TheoremStatement: Literature-sourcedProof: 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.

The limit superior is itself a subsequential limit in $\overline{R}$ and is the greatest one

Statement

Let $(x_{k})$ be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) and write $Λ : = {lim sup}_{k} x_{k} \in \overline{R}$ (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 $\overline{R}$ ). Then, with the extended subsequential limit set $\overset{}{SL} (x)$ of Convergence in $\overline{R}$ and the extended subsequential limit set: $L \in \overline{R}$ is an extended subsequential limit when some subsequence converges to $L$ , or diverges to $L = \pm \infty$ :

$Λ \in \overset{}{SL} (x)$ : there is a strictly increasing $n : N \to N$ such that $(x_{n_{j}})$ converges to $Λ$ in $\overline{R}$ ;
$L \leq Λ$ for every $L \in \overset{}{SL} (x)$ .

So $\overset{}{SL} (x)$ is nonempty and has a greatest element, and that element is ${lim sup}_{k} x_{k}$ . In particular every sequence of reals whatever has a subsequence that converges in $\overline{R}$ .

The extended set is the right home for this statement, and the real set is not. The finite subsequential limit set $SL (x)$ of Subsequential limit of a real sequence, and the subsequential limit set may be empty, and when it is not it may have a greatest element different from ${lim sup}_{k} x_{k}$ ; both failures are exhibited by the dedicated counterexample on the companion page. What is true for $SL (x)$ follows: when $Λ$ is a real number, claim 1 puts it in $SL (x)$ , since the two sets agree on $R$ (Convergence in $\overline{R}$ and the extended subsequential limit set: $L \in \overline{R}$ is an extended subsequential limit when some subsequence converges to $L$ , or diverges to $L = \pm \infty$ ), and claim 2 then makes it the greatest element there too.

Facts & Assumptions

Given: A sequence $(x_{k})$ of reals, its tail ranges $T_{n} = {x_{k} : k \geq n}$ , the extended tail suprema $s_{n} = \sup T_{n}$ , and $Λ : = {lim sup}_{k} x_{k} = \inf {s_{n} : n \in N}$ (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 $\overline{R}$ ).

[L1]

All of $s_{n}$ and $Λ$ exist in $\overline{R}$ , with $s_{n}$ the least upper bound of $T_{n}$ and $Λ$ the greatest lower bound of ${s_{n}}$ (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 total, so the failure of $a \leq b$ is $b < a$ ; $- \infty$ is least and $+ \infty$ greatest; every real is $< + \infty$ and $> - \infty$ ; and on $R$ the order is that 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]

Epsilon characterisation for a real $Λ$ : for every real $η > 0$ one has $x_{k} < Λ + η$ eventually and $x_{k} > Λ - η$ frequently (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).

[L4]

Recursion theorem: for a set $A$ , an element $a \in A$ and a function $f : A \to A$ there is a unique $g : N \to A$ with $g_{0} = a$ and $g_{j + 1} = f (g_{j})$ (The recursion theorem).

[L5]

Well-ordering principle: every nonempty subset of $N$ has a least element (The well-ordering principle).

[L6]

Index maps: if $n_{j} < n_{j + 1}$ for every $j$ then $n$ is strictly increasing, and then $n_{j} \geq j$ for every $j$ ; the composite $(x_{n_{j}})$ is a subsequence (A strictly increasing index map satisfies $n_{k} \geq k$ , Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L7]

Convergence in $\overline{R}$ and the extended subsequential limit set (Convergence in $\overline{R}$ and the extended subsequential limit set: $L \in \overline{R}$ is an extended subsequential limit when some subsequence converges to $L$ , or diverges to $L = \pm \infty$ ); convergence to a real, for which it suffices to produce a threshold for every real $ε > 0$ (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences); divergence to $\pm \infty$ (Divergence to $+ \infty$ and to $- \infty$ ); and $∣ a - b ∣ < c$ if and only if $b - c < a < b + c$ for $c > 0$ (Basic properties of the absolute value).

[L8]

If ${lim sup}_{k} x_{k} = - \infty$ then ${lim inf}_{k} x_{k} = - \infty$ and $x_{k} \to - \infty$ ( $lim inf x_{k} \leq lim sup x_{k}$ for every real sequence, 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 $\pm \infty$ ).

[L9]

Limits preserve non-strict inequalities: if $y_{j} \leq c$ for all large $j$ and $y_{j} \to y$ in $R$ , then $y \leq c$ (Limits preserve non-strict inequalities).

[L10]

Archimedean facts: for every real $M$ there is a natural $p \geq 1$ with $M < p \cdot 1_{R}$ , and for every real $η > 0$ a natural $m \geq 1$ with $1 / m < η$ ; the canonical naturals satisfy $0 \leq n \cdot 1_{R}$ and are increasing in $n$ , and $0 < a \leq b$ gives $0 < 1 / b \leq 1 / a$ (Every complete ordered field is Archimedean, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L11]

Strictly between any two reals lies a rational, hence a real (The rationals embed densely in the reals).

[L12]

The order on $N$ is total and transitive, so any two indices have a common upper bound (Order on the natural numbers, $\leq$ is a linear order on $N$ ).

Proof

technique · constructive

1.1givenL1L2

The element $Λ = {lim sup}_{k} x_{k}$ exists in $\overline{R}$ , and exactly one of the following holds: $Λ$ is a real number, $Λ = + \infty$ , or $Λ = - \infty$ .

1.2givenL1L2

Suppose $Λ = + \infty$ . Since $Λ$ is a lower bound of ${s_{n}}$ , every $n$ has $+ \infty \leq s_{n}$ and so $s_{n} = + \infty$ . Consequently, for every $n \in N$ and every real $M$ there is $k \geq n$ with $x_{k} > M$ : otherwise $M$ would be an upper bound of $T_{n}$ and leastness would give $s_{n} \leq M$ , contradicting $M < + \infty$ .

1.3givenL3L7L12

Suppose $Λ$ is real. Then for every $n \in N$ and every real $η > 0$ there is $k \geq n$ with $∣ x_{k} - Λ ∣ < η$ : by [L3] fix $K$ with $x_{k} < Λ + η$ for all $k \geq K$ , let $K^{'}$ be an index at least as large as both $n$ and $K$ , and use that $x_{k} > Λ - η$ frequently to obtain $k \geq K^{'}$ with $x_{k} > Λ - η$ ; that $k$ satisfies $k \geq K$ , hence also $x_{k} < Λ + η$ , and $k \geq n$ .

1.4givenL6L7L8

Suppose $Λ = - \infty$ . Then $x_{k} \to - \infty$ by [L8], and the identity map $j \mapsto j$ is strictly increasing, so the subsequence $(x_{j})$ of $(x_{k})$ converges to $- \infty$ in $\overline{R}$ and $Λ \in \overset{}{SL} (x)$ .

1.5givenL6L7

Let $L \in \overset{}{SL} (x)$ be arbitrary and fix a strictly increasing $n : N \to N$ such that $(x_{n_{j}})$ converges to $L$ in $\overline{R}$ ; then $n_{j} \geq j$ for every $j$ .

2.1step 1.2L5construct

In the case $Λ = + \infty$ , define $f : N \to N$ by letting $f (n)$ be the least element of $E_{n} : = {k \in N : k > n and x_{k} > n \cdot 1_{R}}$ , which is nonempty by step 1.2 applied with the index $n + 1$ and the real $M = n \cdot 1_{R}$ , and let $a$ be the least element of ${k : x_{k} > 0}$ , nonempty by step 1.2 with $n = 0$ and $M = 0$ . Then $f (n) > n$ and $x_{f (n)} > n \cdot 1_{R}$ for every $n$ .

2.2step 1.3L5L10construct

In the case $Λ$ real, define $g : N \to N$ by letting $g (n)$ be the least element of $F_{n} : = {k \in N : k > n and ∣ x_{k} - Λ ∣ < 1 / (n + 1)}$ , which is nonempty by step 1.3 applied with the index $n + 1$ and $η = 1 / (n + 1) > 0$ , and let $b$ be the least element of ${k : ∣ x_{k} - Λ ∣ < 1}$ , nonempty by step 1.3 with $n = 0$ and $η = 1$ . Then $g (n) > n$ and $∣ x_{g (n)} - Λ ∣ < 1 / (n + 1)$ for every $n$ .

2.3step 1.5L2

If $L = - \infty$ then $L \leq Λ$ , since $- \infty$ is the least element of $\overline{R}$ .

2.4step 1.5L1L2L7L12

If $L = + \infty$ , then for every real $M$ there is $J$ with $x_{n_{j}} > M$ for all $j \geq J$ . Fix $n \in N$ and a real $M$ , and take $j$ at least as large as both $J$ and $n$ ; then $n_{j} \geq j \geq n$ , so $x_{n_{j}} \in T_{n}$ and $M < x_{n_{j}} \leq s_{n}$ . As $M$ was an arbitrary real, $s_{n}$ is neither real nor $- \infty$ , so $s_{n} = + \infty$ ; as $n$ was arbitrary, $Λ = \inf {s_{n}} = + \infty$ and $L \leq Λ$ .

2.5step 1.5step 1.1L1L2L9L11

If $L$ is real, suppose for the sake of the comparison that $Λ < L$ . By step 1.1 the element $Λ$ is then real or $- \infty$ ; choose a real $c$ with $Λ < c < L$ , taking a rational strictly between $Λ$ and $L$ in the first case and $c : = L - 1$ in the second. Since $Λ$ is the greatest lower bound of ${s_{n}}$ and $Λ < c$ , the element $c$ is not a lower bound, so there is $n$ with $s_{n} < c$ , and then $x_{k} \leq s_{n} < c$ for every $k \geq n$ . For $j \geq n$ we have $n_{j} \geq j \geq n$ , hence $x_{n_{j}} \leq c$ , so $L \leq c$ by [L9], contradicting $c < L$ . By totality $L \leq Λ$ .

3.1step 2.1L4L6L10

In the case $Λ = + \infty$ , the recursion theorem applied to $N$ , the element $a$ and the function $f$ gives $n : N \to N$ with $n_{0} = a$ and $n_{j + 1} = f (n_{j})$ . Then $n_{j} < n_{j + 1}$ for every $j$ , so $n$ is strictly increasing and $n_{j} \geq j$ ; and $x_{n_{j + 1}} > n_{j} \cdot 1_{R} \geq j \cdot 1_{R}$ for every $j$ .

3.2step 2.2L4L6L10

In the case $Λ$ real, the recursion theorem applied to $N$ , the element $b$ and the function $g$ gives $n : N \to N$ with $n_{0} = b$ and $n_{j + 1} = g (n_{j})$ . Then $n$ is strictly increasing with $n_{j} \geq j$ , and $∣ x_{n_{j + 1}} - Λ ∣ < 1 / (n_{j} + 1) \leq 1 / (j + 1)$ for every $j$ .

4.1step 3.1L7L10L12

In the case $Λ = + \infty$ , the subsequence $(x_{n_{j}})$ diverges to $+ \infty$ : given a real $M$ , take a natural $p \geq 1$ with $M < p \cdot 1_{R}$ ; every $j \geq p + 1$ satisfies $j - 1 \geq p$ , so step 3.1 applied at $j - 1$ gives $x_{n_{j}} > (j - 1) \cdot 1_{R} \geq p \cdot 1_{R} > M$ . Hence $(x_{n_{j}})$ converges to $+ \infty = Λ$ in $\overline{R}$ and $Λ \in \overset{}{SL} (x)$ .

4.2step 3.2L7L10

In the case $Λ$ real, the subsequence $(x_{n_{j}})$ converges to $Λ$ : given a real $ε > 0$ , take a natural $m \geq 1$ with $1 / m < ε$ ; every $j \geq m$ satisfies $j \geq 1$ , so step 3.2 applied at $j - 1$ gives $∣ x_{n_{j}} - Λ ∣ < 1 / j \leq 1 / m < ε$ . Producing such a threshold for every real $ε > 0$ establishes convergence, so $(x_{n_{j}})$ converges to $Λ$ in $\overline{R}$ and $Λ \in \overset{}{SL} (x)$ .

5.1step 4.1step 4.2step 1.4L7

The three cases of step 1.1 are exhaustive, and each produces a subsequence converging to $Λ$ in $\overline{R}$ : step 4.1 when $Λ = + \infty$ , step 4.2 when $Λ$ is real, and step 1.4 when $Λ = - \infty$ . So $Λ \in \overset{}{SL} (x)$ , which is claim 1.

6.1step 5.1step 2.3step 2.4step 2.5L2discharge-construct∎

Steps 2.3, 2.4 and 2.5 cover the three possibilities for an arbitrary $L \in \overset{}{SL} (x)$ and give $L \leq Λ$ in each, which is claim 2. With claim 1 this makes $\overset{}{SL} (x)$ nonempty with greatest element $Λ = {lim sup}_{k} x_{k}$ .

Remarks

The construction uses no choice. Both index maps are built by taking a least element (The well-ordering principle) of an explicitly described nonempty set of naturals, so the functions $f$ and $g$ are defined outright and The recursion theorem then produces the index map. This is the same device as in Every real sequence has a monotone subsequence (the peak / rising-sun lemma), and for the same reason: a subsequence selected by repeated arbitrary choices would need a choice principle, and none is needed here.
Why the recursion threshold is indexed by the previous index rather than by the step number. The recursion theorem produces a function of one variable, so the state carried from one step to the next is the index $n_{j}$ alone. Demanding $x_{n_{j + 1}} > n_{j}$ rather than $x_{n_{j + 1}} > j$ keeps that single-variable form, and $n_{j} \geq j$ (A strictly increasing index map satisfies $n_{k} \geq k$ ) then upgrades the bound to the one actually wanted. The same trick fixes the accuracy in the finite case at $1 / (n_{j} + 1) \leq 1 / (j + 1)$ .
Claim 2 is where the $lim sup$ earns the word "greatest". A subsequence cannot do better than the tail suprema allow: past any index $n$ , every term of the sequence, and so every term of any subsequence, is at most $s_{n}$ , and $Λ$ is the infimum of those. That is the entire content of step 2.5, and the strictness of the inequality $Λ < c$ is what gives the contradiction, since a limit inherits only the non-strict inequality (Limits preserve non-strict inequalities).
Both failures of the real version really occur, and A sequence with $lim sup = + \infty$ : the greatest subsequential limit exists only in $\overline{R}$ ↗ on the companion page is the witness: there $SL (x)$ is nonempty with greatest element $0$ while ${lim sup}_{k} x_{k} = + \infty$ .
The dual statement is The limit inferior is the least subsequential limit in $\overline{R}$ , obtained from this theorem by reflection rather than by repeating the construction.

The limit superior is itself a subsequential limit in $\overline{R}$ and is the greatest one

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The limit superior is itself a subsequential limit in R‾R and is the greatest one

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The limit superior is itself a subsequential limit in $\overline{R}$ and is the greatest one