LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

How statement and proof provenance work

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.

$lim sup (- x_{k}) = - lim inf (x_{k})$ , with the reflection of $\overline{R}$ exchanging $\pm \infty$

Statement

Write $- A : = {- a : a \in A}$ for $A \subseteq \overline{R}$ , with the reflection of The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined, which fixes no point of ${- \infty, + \infty}$ but exchanges the two.

Reflection exchanges the extended bounds. For every $A \subseteq \overline{R}$ , $\sup (- A) = - \inf A and \inf (- A) = - \sup A,$ with the bounds of 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 $R$ and no hypothesis on $A$ .
Reflection exchanges $lim sup$ and $lim inf$ . For every sequence $(x_{k})$ of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences), ${lim sup}_{k} (- x_{k}) = - {lim inf}_{k} x_{k} and {lim inf}_{k} (- x_{k}) = - {lim sup}_{k} x_{k},$ with $lim sup$ and $lim inf$ as in 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 $\overline{R}$ .

Claim 2 is what turns every statement about $lim sup$ on this page into its dual about $lim inf$ without a second proof, exactly as the identity $\inf S = - \sup (- S)$ does in $R$ . The novelty is only that the reflection now has to move the two new points, and it does: $- (+ \infty) = - \infty$ .

Facts & Assumptions

Given: A sequence $(x_{k})$ of reals, the reflected sequence $y_{k} : = - x_{k}$ , and for $A \subseteq \overline{R}$ the reflected set $- A = {- a : a \in A}$ .

[L1]

Reflection on $\overline{R}$ : the map $a \mapsto - a$ satisfies $- (- a) = a$ and $a \leq b$ if and only if $- b \leq - a$ , for all $a, b \in \overline{R}$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

[L2]

Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , with no hypothesis on the subset (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 $R$ ).

[L3]

Least upper bound and greatest lower bound in a poset, and their uniqueness (Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set).

[L4]

Tail ranges $T_{n} = {x_{k} : k \geq n}$ , the extended tail bounds $s_{n} = \sup T_{n}$ and $i_{n} = \inf T_{n}$ , and ${lim sup}_{k} x_{k} = \inf {s_{n}}$ , ${lim inf}_{k} x_{k} = \sup {i_{n}}$ (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 $\overline{R}$ ).

[L5]

All four families exist for every sequence (The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence).

Proof

technique · direct

1.1

Let $A \subseteq \overline{R}$ be arbitrary. Since $- (- a) = a$ for every $a$ , the map $a \mapsto - a$ carries $A$ onto $- A$ and $- A$ onto $A$ , so $- (- A) = A$ ; and by [L2] each of $\sup A$ , $\inf A$ , $\sup (- A)$ , $\inf (- A)$ exists.

givenL1L2

1.2

Let $T_{n}$ and $T_{n}^{'}$ be the tail ranges of $(x_{k})$ and of $(y_{k}) = (- x_{k})$ . Since $y_{k} = - x_{k}$ , the set $T_{n}^{'} = {y_{k} : k \geq n}$ is exactly $- T_{n}$ .

givenL4

2.1

The element $- \inf A$ is an upper bound of $- A$ : for $a \in A$ we have $\inf A \leq a$ , hence $- a \leq - \inf A$ by [L1], and every element of $- A$ is such a $- a$ . If $v$ is any upper bound of $- A$ , then for $a \in A$ we get $- a \leq v$ , hence $- v \leq a$ by [L1], so $- v$ is a lower bound of $A$ and therefore $- v \leq \inf A$ , which gives $- \inf A \leq v$ by [L1] again. So $- \inf A$ is the least upper bound of $- A$ , that is $\sup (- A) = - \inf A$ .

step 1.1L1L2L3

3.1

Applying the identity just proved to the set $- A$ in place of $A$ , and using $- (- A) = A$ , gives $\sup A = - \inf (- A)$ ; reflecting both sides and using $- (- a) = a$ yields $\inf (- A) = - \sup A$ . Claim 1 is proved.

step 2.1step 1.1L1

4.1

By claim 1 applied to $T_{n}$ , the $n$ -th tail supremum of $(y_{k})$ is $\sup T_{n}^{'} = \sup (- T_{n}) = - i_{n}$ , and its $n$ -th tail infimum is $\inf (- T_{n}) = - s_{n}$ .

step 1.2step 2.1step 3.1L4

5.1

Hence the family of tail suprema of $(y_{k})$ is ${- i_{n} : n \in N} = - {i_{n} : n \in N}$ , so claim 1 applied to ${i_{n}}$ gives ${lim sup}_{k} (- x_{k}) = \inf (- {i_{n}}) = - \sup {i_{n}} = - {lim inf}_{k} x_{k}$ .

step 4.1step 3.1L4L5

6.1

The same identity applied to the sequence $(y_{k})$ , whose reflection is $(- y_{k}) = (x_{k})$ by [L1], reads ${lim sup}_{k} x_{k} = - {lim inf}_{k} (- x_{k})$ ; reflecting both sides gives ${lim inf}_{k} (- x_{k}) = - {lim sup}_{k} x_{k}$ . Both parts of claim 2 are proved.

step 5.1L1∎

Remarks

Claim 1 needs no hypothesis, and that is the whole gain over $R$ . The corresponding real statement, $\inf S = - \sup (- S)$ (Every nonempty set bounded below has an infimum), carries the hypotheses that $S$ be nonempty and bounded below, because otherwise neither side denotes anything. Here both sides always denote, so the identity is unconditional and can be applied to the family ${i_{n}}$ without first checking that it is bounded, which for an unbounded sequence it is not.
The reflection is an order anti-isomorphism, not merely a bijection. What step 2.1 uses is that $a \mapsto - a$ is a bijection and reverses the order, both recorded in The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined. A bijection alone would not exchange bounds, and an order-reversing map that is not injective would not carry least upper bounds to greatest lower bounds.
Consequences on this page. The limit inferior is the least subsequential limit in $\overline{R}$ is The limit superior is itself a subsequential limit in $\overline{R}$ and is the greatest one read through this lemma, the $lim inf$ half 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 its $lim sup$ half read the same way, and the $- \infty$ case of 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 $\pm \infty$ is its $+ \infty$ case.

Depends on

Used by

Dependency tree · two levels

20 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)
Infimum and supremum (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)
M. Boyle, Liminf and limsup notes (standard reference, not scraped)
Extended real number line (Wikipedia) (standard reference, not scraped)

lim sup⁡(−xk)=−lim inf⁡(xk)limsup(−xk​)=−liminf(xk​), with the reflection of R‾R exchanging ±∞±∞