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

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The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence

Statement

Let $(x_{k})$ be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences), with tail ranges $T_{n}$ and extended tail bounds $s_{n} = \sup T_{n}$ , $i_{n} = \inf T_{n}$ 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 $\overline{R}$ .

Monotonicity of the extended bounds under inclusion. If $A \subseteq B \subseteq \overline{R}$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined) then $\sup A \leq \sup B and \inf B \leq \inf A,$ the four quantities being the extended 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 $R$ . No hypothesis is placed on $A$ or $B$ ; in particular $A$ may be empty.
The tail bounds are monotone. $T_{m} \subseteq T_{n}$ whenever $n \leq m$ , and hence $s_{m} \leq s_{n} and i_{n} \leq i_{m} (n \leq m) .$ In particular $s_{n + 1} \leq s_{n}$ and $i_{n} \leq i_{n + 1}$ for every $n$ , and $i_{n} \leq s_{n}$ for every $n$ .
Existence. ${lim sup}_{k} x_{k}$ and ${lim inf}_{k} x_{k}$ exist in $\overline{R}$ for every sequence of reals, bounded or not.

Claim 1 is the tool the rest of this page uses whenever two extended suprema are compared. It is proved here, from the definition of a least upper bound, rather than quoted from the suprema page, for the reason given in the remarks below.

Facts & Assumptions

Given: A sequence $(x_{k})$ of reals, its tail ranges $T_{n} = {x_{k} : k \geq n}$ , and the extended bounds $s_{n} = \sup T_{n}$ , $i_{n} = \inf T_{n}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, 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]

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 $R$ ).

[L2]

Least upper bound and greatest lower bound in a poset: $\sup A$ is an upper bound of $A$ that is $\leq$ every upper bound of $A$ , and $\inf A$ is a lower bound that is $\geq$ every lower bound; each is unique when it exists (Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set).

[L3]

$(\overline{R}, \leq)$ is a totally ordered set, so its order is reflexive and transitive (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).

[L4]

The order on $N$ is total and transitive (Order on the natural numbers, $\leq$ is a linear order on $N$ ).

Proof

technique · direct

1.1

Let $A \subseteq B \subseteq \overline{R}$ be arbitrary. By [L1] the four elements $\sup A$ , $\sup B$ , $\inf A$ , $\inf B$ of $\overline{R}$ all exist and are uniquely determined.

givenL1L2

1.2

Let $n \leq m$ in $N$ . Every element of $T_{m}$ has the form $x_{k}$ with $k \geq m$ , and then $k \geq n$ by transitivity, so $x_{k} \in T_{n}$ ; hence $T_{m} \subseteq T_{n}$ .

givenL4

1.3

For every $n$ the tail range $T_{n}$ contains $x_{n}$ , so $i_{n} \leq x_{n}$ because $i_{n}$ is a lower bound of $T_{n}$ , and $x_{n} \leq s_{n}$ because $s_{n}$ is an upper bound of $T_{n}$ ; transitivity gives $i_{n} \leq s_{n}$ .

givenL1L2L3

2.1

Since $\sup B$ is an upper bound of $B$ and $A \subseteq B$ , every element of $A$ is $\leq \sup B$ , so $\sup B$ is an upper bound of $A$ ; as $\sup A$ is the least of the upper bounds of $A$ , this gives $\sup A \leq \sup B$ . Dually $\inf B$ is a lower bound of $B$ , hence of $A$ , and as $\inf A$ is the greatest of the lower bounds of $A$ this gives $\inf B \leq \inf A$ . Claim 1 is proved.

step 1.1L1L2

3.1

Applying claim 1 to the inclusion $T_{m} \subseteq T_{n}$ valid for $n \leq m$ gives $s_{m} \leq s_{n}$ and $i_{n} \leq i_{m}$ ; the special case $m = n + 1$ gives $s_{n + 1} \leq s_{n}$ and $i_{n} \leq i_{n + 1}$ . Together with $i_{n} \leq s_{n}$ this is claim 2.

step 1.2step 1.3step 2.1

4.1

The families ${s_{n} : n \in N}$ and ${i_{n} : n \in N}$ are subsets of $\overline{R}$ , so [L1] applies to them with no hypothesis, and ${lim sup}_{k} x_{k} = \inf {s_{n}}$ and ${lim inf}_{k} x_{k} = \sup {i_{n}}$ exist in $\overline{R}$ for every sequence of reals. This is claim 3.

step 3.1L1L2∎

Remarks

The monotonicity is where the two operations of the definition interlock. Because $(s_{n})$ is nonincreasing, the outer infimum in ${lim sup}_{k} x_{k} = \inf_{n} s_{n}$ is an infimum of a decreasing family, so it is the value the tail suprema are pressing down towards; and because $(i_{n})$ is nondecreasing, ${lim inf}_{k} x_{k} = \sup_{n} i_{n}$ is the value the tail infima are pressing up towards. Nothing in this lemma says the pressing converges, and for an unbounded sequence there is nothing in $R$ for it to converge to; the exact statement is 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.
Why the word "nonincreasing" is spelled out rather than cited. Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences defines monotone for sequences of reals, and $(s_{n})$ takes values in $\overline{R}$ , so the definition does not apply to it. Claim 2 is therefore stated as the inequality it is. When $(x_{k})$ is bounded every $s_{n}$ is real (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 $\overline{R}$ ) and $(s_{n})$ is then a nonincreasing sequence of reals in the sense of Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, but no proof on this page needs that reading.
Claim 1 is not Monotonicity of the supremum under inclusion. That lemma is the same one-line argument carried out in $R$ , and its statement carries the hypotheses that the smaller set be nonempty and the larger one bounded above, without which neither supremum denotes anything. Those are exactly the hypotheses that the extended 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 $R$ dispense with, so the extended statement is not an instance of the real one and is proved from the definition of a least upper bound instead.
Claim 1 costs nothing and is used everywhere. It is the one-line poset argument: the larger set's supremum bounds the smaller set, and leastness does the rest. It is stated as part of this lemma rather than as an item of its own because it is used only in company with the tail bounds.

Depends on

Used by

Dependency tree · two levels

27 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)
N. Donaldson, Math 140A: Real Analysis notes (standard reference, not scraped)

The tail suprema of any real sequence are nonincreasing in R‾R, so the limit superior exists for every sequence