LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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If $x_k \le y_k$ eventually then $\limsup x_k \le \limsup y_k$ and $\liminf x_k \le \liminf y_k$

Statement

Let $(x_k)$ and $(y_k)$ be sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) with $x_k \le y_k$ eventually, that is for all $k$ from some index on. Then

$\limsup_{k} x_k \;\le\; \limsup_{k} y_k \qquad \text{and} \qquad \liminf_{k} x_k \;\le\; \liminf_{k} y_k$

in $\overline{\mathbb{R}}$ (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 $\overline{\mathbb{R}}$ , The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$ , 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 $K \in \mathbb{N}$ with $x_k \le y_k$ for every $k \ge K$ ; the tail ranges $T_n(x) = \{x_k : k \ge n\}$ 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 $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$ ).

[L1]

All tail bounds and both of $\limsup$ , $\liminf$ exist in $\overline{\mathbb{R}}$ ; $s_n$ is the least upper bound of the tail range and $i_n$ its greatest lower bound; $\limsup_k y_k$ is the greatest lower bound of $\{s_n(y)\}$ and $\liminf_k y_k$ the least upper bound of $\{i_n(y)\}$ ; and $s_m \le s_n$ , $i_n \le i_m$ whenever $n \le m$ (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 $\mathbb{R}$ , Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set).

[L2]

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

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 $\mathbb{N}$ is total, so every $n$ satisfies $n \ge K$ or $n < K$ , and in the latter case $n \le K$ (Order on the natural numbers, $\le$ is a linear order on $\mathbb{N}$ ).

Proof

technique · direct

1.1

By hypothesis fix $K \in \mathbb{N}$ with $x_k \le y_k$ for every $k \ge K$ .

givenL3

2.1

Let $n \ge K$ . Every $k \ge n$ satisfies $k \ge K$ , so $x_k \le y_k \le s_n(y)$ , and therefore $s_n(y)$ is an upper bound of $T_n(x)$ , whence $s_n(x) \le s_n(y)$ by leastness. Dually $i_n(x) \le x_k \le y_k$ for every $k \ge n$ , so $i_n(x)$ is a lower bound of $T_n(y)$ and $i_n(x) \le i_n(y)$ by greatest-lower-boundedness.

step 1.1L1L2L4

3.1

For every $n \in \mathbb{N}$ one has $\limsup_k x_k \le s_n(y)$ . If $n \ge K$ this is $\limsup_k x_k \le s_n(x) \le s_n(y)$ , the first inequality because $\limsup_k x_k$ is a lower bound of $\{s_m(x)\}$ . If $n < K$ then $n \le K$ , so $s_K(y) \le s_n(y)$ , and $\limsup_k x_k \le s_K(x) \le s_K(y) \le s_n(y)$ .

step 2.1L1L2L4

3.2

For every $n \in \mathbb{N}$ one has $i_n(x) \le \liminf_k y_k$ . If $n \ge K$ this is $i_n(x) \le i_n(y) \le \liminf_k y_k$ , the second inequality because $\liminf_k y_k$ is an upper bound of $\{i_m(y)\}$ . If $n < K$ then $n \le K$ , so $i_n(x) \le i_K(x) \le i_K(y) \le \liminf_k y_k$ .

step 2.1L1L2L4

4.1

By step 3.1 the element $\limsup_k x_k$ is a lower bound of $\{s_n(y) : n \in \mathbb{N}\}$ , whose greatest lower bound is $\limsup_k y_k$ , so $\limsup_k x_k \le \limsup_k y_k$ . By step 3.2 the element $\liminf_k y_k$ is an upper bound of $\{i_n(x) : n \in \mathbb{N}\}$ , whose least upper bound is $\liminf_k x_k$ , so $\liminf_k x_k \le \liminf_k y_k$ .

step 3.1step 3.2L1∎

Remarks

"Eventually" is enough, and the proof shows why. Only tails with $n \ge K$ 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 $\overline{\mathbb{R}}$ , 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 $\limsup_k x_k \le \limsup_k y_k$ ; 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 $\limsup$ and $\liminf$ , 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 $\limsup_k x_k$ with $\liminf_k y_k$ .

Depends on

Used by

For aₖ > 0: liminf aₖ₊₁/aₖ ≤ liminf aₖ^1/k ≤ limsup aₖ^1/k ≤ limsup aₖ₊₁/aₖ Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 54 results over 18 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)

If xk≤ykx_k \le y_kxk​≤yk​ eventually then lim sup⁡xk≤lim sup⁡yk\limsup x_k \le \limsup y_klimsupxk​≤limsupyk​ and lim inf⁡xk≤lim inf⁡yk\liminf x_k \le \liminf y_kliminfxk​≤liminfyk​