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

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If $x_{k} \leq y_{k}$ eventually then $lim sup x_{k} \leq lim sup y_{k}$ and $lim inf x_{k} \leq lim inf y_{k}$

Statement

Let $(x_{k})$ and $(y_{k})$ be sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) with $x_{k} \leq y_{k}$ eventually, that is for all $k$ from some index on. Then

${lim sup}_{k} x_{k} \leq {lim sup}_{k} y_{k} and {lim inf}_{k} x_{k} \leq {lim inf}_{k} y_{k}$

in $\overline{R}$ (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 $\overline{R}$ , The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined). No boundedness or convergence hypothesis is placed on either sequence.

Facts & Assumptions

Given: Sequences $(x_{k})$ and $(y_{k})$ of reals and an index $K \in N$ with $x_{k} \leq y_{k}$ for every $k \geq K$ ; the tail ranges $T_{n} (x) = {x_{k} : k \geq n}$ and $T_{n} (y)$ , and the extended tail bounds $s_{n} (x) = \sup T_{n} (x)$ , $i_{n} (x) = \inf T_{n} (x)$ and likewise for $y$ (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]

All tail bounds and both of $lim sup$ , $lim inf$ exist in $\overline{R}$ ; $s_{n}$ is the least upper bound of the tail range and $i_{n}$ its greatest lower bound; ${lim sup}_{k} y_{k}$ is the greatest lower bound of ${s_{n} (y)}$ and ${lim inf}_{k} y_{k}$ the least upper bound of ${i_{n} (y)}$ ; and $s_{m} \leq s_{n}$ , $i_{n} \leq i_{m}$ whenever $n \leq m$ (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).

[L2]

The order on $\overline{R}$ is transitive and 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]

A property holds eventually when it holds for all indices from some index on (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L4]

The order on $N$ is total, so every $n$ satisfies $n \geq K$ or $n < K$ , and in the latter case $n \leq K$ (Order on the natural numbers, $\leq$ is a linear order on $N$ ).

Proof

technique · direct

1.1

By hypothesis fix $K \in N$ with $x_{k} \leq y_{k}$ for every $k \geq K$ .

givenL3

2.1

Let $n \geq K$ . Every $k \geq n$ satisfies $k \geq K$ , so $x_{k} \leq y_{k} \leq s_{n} (y)$ , and therefore $s_{n} (y)$ is an upper bound of $T_{n} (x)$ , whence $s_{n} (x) \leq s_{n} (y)$ by leastness. Dually $i_{n} (x) \leq x_{k} \leq y_{k}$ for every $k \geq n$ , so $i_{n} (x)$ is a lower bound of $T_{n} (y)$ and $i_{n} (x) \leq i_{n} (y)$ by greatest-lower-boundedness.

step 1.1L1L2L4

3.1

For every $n \in N$ one has ${lim sup}_{k} x_{k} \leq s_{n} (y)$ . If $n \geq K$ this is ${lim sup}_{k} x_{k} \leq s_{n} (x) \leq s_{n} (y)$ , the first inequality because ${lim sup}_{k} x_{k}$ is a lower bound of ${s_{m} (x)}$ . If $n < K$ then $n \leq K$ , so $s_{K} (y) \leq s_{n} (y)$ , and ${lim sup}_{k} x_{k} \leq s_{K} (x) \leq s_{K} (y) \leq s_{n} (y)$ .

step 2.1L1L2L4

3.2

For every $n \in N$ one has $i_{n} (x) \leq {lim inf}_{k} y_{k}$ . If $n \geq K$ this is $i_{n} (x) \leq i_{n} (y) \leq {lim inf}_{k} y_{k}$ , the second inequality because ${lim inf}_{k} y_{k}$ is an upper bound of ${i_{m} (y)}$ . If $n < K$ then $n \leq K$ , so $i_{n} (x) \leq i_{K} (x) \leq i_{K} (y) \leq {lim inf}_{k} y_{k}$ .

step 2.1L1L2L4

4.1

By step 3.1 the element ${lim sup}_{k} x_{k}$ is a lower bound of ${s_{n} (y) : n \in N}$ , whose greatest lower bound is ${lim sup}_{k} y_{k}$ , so ${lim sup}_{k} x_{k} \leq {lim sup}_{k} y_{k}$ . By step 3.2 the element ${lim inf}_{k} y_{k}$ is an upper bound of ${i_{n} (x) : n \in N}$ , whose least upper bound is ${lim inf}_{k} x_{k}$ , so ${lim inf}_{k} x_{k} \leq {lim inf}_{k} y_{k}$ .

step 3.1step 3.2L1∎

Remarks

"Eventually" is enough, and the proof shows why. Only tails with $n \geq K$ are compared directly; the finitely many earlier tail bounds are absorbed by monotonicity of the tail bounds (The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence), which lets $s_{K} (y)$ stand in for every earlier $s_{n} (y)$ . No appeal to Convergence depends only on the tail is needed, since neither quantity is defined as a limit.
The comparison does not become strict. From $x_{k} < y_{k}$ for every $k$ one gets only ${lim sup}_{k} x_{k} \leq {lim sup}_{k} y_{k}$ ; the sequences $x_{k} = 0$ and $y_{k} = 1 / (k + 1)$ have equal limits and hence equal limit superiors. This is the same phenomenon as for limits (Limits preserve non-strict inequalities).
Both conclusions have the same direction. It is the inner operation that differs between $lim sup$ and $lim inf$ , and both a supremum and an infimum are monotone in the set, so a pointwise inequality pushes both quantities the same way. What fails to be monotone is the gap between them: nothing here compares ${lim sup}_{k} x_{k}$ with ${lim inf}_{k} y_{k}$ .

Depends on

Used by

Dependency tree · two levels

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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)

If xk≤ykxk​≤yk​ eventually then lim sup⁡xk≤lim sup⁡yklimsupxk​≤limsupyk​ and lim inf⁡xk≤lim inf⁡ykliminfxk​≤liminfyk​