k x_k$ real implies that for every real $\varepsilon > 0$ one has $x_k < \Lambda + \varepsilon$ eventually (For finite $L$ : $L = \limsup x_k$ iff for every $\varepsilon > 0$ one has $x_k < L + \varepsilon$ eventually and $x_k > L - \varepsilon$ frequently).

[L5]

$\limsup_k z_k = -\infty$ implies $z_k \to -\infty$ ; and $z_k \to -\infty$ implies $\limsup_k z_k = -\infty$ ( $\liminf x_k \le \limsup x_k$ for every real sequence, A real sequence converges to $L \in \mathbb{R}$ iff $\liminf x_k = \limsup x_k = L$ , and diverges to $\pm\infty$ iff both equal $\pm\infty$ , Divergence to $+\infty$ and to $-\infty$ ).

[L6]

Reflection: $\limsup_k(-z_k) = -\liminf_k z_k$ and $\liminf_k(-z_k) = -\limsup_k z_k$ ( $\limsup(-x_k) = -\liminf(x_k)$ , with the reflection of $\overline{\mathbb{R}}$ exchanging $\pm\infty$ ).

[L7]

Order arithmetic in $\mathbb{R}$ : 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 $a \le b$ if and only if $-b \le -a$ : 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 $\delta > 0$ there is a natural $m \ge 1$ with $1/m < \delta$ ; for a natural $m \ge 1$ the element $2m$ is a natural $\ge 1$ with $(2m) \cdot 1_{\mathbb{R}} = 2\big(m \cdot 1_{\mathbb{R}}\big) > 0$ , so $1/(2m) > 0$ and $1/(2m) + 1/(2m) = 1/m$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , 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, $\le$ is a linear order on $\mathbb{N}$ , Order on the natural numbers).

Proof

technique · direct

1.1

Since $\Lambda + M$ is defined, exactly one of the following three situations holds: at least one of $\Lambda$ , $M$ equals $+\infty$ , and then the other is $\ne -\infty$ ; 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 $\Lambda = -\infty$ .

givenL2L3

2.1

In the first situation $\Lambda + M = +\infty$ by the addition table, and every element of $\overline{\mathbb{R}}$ is $\le +\infty$ , so $\limsup_k(x_k + y_k) \le \Lambda + M$ .

step 1.1L2L3

2.2

In the second situation let $\delta > 0$ be an arbitrary real, take a natural $m \ge 1$ with $1/m < \delta$ and put $\varepsilon := 1/(2m) > 0$ , so that $\varepsilon + \varepsilon = 1/m < \delta$ . By [L4] there are thresholds beyond which $x_k < \Lambda + \varepsilon$ and beyond which $y_k < M + \varepsilon$ ; beyond the larger of them both hold, so adding the two inequalities gives $x_k + y_k < \Lambda + M + \varepsilon + \varepsilon$ for all $k \ge N$ , where $N$ is that larger threshold. Hence $\Lambda + M + \varepsilon + \varepsilon$ is an upper bound of the $N$ -th tail range of $(x_k + y_k)$ , so the $N$ -th tail supremum is $\le \Lambda + M + \varepsilon + \varepsilon$ , and therefore $\limsup_k(x_k + y_k) \le \Lambda + M + \varepsilon + \varepsilon < \Lambda + M + \delta$ .

step 1.1L1L2L4L7L8L9

2.3

In the third situation, with $\Lambda = -\infty$ , first note that there is a real $B$ with $y_k < B$ eventually: if $M$ is real, [L4] with $\varepsilon = 1$ 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 $\Lambda = -\infty$ 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 $\limsup_k(x_k + y_k) = -\infty = \Lambda + M$ by [L5] and the addition table.

step 1.1L3L4L5L7L9

3.1

In the second situation the conclusion follows from step 2.2: taking $\delta = 1$ shows $\limsup_k(x_k + y_k) \le \Lambda + M + 1$ , a real number, so the left-hand side is not $+\infty$ ; if it is $-\infty$ then it is $\le \Lambda + M$ because $-\infty$ is least; and if it is a real $S$ with $S > \Lambda + M$ , then $\delta_0 := S - (\Lambda + M) > 0$ and step 2.2 applied with $\delta = \delta_0$ gives $S < \Lambda + M + \delta_0 = S$ , which is impossible. So $\limsup_k(x_k + y_k) \le \Lambda + M$ 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 $\lambda + \mu$ is defined. By [L6] the reflected sequences have $\limsup_k(-x_k) = -\lambda$ and $\limsup_k(-y_k) = -\mu$ , and $(-\lambda) + (-\mu) = -(\lambda + \mu)$ is defined exactly when $\lambda + \mu$ is, by [L3]. Claim 1 applied to $(-x_k)$ and $(-y_k)$ , whose termwise sum is $(-(x_k + y_k))$ , therefore gives $-\liminf_k(x_k + y_k) = \limsup_k\big(-(x_k+y_k)\big) \le (-\lambda) + (-\mu) = -(\lambda + \mu)$ ; reflecting this inequality reverses it into $\liminf_k(x_k + y_k) \ge \lambda + \mu$ .

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, $s_n(x+y) \le s_n(x) + s_n(y)$ followed by a passage to the infimum, needs the first inequality in $\overline{\mathbb{R}}$ and then still needs an $\varepsilon$ argument to compare $\inf_n\big(s_n(x) + s_n(y)\big)$ with $\Lambda + M$ . 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 $\mathbb{R}$ bounded above, and a tail range of an unbounded sequence is not. The $\varepsilon$ argument is therefore made directly, once.
Both halves of the $\varepsilon$ split are reciprocals of natural numbers, not halvings in $\mathbb{R}$ . Choosing $m$ with $1/m < \delta$ and then working with $1/(2m)$ 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 $\limsup(x_k + y_k) = 0 < 2 = \limsup x_k + \limsup y_k$ ↗ 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 73 results over 21 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)
J. Lebl, Basic Analysis I, §2.3 (standard reference, not scraped)

lim sup⁡(xk+yk)≤lim sup⁡xk+lim sup⁡yk\limsup(x_k + y_k) \le \limsup x_k + \limsup y_klimsup(xk​+yk​)≤limsupxk​+limsupyk​ whenever the right-hand side is defined in R‾\overline{\mathbb{R}}R, and dually for lim inf⁡\liminfliminf