Session-authored (Fable 5 assisted)

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.

✓ 17 results · all verified · 15 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 2 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Equivalent Forms of Completeness

1 · Prerequisites

2 · Summary

Objective. Every earlier page of this track took the least-upper-bound property as the axiom of $R$ and derived the monotone convergence theorem, the nested interval property, Bolzano-Weierstrass and the Cauchy criterion from it. This page asks the converse question, and asks it for an arbitrary ordered field rather than for $R$ : which of those four could have been the axiom instead? The answer is that with one hypothesis added in two places they are all the same statement, and that the hypothesis in question is not removable. The page then turns to what a limit still means when a sequence has none, and develops Cesaro means, Stolz-Cesaro and the Silverman-Toeplitz characterisation of the weightings that preserve limits.

Everything here is stated in an arbitrary ordered field, and that is a constraint, not a flourish. The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness fixes the five properties (LUB), (MCT), (NIP), (BW) and (CC) for an ordered field $F$ , reading every $ε$ inside $F$ itself, as Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field requires. The consequence is that none of the sequence lemmas proved earlier for $R$ may be cited here: a theorem about sequences of reals is a theorem about $R$ , and the observation that its proof would transfer is a statement about the proof, not a licence. Sequence basics in an arbitrary ordered field: limits are unique, limits preserve non-strict inequalities, convergent sequences are Cauchy, Cauchy sequences are bounded, and a Cauchy sequence with a convergent subsequence converges exists for exactly this reason. It proves, from the ordered field axioms alone, that limits are unique, that they preserve non-strict inequalities, that convergent sequences are Cauchy, that Cauchy sequences are bounded, and that a Cauchy sequence with a convergent subsequence converges. Four of the implications below rest on it, and it deliberately contains no arithmetic of limits, since no such result is available for a general ordered field anywhere in this library.

The nested interval property is stated in its shrinking form, with the lengths required to tend to $0$ in the order of $F$ . This is the form satisfied by $R ((t^{- 1}))$ ( $R ((t^{- 1}))$ has the nested interval property for lengths tending to $0$ ) and the form used by the bisection argument (Nested intervals plus the Archimedean property imply Bolzano-Weierstrass, by repeated bisection).

The equivalence, and its cycle. For an ordered field the five completeness properties are equivalent, provided the Archimedean property is assumed alongside nested intervals and Cauchy completeness proves that for an ordered field $F$ the following are equivalent: (LUB); (ARCH) with (NIP); (BW); (ARCH) with (CC); (MCT). Seven lemmas close the cycle $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 1$ . An ordered field with the least-upper-bound property has the nested interval property and is Archimedean takes the supremum of the left endpoints, and gets the Archimedean property from Every complete ordered field is Archimedean on the way. Nested intervals plus the Archimedean property imply Bolzano-Weierstrass, by repeated bisection is the bisection argument: an interval is halved forever, keeping a half the sequence visits infinitely often, and the Archimedean property is what makes the halved lengths tend to $0$ in $F$ . Bolzano-Weierstrass implies Cauchy completeness in any ordered field is pure bookkeeping, because both of its ingredients live in Sequence basics in an arbitrary ordered field: limits are unique, limits preserve non-strict inequalities, convergent sequences are Cauchy, Cauchy sequences are bounded, and a Cauchy sequence with a convergent subsequence converges. Cauchy completeness plus the Archimedean property imply the monotone convergence property shows a bounded nondecreasing sequence is Cauchy, by extracting increments of a fixed size and adding them up until they pass the bound. The monotone convergence property plus the Archimedean property imply the least-upper-bound property bisects between an upper bound and a non-upper bound and identifies the supremum as the common limit of the two bracketing sequences. Both bisections are genuine recursions with no choice: the rule that decides which half to keep is a definite condition, and every index taken is the least admissible one.

Three of the five carry the Archimedean property and two do not, and the difference is the whole subtlety of the page. Bolzano-Weierstrass alone forces the Archimedean property, so it needs no separate Archimedean hypothesis and The monotone convergence property alone forces the Archimedean property, so it carries no separate Archimedean hypothesis show that (BW) and (MCT) each force the Archimedean property on their own, as (LUB) does; so those three need no such hypothesis attached. (NIP) and (CC) do, and FALSE: the nested interval property alone implies the least-upper-bound property and FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property say why: the formal Laurent series field $R ((t^{- 1}))$ , built on the prerequisite page, has both of them and lacks least upper bounds. Which of the five completeness properties carry the Archimedean property on their own, and which must be handed it sorts the five and explains the pattern: the three that carry it quantify over an object assumed only to be bounded, and can therefore be tested against the canonical naturals themselves, while the two that do not quantify over data already forced together.

When a sequence has no limit, one may still average it. The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability introduces $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ , indexed from $0$ so that $(σ_{n})$ is a genuine sequence on $N$ . If $x_{k} \to L$ then $σ_{n} \to L$ : convergence implies $(C, 1)$ -summability to the same value proves that averaging never changes a limit that exists, by the head-and-tail estimate that every regularity proof repeats; and FALSE: if the Cesaro means of a sequence converge then the sequence converges proves that it can create one where none existed, the alternating sequence being $(C, 1)$ -summable to $0$ and divergent. So Cesaro summability is strictly weaker than convergence and consistent with it.

Stolz-Cesaro is the discrete l'Hopital rule. Stolz-Cesaro, $\infty / \infty$ form: if $b_{k}$ is strictly increasing and unbounded and $(a_{k + 1} - a_{k}) / (b_{k + 1} - b_{k}) \to L$ then $a_{k} / b_{k} \to L$ proves that if $(b_{k})$ increases without bound then the behaviour of $a_{k} / b_{k}$ is governed by the difference quotient $(a_{k + 1} - a_{k}) / (b_{k + 1} - b_{k})$ ; the argument telescopes the increments and lets the growing denominator wash out a fixed head. It is stated for a tail, because nothing forbids $b_{0} = 0$ and the two applications on the companion page have exactly that. Stolz-Cesaro, $0 / 0$ form: if $b_{k}$ is strictly decreasing to $0$ , $a_{k} \to 0$ , and the difference quotient converges, then $a_{k} / b_{k}$ converges to the same value is the companion $0 / 0$ form, proved by the same telescoping device rather than deduced from the theorem, with the limit taken at the far end of the telescope instead.

Which weightings preserve limits. A summability (Toeplitz) matrix, the transformed sequence $y_{n} = \sum_{k} c_{n, k} x_{k}$ , and regularity replaces the equal Cesaro weights by an arbitrary array $c_{n, k}$ , with finitely many nonzero entries per row so that every $y_{n} = \sum_{k} c_{n, k} x_{k}$ is a finite sum; series are not available until the next page of this track, and the restriction is what keeps the definition meaningful here. Such a matrix is regular when it never changes an existing limit. A summability matrix with only finitely many nonzero entries per row is regular iff each column tends to $0$ , the row sums tend to $1$ , and the row absolute sums are uniformly bounded characterises regularity by three conditions: null columns, row sums tending to $1$ , and uniformly bounded row absolute sums. Sufficiency is the head-and-tail estimate again. Necessity of the first two conditions is read off two particular inputs, a single nonzero coordinate and the constant sequence $1$ . Necessity of the third is the one genuinely hard argument on the page, a gliding hump: a sequence is built in blocks whose signs are chosen against one row at a time, so that its transform exceeds any prescribed value. The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem then recovers the Cesaro mean theorem from the characterisation, by checking those three conditions for the Cesaro matrix.

What this page does not do. It proves nothing new about $R$ : every implication here was already available there, and nothing above may be cited as a fact about $R$ that is not proved on an earlier page. It also proves no Tauberian theorem, that is, no converse to Cesaro summability under an extra side condition; and it defines no series, so the classical reading of Cesaro summation as a method for divergent series is only mentioned, never used.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)verified 2026-08-09 (gpt-5.6-terra-codex-subscription)Open item page →

The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness

Definition

Throughout, $F$ is an ordered field (Ordered field) with its order and its absolute value. Sequences in $F$ , and the notions of convergence in $F$ , Cauchyness in $F$ , boundedness, nondecreasing and nonincreasing, subsequence, closed interval $[a, b]_{F}$ , nesting, and lengths tending to $0$ in $F$ , are the ones fixed once and for all in Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field. They are not restated here and they are never read in $R$ : every $ε$ below ranges over the positive elements of $F$ itself.

A sequence $(x_{k})$ in $F$ is bounded above when there is $B \in F$ with $x_{k} \leq B$ for every $k \in N$ , and a subset $S \subseteq F$ is bounded above when there is $B \in F$ with $s \leq B$ for every $s \in S$ (Complete ordered field (least-upper-bound property), Upper bound, least upper bound, and strict upper bound).

The following are five properties that $F$ may or may not have.

(LUB), the least-upper-bound property. Every nonempty $S \subseteq F$ that is bounded above has a least upper bound in $F$ . This is exactly the condition that makes $F$ a complete ordered field (Complete ordered field (least-upper-bound property)), and the two names are used interchangeably here.
(MCT), the monotone convergence property. Every nondecreasing sequence in $F$ that is bounded above converges in $F$ .
(NIP), the nested interval property. For every nested sequence $(I_{k})_{k \in N}$ of closed intervals $I_{k} = [a_{k}, b_{k}]_{F}$ of $F$ whose lengths tend to $0$ in $F$ , the intersection

$⋂_{k \in N} I_{k}$

is nonempty.
(BW), the Bolzano-Weierstrass property. Every bounded sequence in $F$ has a subsequence that converges in $F$ .
(CC), Cauchy completeness. Every Cauchy sequence in $F$ converges in $F$ .

Alongside these we use the Archimedean property (ARCH) of Archimedean ordered field: for every $x \in F$ there is a natural number $n$ with $x < n \cdot 1_{F}$ .

Remarks

(NIP) is stated in the shrinking form because that is the form both satisfied by the formal Laurent series field and used by the bisection theorem. The field $R ((t^{- 1}))$ is Cauchy complete without having least upper bounds (Every Cauchy sequence in $R ((t^{- 1}))$ converges: $K$ is sequentially Cauchy complete, $R ((t^{- 1}))$ does not have the least-upper-bound property; its canonical naturals have no supremum), and it satisfies shrinking (NIP) ( $R ((t^{- 1}))$ has the nested interval property for lengths tending to $0$ ). The same shrinking condition is exactly what the bisection argument of Nested intervals plus the Archimedean property imply Bolzano-Weierstrass, by repeated bisection produces.
"Lengths tend to $0$ " is read in $F$ . For a non-Archimedean $F$ this is strictly stronger than the same words read in $R$ through some identification of the rational scalars, and the difference is not academic: the remarks of $R ((t^{- 1}))$ has the nested interval property for lengths tending to $0$ exhibit intervals in $R ((t^{- 1}))$ whose lengths are the real constants $2 / (n + 1)$ , which tend to $0$ in the ordinary real sense and do not tend to $0$ in the order of that field.
Boundedness of a sequence is two-sided, boundedness above is not. Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field calls $(x_{k})$ bounded when $∣ x_{k} ∣ \leq M$ for every $k$ , which is the hypothesis of (BW); (MCT) asks only for the one-sided bound $x_{k} \leq B$ , which for a nondecreasing sequence is the only side in question, since $x_{0} \leq x_{k}$ always.
(MCT) is stated for nondecreasing sequences only. The nonincreasing case is not a separate assumption: if $(x_{k})$ is nonincreasing and bounded below by $b$ then $(- x_{k})$ is nondecreasing and bounded above by $- b$ , and $x_{k} \to L$ exactly when $- x_{k} \to - L$ , because $∣ (- x_{k}) - (- L) ∣ = ∣ x_{k} - L ∣$ . That reduction is used in the proof of The monotone convergence property plus the Archimedean property imply the least-upper-bound property.
Nothing here presumes that any of the five holds. They are predicates on an ordered field, and the point of the page they open is that in the presence of (ARCH) they are all the same predicate (For an ordered field the five completeness properties are equivalent, provided the Archimedean property is assumed alongside nested intervals and Cauchy completeness), while without it two of them are strictly weaker (FALSE: the nested interval property alone implies the least-upper-bound property, FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property).
(CC) is this library's third rendering of "Cauchy complete", and for $R$ all three agree. Cauchy sequence in a metric space and Complete metric space: every Cauchy sequence converges in the space read Cauchyness and completeness in a metric space, and the case $n = 1$ of $R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ proves $(R, ∣ x - y ∣)$ complete; Limits and Cauchy sequences of reals reads both notions for real sequences, with $ε$ ranging over the positive rationals; the present definition reads them in an ordered field $F$ . For $F = R$ under the metric $d (x, y) = ∣ x - y ∣$ of The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded the three unfold to the same quantified statement: below every positive real lies a positive rational (The rationals embed densely in the reals), so the two ranges of $ε$ pick out the same Cauchy sequences and the same convergent ones. So " $R$ satisfies (CC)" is a statement this library has already proved twice, as The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges and as the case $n = 1$ of $R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ . The parallel stops at $R$ . The absolute value of an ordered field takes its values in $F$ , while a metric is required to take its values in $R$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), so for a non-Archimedean $F$ the map $(x, y) \mapsto ∣ x - y ∣$ is not a metric in this library's sense and the metric development says nothing about it. That is why Sequence basics in an arbitrary ordered field: limits are unique, limits preserve non-strict inequalities, convergent sequences are Cauchy, Cauchy sequences are bounded, and a Cauchy sequence with a convergent subsequence converges had to be proved from the order axioms alone, although its Cauchy clauses reappear for metric spaces as Every convergent sequence in a metric space is Cauchy, Every Cauchy sequence in a metric space is bounded and A Cauchy sequence in a metric space with a convergent subsequence converges to that subsequence’s limit. Neither development generalises the other; the comparison above identifies their agreement at $R$ without claiming that $R$ is their only overlap.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Sequence basics in an arbitrary ordered field: limits are unique, limits preserve non-strict inequalities, convergent sequences are Cauchy, Cauchy sequences are bounded, and a Cauchy sequence with a convergent subsequence converges

Statement

Let $F$ be an ordered field (Ordered field) and let $(x_{k})$ , $(y_{k})$ be sequences in $F$ , with convergence in $F$ , Cauchyness in $F$ , boundedness and subsequences as in Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field. Then:

Limits are unique. If $x_{k} \to L$ and $x_{k} \to L^{'}$ in $F$ , then $L = L^{'}$ . A convergent sequence therefore has exactly one limit in $F$ and the notation $\lim_{k} x_{k}$ denotes it unambiguously. This is the licence under which the remaining clauses are written as equations between limits, and it is not new here: Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field already establishes it, in an arbitrary ordered field and with no completeness or Archimedean hypothesis. It is restated as clause 1 so that this lemma is self-contained as the citation target of the whole abstract chain on this page.
Limits preserve non-strict inequalities. If $(x_{k})$ and $(y_{k})$ both converge in $F$ and $x_{k} \leq y_{k}$ for every $k$ , then

$\lim_{k} x_{k} \leq \lim_{k} y_{k} .$
Convergent implies Cauchy. If $(x_{k})$ converges in $F$ , it is Cauchy in $F$ .
Cauchy implies bounded. If $(x_{k})$ is Cauchy in $F$ , it is bounded.
A Cauchy sequence with a convergent subsequence converges. If $(x_{k})$ is Cauchy in $F$ and some subsequence $(x_{n_{j}})$ converges in $F$ , then $(x_{k})$ converges in $F$ as well, and

$\lim_{k} x_{k} = \lim_{j} x_{n_{j}} .$

Both sides are asserted to exist: the right-hand side by hypothesis, the left-hand side as part of the conclusion.

Why this is a separate item. Each of the five is proved in this library for sequences of reals, and none of those proofs may be cited here. Conventions for sequences: indexing, eventually, $\lim$ , and rational $ε$ is explicit about it: a theorem about sequences of reals is a theorem about $R$ , and the fact that its argument would transfer to an arbitrary ordered field is a statement about the argument, not a licence to cite the result. The five are collected here, proved from the ordered field axioms alone, so that the completeness equivalences of this page have one place to cite instead of five inline reconstructions.

Facts & Assumptions

Given: An ordered field $F$ and sequences $(x_{k})$ , $(y_{k})$ in $F$ . Each of the five claims is proved under its own stated hypotheses; nothing is assumed of $(x_{k})$ or $(y_{k})$ outside the claim being proved.

[L1]

Sequences in an ordered field: $(x_{k})$ converges to $L$ in $F$ when for every $ε > 0$ in $F$ there is $N \in N$ with $∣ x_{k} - L ∣ < ε$ for all $k \geq N$ ; $(x_{k})$ is Cauchy in $F$ when for every $ε > 0$ in $F$ there is $N \in N$ with $∣ x_{k} - x_{l} ∣ < ε$ for all $k, l \geq N$ ; $(x_{k})$ is bounded when there is $M \in F$ with $∣ x_{k} ∣ \leq M$ for every $k$ ; and a subsequence of $(x_{k})$ is a sequence $(x_{n_{j}})_{j \in N}$ for a strictly increasing $n : N \to N$ (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field).

[L2]

Triangle inequality: $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ for $u, v \in F$ (The triangle inequality).

[L3]

Absolute value: $∣ u ∣ \geq 0$ ; $∣ u ∣ = 0$ if and only if $u = 0$ ; $∣ - u ∣ = ∣ u ∣$ ; and $u \leq ∣ u ∣$ (Basic properties of the absolute value).

[L4]

Order in $F$ : exactly one of $u < v$ , $u = v$ , $v < u$ holds, so the order is total, and both $<$ and $\leq$ are transitive; adding a constant preserves the strict order and two strict inequalities may be added (Order is preserved by adding a constant and by adding inequalities); the nonstrict forms of those two, used below, are the strict forms together with the equality cases, which trichotomy settles (Ordered field).

[L5]

Halving: $0 < 1_{F}$ (The multiplicative identity is positive), so $2 \cdot 1_{F} = 1_{F} + 1_{F} > 0$ (Canonical naturals are positive and strictly increasing) and $2 \cdot 1_{F}$ is nonzero, hence invertible with $(2 \cdot 1_{F})^{- 1} > 0$ (Inverses of positives are positive, and reciprocation reverses order). Writing $ε / 2$ for $ε \cdot (2 \cdot 1_{F})^{- 1}$ , an $ε > 0$ gives $ε / 2 > 0$ and $ε / 2 + ε / 2 = ε$ (Ordered field).

[L6]

Induction principle on $N$ (The principle of mathematical induction).

[L7]

Growth of an index map: a strictly increasing $n : N \to N$ satisfies $n_{j} \geq j$ for every $j$ (A strictly increasing index map satisfies $n_{k} \geq k$ ).

[L8]

The order on $N$ is total and transitive, so of any two indices one is $\geq$ the other, and every index $k$ satisfies $k \leq N$ or $k \geq N$ ( $\leq$ is a linear order on $N$ ).

Proof

technique · direct

1.1

If $d \in F$ satisfies $d < ε$ for every $ε > 0$ in $F$ , then $d \leq 0$ : were $d > 0$ , the instance $ε = d$ would give $d < d$ , which trichotomy forbids, so $d > 0$ fails and totality leaves $d \leq 0$ .

L4algebra

1.2

For every $ε > 0$ in $F$ one has $ε / 2 > 0$ and $ε / 2 + ε / 2 = ε$ .

1.3

Claim 1. Assume $x_{k} \to L$ and $x_{k} \to L^{'}$ , and let $ε > 0$ in $F$ be arbitrary; choose $N_{1}$ with $∣ x_{k} - L ∣ < ε / 2$ for $k \geq N_{1}$ , choose $N_{2}$ with $∣ x_{k} - L^{'} ∣ < ε / 2$ for $k \geq N_{2}$ , and let $N$ be whichever of $N_{1}, N_{2}$ is the larger.

L1L8choose

1.4

Claim 2. Assume $x_{k} \to L$ , $y_{k} \to M$ and $x_{k} \leq y_{k}$ for every $k$ , and let $ε > 0$ in $F$ be arbitrary; choose $N_{1}$ with $∣ x_{k} - L ∣ < ε / 2$ for $k \geq N_{1}$ , choose $N_{2}$ with $∣ y_{k} - M ∣ < ε / 2$ for $k \geq N_{2}$ , and let $N$ be the larger of the two.

L1L8choose

1.5

Claim 3. Assume $x_{k} \to L$ and let $ε > 0$ in $F$ be arbitrary; choose $N$ with $∣ x_{k} - L ∣ < ε / 2$ for all $k \geq N$ .

L1choose

1.6

Claim 4. For every $n \in N$ there is $B \in F$ with $∣ x_{j} ∣ \leq B$ for all $j \leq n$ , by induction on $n$ : for $n = 0$ take $B = ∣ x_{0} ∣$ ; and given such a $B$ for $n$ , totality of the order on $F$ gives either $∣ x_{n + 1} ∣ \leq B$ , in which case the same $B$ serves for $n + 1$ , or $B < ∣ x_{n + 1} ∣$ , in which case $∣ x_{n + 1} ∣$ serves for $n + 1$ by transitivity.

L1L4L6

1.7

Claim 4, continued. Assume $(x_{k})$ is Cauchy; since $1_{F} > 0$ , choose $N$ with $∣ x_{k} - x_{l} ∣ < 1_{F}$ for all $k, l \geq N$ , so that for $k \geq N$ one has $∣ x_{k} ∣ = ∣ (x_{k} - x_{N}) + x_{N} ∣ \leq ∣ x_{k} - x_{N} ∣ + ∣ x_{N} ∣ < 1_{F} + ∣ x_{N} ∣$ .

L1L2L4L5choose

1.8

Claim 5. Assume $(x_{k})$ is Cauchy and $x_{n_{j}} \to L$ along a strictly increasing $n$ , and let $ε > 0$ in $F$ be arbitrary; choose $N_{1}$ with $∣ x_{k} - x_{l} ∣ < ε / 2$ for $k, l \geq N_{1}$ , choose $N_{2}$ with $∣ x_{n_{j}} - L ∣ < ε / 2$ for $j \geq N_{2}$ , and let $N$ be the larger of the two, so that $n_{N} \geq N \geq N_{1}$ and $N \geq N_{2}$ .

L1L7L8choose

2.1

For every $k \geq N$ in the situation of step 1.3: $∣ L - L^{'} ∣ = ∣ (L - x_{k}) + (x_{k} - L^{'}) ∣ \leq ∣ L - x_{k} ∣ + ∣ x_{k} - L^{'} ∣ = ∣ x_{k} - L ∣ + ∣ x_{k} - L^{'} ∣ < ε / 2 + ε / 2 = ε$ .

step 1.2step 1.3L2L3L4

2.2

For every $k \geq N$ in the situation of step 1.4: $L - M = (L - x_{k}) + (x_{k} - y_{k}) + (y_{k} - M)$ , where $L - x_{k} \leq ∣ L - x_{k} ∣ < ε / 2$ and $y_{k} - M \leq ∣ y_{k} - M ∣ < ε / 2$ and $x_{k} - y_{k} \leq 0$ ; adding, $L - M < ε$ .

step 1.2step 1.4L3L4

2.3

For all $k, l \geq N$ in the situation of step 1.5: $∣ x_{k} - x_{l} ∣ = ∣ (x_{k} - L) + (L - x_{l}) ∣ \leq ∣ x_{k} - L ∣ + ∣ x_{l} - L ∣ < ε / 2 + ε / 2 = ε$ .

step 1.2step 1.5L2L3L4

2.4

In the situation of steps 1.6 and 1.7, let $B$ be a bound for $∣ x_{j} ∣$ over $j \leq N$ and set $M : = B + 1_{F} + ∣ x_{N} ∣$ ; then $B \geq ∣ x_{0} ∣ \geq 0$ and $1_{F} + ∣ x_{N} ∣ > 0$ , so $M \geq B$ and $M \geq 1_{F} + ∣ x_{N} ∣$ , whence $∣ x_{k} ∣ \leq B \leq M$ for $k \leq N$ and $∣ x_{k} ∣ < 1_{F} + ∣ x_{N} ∣ \leq M$ for $k \geq N$ ; as every index satisfies $k \leq N$ or $k \geq N$ , $(x_{k})$ is bounded.

step 1.6step 1.7L1L3L4L8

2.5

For every $k \geq N$ in the situation of step 1.8: $∣ x_{k} - L ∣ = ∣ (x_{k} - x_{n_{N}}) + (x_{n_{N}} - L) ∣ \leq ∣ x_{k} - x_{n_{N}} ∣ + ∣ x_{n_{N}} - L ∣ < ε / 2 + ε / 2 = ε$ , the first summand being covered because $k \geq N \geq N_{1}$ and $n_{N} \geq N_{1}$ .

step 1.2step 1.8L2L4

3.1

By step 2.1 the element $∣ L - L^{'} ∣$ is below every $ε > 0$ , so $∣ L - L^{'} ∣ \leq 0$ ; with $∣ L - L^{'} ∣ \geq 0$ this forces $∣ L - L^{'} ∣ = 0$ and hence $L = L^{'}$ , which is claim 1.

step 1.1step 2.1L3L4

3.2

By step 2.2 the element $L - M$ is below every $ε > 0$ , so $L - M \leq 0$ , that is $L \leq M$ , which is claim 2.

step 1.1step 2.2L4

3.3

Step 2.3 produced, for an arbitrary $ε > 0$ , an $N$ beyond which all pairs are within $ε$ , so $(x_{k})$ is Cauchy in $F$ , which is claim 3.

step 2.3L1

4.1

Step 2.5 produced, for an arbitrary $ε > 0$ , an $N$ beyond which $∣ x_{k} - L ∣ < ε$ , so $(x_{k})$ converges in $F$ with $x_{k} \to L$ ; since also $x_{n_{j}} \to L$ , step 3.1 identifies both limits as $L$ and gives $\lim_{k} x_{k} = \lim_{j} x_{n_{j}}$ , which is claim 5.

step 2.5step 3.1L1

5.1

Claims 1, 2, 3, 4 and 5 are steps 3.1, 3.2, 3.3, 2.4 and 4.1 respectively, so all five hold.

step 2.4step 3.1step 3.2step 3.3step 4.1∎

Remarks

Nothing above uses the Archimedean property, and nothing above uses completeness. The five claims hold in every ordered field, including $R (t)$ and $R ((t^{- 1}))$ . That is what makes them safe to use on both sides of every implication proved on this page.
Claim 2 is genuinely non-strict. From $x_{k} < y_{k}$ at every index one gets only $L \leq M$ : the sequences $x_{k} = 0$ and $y_{k} = ε / (k + 1)$ in an Archimedean $F$ have $x_{k} < y_{k}$ and equal limits. The real-number version of this warning is recorded at Limits preserve non-strict inequalities.
There is deliberately no arithmetic clause here. Nothing above lets one add, multiply or divide two limits in a general ordered field, and no item in this library does: Algebra of limits: sums, scalar multiples, products and quotients is stated for sequences of reals, and by the rule recalled above it may not be cited for a general $F$ . No proof on this page needs such a clause; every abstract argument here works with the defining $ε$ and $N$ directly, or with clauses 1 to 5.
Claim 4 avoids any appeal to a maximum of a finite set. The library's finite-maximum lemma Every nonempty finite set of reals has a maximum and a minimum is stated for $R$ , so it is unavailable here for the same reason the other four real-valued lemmas are; step 1.6 replaces it by an induction that uses nothing but totality of the order of $F$ .

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An ordered field with the least-upper-bound property has the nested interval property and is Archimedean

Statement

Let $F$ be an ordered field with the least-upper-bound property (LUB) of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness. Then:

$F$ is Archimedean (Archimedean ordered field);
$F$ has the nested interval property (NIP).

The intersection point produced in claim 2 is the supremum of the left endpoints, and the proof does not use the hypothesis that the lengths tend to $0$ : an ordered field with (LUB) satisfies the unrestricted nested interval property, of which (NIP) as defined is a special case.

Facts & Assumptions

Given: An ordered field $F$ with the least-upper-bound property, and a nested sequence $(I_{k})_{k \in N}$ of closed intervals $I_{k} = [a_{k}, b_{k}]_{F}$ of $F$ , so that $a_{k} \leq b_{k}$ for every $k$ and $I_{k + 1} \subseteq I_{k}$ for every $k$ .

[L1]

Least upper bounds: every nonempty $S \subseteq F$ bounded above has a least upper bound $\sup S \in F$ ; a least upper bound is an upper bound and is $\leq$ every upper bound (Complete ordered field (least-upper-bound property), Upper bound, least upper bound, and strict upper bound).

[L2]

Every complete ordered field is Archimedean (Every complete ordered field is Archimedean).

[L3]

The properties (LUB), (NIP) and the Archimedean property, as fixed in The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness and Archimedean ordered field; (LUB) for $F$ is by definition the statement that $F$ is a complete ordered field (Complete ordered field (least-upper-bound property)).

[L4]

Closed intervals and nesting in $F$ : $[a, b]_{F} = {x \in F : a \leq x \leq b}$ for $a \leq b$ , and $(I_{k})$ is nested when $I_{k + 1} \subseteq I_{k}$ for every $k$ (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field).

[L5]

The order of $F$ is total and transitive (Ordered field).

[L6]

Induction principle on $N$ (The principle of mathematical induction), and the order on $N$ is total, so of any two indices one is the larger ( $\leq$ is a linear order on $N$ ).

Proof

technique · direct

1.1

Having (LUB) is by definition being a complete ordered field, so $F$ is a complete ordered field.

L1L3

1.2

For every $k$ : $a_{k + 1}$ and $b_{k + 1}$ lie in $I_{k + 1}$ , hence in $I_{k}$ , so $a_{k} \leq a_{k + 1} \leq b_{k + 1} \leq b_{k}$ .

L4L5

2.1

$F$ is Archimedean, which is claim 1.

step 1.1L2

2.2

By induction on the difference of the indices, $a_{j} \leq a_{m}$ and $b_{m} \leq b_{j}$ whenever $j \leq m$ .

step 1.2L5L6

3.1

For all $j, l \in N$ one has $a_{j} \leq b_{l}$ : letting $m$ be the larger of $j$ and $l$ , $a_{j} \leq a_{m} \leq b_{m} \leq b_{l}$ .

step 1.2step 2.2L5L6

4.1

The set $A : = {a_{k} : k \in N}$ is nonempty and is bounded above by $b_{0}$ , so $c : = \sup A$ exists in $F$ .

step 3.1L1

5.1

For every $k$ : $a_{k} \leq c$ because $c$ is an upper bound of $A$ ; and $c \leq b_{k}$ because $b_{k}$ is an upper bound of $A$ by step 3.1 while $c$ is the least such.

step 3.1step 4.1L1

6.1

So $c \in [a_{k}, b_{k}]_{F}$ for every $k$ , the intersection of the $I_{k}$ is nonempty, and $F$ has (NIP), which with step 2.1 gives both claims.

step 2.1step 5.1L3L4∎

Remarks

Where the lengths would be used. They are not used at all above. Their role is to force the intersection to be a single point: if the lengths tend to $0$ and $c, c^{'}$ both lie in every $I_{k}$ then $∣ c - c^{'} ∣ \leq b_{k} - a_{k}$ for every $k$ , so $∣ c - c^{'} ∣$ is below every positive element of $F$ and therefore $0$ . Uniqueness is not part of (NIP) as defined in The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness and is not needed anywhere on this page.
The converse of claim 1 fails, and that is the point of two items later on this page. FALSE: the nested interval property alone implies the least-upper-bound property shows that (NIP) does not imply (LUB), and FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property shows the same for (CC); in both the witness is a non-Archimedean field, so neither carries the Archimedean property that (LUB) carries here.

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Nested intervals plus the Archimedean property imply Bolzano-Weierstrass, by repeated bisection

Statement

Let $F$ be an ordered field that is Archimedean (Archimedean ordered field) and has the nested interval property (NIP) of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness. Then $F$ has the Bolzano-Weierstrass property (BW): every bounded sequence in $F$ has a subsequence converging in $F$ .

Say that a set $E \subseteq N$ is cofinal when for every $K \in N$ there is $k \geq K$ with $k \in E$ . The construction below bisects a bracketing interval, keeping at each stage a half that the sequence visits cofinally often, and reads the limit off (NIP).

Facts & Assumptions

Given: An Archimedean ordered field $F$ with (NIP), and a bounded sequence $(x_{k})$ in $F$ , so that $∣ x_{k} ∣ \leq M_{0}$ for every $k$ and some $M_{0} \in F$ .

[L1]

The properties (NIP) and (BW) of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness.

[L2]

Sequences in an ordered field: boundedness, $[a, b]_{F} = {x \in F : a \leq x \leq b}$ for $a \leq b$ , nesting, lengths tending to $0$ in $F$ , convergence in $F$ , and subsequences along a strictly increasing index map (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field).

[L3]

Archimedean property: for every $x \in F$ there is a natural number $n$ with $x < n \cdot 1_{F}$ (Archimedean ordered field); and the canonical naturals satisfy $n \cdot 1_{F} > 0$ for $n \geq 1$ and $n \cdot 1_{F} \leq m \cdot 1_{F}$ whenever $n \leq m$ (Canonical naturals are positive and strictly increasing).

[L4]

Recursion theorem (The recursion theorem).

[L5]

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

[L6]

Consecutive comparisons suffice for strict increase: if $n_{j} < n_{j + 1}$ for every $j$ then $n$ is strictly increasing (A strictly increasing index map satisfies $n_{k} \geq k$ ).

[L7]

Powers and Bernoulli: $a^{0} = 1$ and $a^{n + 1} = a^{n} a$ (Integer powers $a^{m}$ ); and $(1_{F} + x)^{n} \geq 1_{F} + n \cdot x$ for $x \geq - 1_{F}$ (Bernoulli's inequality $(1 + x)^{n} \geq 1 + n x$ ).

[L8]

Order arithmetic: $0 < 1_{F}$ (The multiplicative identity is positive); adding a constant preserves the strict order and strict inequalities add (Order is preserved by adding a constant and by adding inequalities), the nonstrict forms following with the equality cases; $a > 0$ gives $a^{- 1} > 0$ , and $0 < a < b$ gives $0 < b^{- 1} < a^{- 1}$ (Inverses of positives are positive, and reciprocation reverses order); the order is total and transitive and sums and products of positives are positive (Ordered field).

[L9]

Absolute value: $∣ u ∣ \geq 0$ , and $∣ u ∣$ equals $u$ or $- u$ , so $∣ u ∣ \leq c$ whenever both $u \leq c$ and $- u \leq c$ (Basic properties of the absolute value).

[L10]

Induction principle (The principle of mathematical induction) and totality of the order on $N$ ( $\leq$ is a linear order on $N$ ).

Proof

technique · constructive

1.1

Since $M_{0} \geq ∣ x_{0} ∣ \geq 0$ , the element $M : = M_{0} + 1_{F}$ satisfies $M > 0$ and $∣ x_{k} ∣ \leq M_{0} < M$ for every $k$ , so $- M \leq x_{k} \leq M$ and $x_{k} \in [- M, M]_{F}$ for every $k$ .

L2L8L9construct

1.2

Writing $m (a, b) : = (a + b) \cdot (2 \cdot 1_{F})^{- 1}$ , define $f : F \times F \to F \times F$ by $f (a, b) : = (a, m (a, b))$ when $a \leq b$ and the set of $k$ with $x_{k} \in [a, m (a, b)]_{F}$ is cofinal, and $f (a, b) : = (m (a, b), b)$ otherwise; the recursion theorem applied to $F \times F$ , the element $(- M, M)$ and $f$ gives a unique $g : N \to F \times F$ with $g (0) = (- M, M)$ and $g (n + 1) = f (g (n))$ , and we write $g (n) = (a_{n}, b_{n})$ and $I_{n} : = [a_{n}, b_{n}]_{F}$ .

L4L8construct

2.1

By induction on $n$ , all of the following hold: $a_{n} \leq b_{n}$ ; $b_{n} - a_{n} = 2 M \cdot ((2 \cdot 1_{F})^{n})^{- 1}$ ; $I_{n + 1} \subseteq I_{n}$ ; and the set $E_{n} : = {k : x_{k} \in I_{n}}$ is cofinal. For $n = 0$ this is step 1.1 together with $b_{0} - a_{0} = 2 M$ and $(2 \cdot 1_{F})^{0} = 1_{F}$ . For the step, put $m : = m (a_{n}, b_{n})$ , so that $a_{n} \leq m \leq b_{n}$ and $m - a_{n} = b_{n} - m = (b_{n} - a_{n}) \cdot (2 \cdot 1_{F})^{- 1}$ ; if the first clause of $f$ applies then $I_{n + 1} = [a_{n}, m]_{F}$ has the four properties by construction, and otherwise there is $K$ with $x_{k} \notin [a_{n}, m]_{F}$ for all $k \geq K$ , so every $k \geq K$ in the cofinal set $E_{n}$ has $m < x_{k} \leq b_{n}$ and hence lies in ${k : x_{k} \in [m, b_{n}]_{F}}$ , which is therefore cofinal as well.

step 1.1step 1.2L2L7L8L9L10

3.1

The lengths $b_{n} - a_{n}$ tend to $0$ in $F$ : given $ε > 0$ , the element $2 M ε^{- 1}$ is positive, so [L3] supplies $n \geq 1$ with $2 M ε^{- 1} < n \cdot 1_{F}$ , and then for every $p \geq n$ Bernoulli at $x = 1_{F}$ gives $(2 \cdot 1_{F})^{p} \geq 1_{F} + p \cdot 1_{F} > p \cdot 1_{F} \geq n \cdot 1_{F} > 2 M ε^{- 1} > 0$ , whence $((2 \cdot 1_{F})^{p})^{- 1} < ε (2 M)^{- 1}$ and $b_{p} - a_{p} = 2 M \cdot ((2 \cdot 1_{F})^{p})^{- 1} < ε$ .

step 2.1L3L7L8

3.2

Since each $E_{j}$ is cofinal, for every $j$ and every $n$ the set ${k \in N : k > n and x_{k} \in I_{j + 1}}$ is nonempty and so has a least element; the recursion theorem applied to $N \times N$ , the element $(0, 0)$ and the map sending $(j, n)$ to $(j + 1, \min {k : k > n, x_{k} \in I_{j + 1}})$ therefore yields indices $n_{0} = 0$ and $n_{j + 1} = \min {k : k > n_{j}, x_{k} \in I_{j + 1}}$ .

step 2.1L4L5construct

4.1

The sequence $(I_{n})$ is nested with lengths tending to $0$ , so (NIP) supplies an element $c$ lying in $I_{n}$ for every $n$ .

step 2.1step 3.1L1L2

4.2

Since $n_{j} < n_{j + 1}$ for every $j$ , the map $j \mapsto n_{j}$ is strictly increasing and $(x_{n_{j}})$ is a subsequence of $(x_{k})$ ; moreover $x_{n_{j}} \in I_{j}$ for every $j$ , the case $j = 0$ being $x_{0} \in I_{0}$ from step 1.1.

step 3.2L2L6

5.1

For every $j$ , both $x_{n_{j}}$ and $c$ lie in $I_{j}$ , so $x_{n_{j}} - c \leq b_{j} - a_{j}$ and $c - x_{n_{j}} \leq b_{j} - a_{j}$ , whence $∣ x_{n_{j}} - c ∣ \leq b_{j} - a_{j}$ .

step 4.1step 4.2L2L8L9

6.1

Given $ε > 0$ in $F$ , step 3.1 supplies $J$ with $b_{j} - a_{j} < ε$ for all $j \geq J$ , so $∣ x_{n_{j}} - c ∣ < ε$ for all $j \geq J$ ; hence $x_{n_{j}} \to c$ in $F$ .

step 3.1step 5.1L2L8

7.1

An arbitrary bounded sequence in $F$ has therefore been given a subsequence converging in $F$ , so $F$ has (BW).

step 6.1L1discharge-construct∎

Remarks

No choice is used. Both recursions are applications of The recursion theorem to functions defined outright: the bisection rule keeps the left half exactly when that half is visited cofinally often, and the index $n_{j + 1}$ is the least admissible one, supplied by The well-ordering principle rather than chosen.
Where each hypothesis enters. (NIP) is used once, at step 4.1. The Archimedean property is used once, at step 3.1, and only to know that the halved lengths get below every positive element of $F$ . Without it the bisection still runs and still produces nested intervals, but their lengths need not tend to $0$ in $F$ , and (NIP) as stated would not apply.
The bracketing interval is widened by $1_{F}$ in step 1.1 so that $M > 0$ even when the sequence is identically $0$ ; the argument of step 3.1 divides by $2 M$ and would otherwise have to treat that case separately.

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Bolzano-Weierstrass alone forces the Archimedean property, so it needs no separate Archimedean hypothesis

Statement

Let $F$ be an ordered field with the Bolzano-Weierstrass property (BW) of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness. Then $F$ is Archimedean (Archimedean ordered field).

Consequently (BW) needs no Archimedean hypothesis attached to it, in contrast with the nested interval property and with Cauchy completeness, which do (FALSE: the nested interval property alone implies the least-upper-bound property, FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property).

Facts & Assumptions

Given: An ordered field $F$ with (BW).

[L1]

The property (BW): every bounded sequence in $F$ has a subsequence converging in $F$ (The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness).

[L2]

Sequences in an ordered field: a sequence is a function $N \to F$ ; it is bounded when $∣ x_{k} ∣ \leq M$ for every $k$ and some $M \in F$ ; a subsequence is taken along a strictly increasing $n : N \to N$ ; convergence and Cauchyness in $F$ are as fixed there (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field).

[L3]

Archimedean property: $F$ is Archimedean when for every $x \in F$ there is a natural number $n$ with $x < n \cdot 1_{F}$ , where $0 \cdot 1_{F} = 0$ and $(n + 1) \cdot 1_{F} = n \cdot 1_{F} + 1_{F}$ (Archimedean ordered field).

[L4]

Canonical naturals: $n \cdot 1_{F} > 0$ for $n \geq 1$ , the map $n \mapsto n \cdot 1_{F}$ is strictly increasing on ${1, 2, 3, \dots}$ , and $(m + n) \cdot 1_{F} = m \cdot 1_{F} + n \cdot 1_{F}$ (Canonical naturals are positive and strictly increasing).

[L5]

Absolute value: $∣ u ∣ = u$ whenever $u \geq 0$ (Basic properties of the absolute value).

[L6]

A sequence converging in $F$ is Cauchy in $F$ (clause 3 of Sequence basics in an arbitrary ordered field: limits are unique, limits preserve non-strict inequalities, convergent sequences are Cauchy, Cauchy sequences are bounded, and a Cauchy sequence with a convergent subsequence converges).

[L7]

Order arithmetic: $0 < 1_{F}$ (The multiplicative identity is positive); the order is total, so the failure of $x < y$ is $y \leq x$ ; adding a constant preserves the order (Order is preserved by adding a constant and by adding inequalities, Ordered field). Here Order is preserved by adding a constant and by adding inequalities states the STRICT forms and only those; the nonstrict forms used below are those together with the equality cases, which trichotomy settles, the order being total (Ordered field).

[L8]

Discreteness of $N$ : $m < p$ if and only if $m + 1 \leq p$ (Discreteness: $σ (n)$ is the immediate successor).

Proof

technique · contradiction

1.1

Suppose $F$ has (BW) and is not Archimedean; then there is $x \in F$ such that $x < n \cdot 1_{F}$ fails for every natural $n$ , that is, $n \cdot 1_{F} \leq x$ for every $n \in N$ .

L3L7assume-contra

1.2

Let $(y_{k})$ be the sequence in $F$ given by $y_{k} : = k \cdot 1_{F}$ , so that $y_{0} = 0$ , $y_{k + 1} = y_{k} + 1_{F}$ , and $y_{k} \geq 0$ for every $k$ .

L2L3L4

2.1

$(y_{k})$ is bounded: $∣ y_{k} ∣ = y_{k} \leq x$ for every $k$ .

step 1.1step 1.2L2L5

3.1

By (BW) there is a strictly increasing $n : N \to N$ and an $L \in F$ with $y_{n_{j}} \to L$ in $F$ .

step 2.1L1L2

4.1

The subsequence $(y_{n_{j}})$ is therefore Cauchy in $F$ , so, $1_{F}$ being positive, there is $J \in N$ with $∣ y_{n_{j}} - y_{n_{i}} ∣ < 1_{F}$ for all $i, j \geq J$ .

step 3.1L2L6L7

5.1

But $n_{J} < n_{J + 1}$ gives $n_{J} + 1 \leq n_{J + 1}$ and hence $y_{n_{J + 1}} \geq y_{n_{J} + 1} = y_{n_{J}} + 1_{F}$ , so $y_{n_{J + 1}} - y_{n_{J}} \geq 1_{F} > 0$ and $∣ y_{n_{J + 1}} - y_{n_{J}} ∣ \geq 1_{F}$ , contradicting step 4.1.

step 1.2step 4.1L4L5L7L8

6.1

The assumption of step 1.1 is therefore untenable, and an ordered field with (BW) is Archimedean.

step 5.1discharge-contradiction∎

Remarks

The witness sequence is the obstruction itself. In a non-Archimedean field the canonical naturals are bounded, so they form a bounded sequence; and no subsequence of them can converge, because consecutive terms of any subsequence stay at distance at least $1_{F}$ . That is the whole argument, and it shows that (BW) fails in every non-Archimedean ordered field, for instance in $R (t)$ (Not every ordered field is Archimedean) and in $R ((t^{- 1}))$ ( $R ((t^{- 1}))$ is non-Archimedean, and the monomials $t^{- k}$ are cofinal below its positive elements).
Note which direction is being used: the sequence is bounded and has no convergent subsequence, so (BW) is contradicted. Nothing here says that $(y_{k})$ fails to be Cauchy for some other reason; it is Cauchy along no subsequence at all.

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Bolzano-Weierstrass implies Cauchy completeness in any ordered field

Statement

No Archimedean hypothesis is needed here, and none is hidden: (BW) already carries the Archimedean property on its own (Bolzano-Weierstrass alone forces the Archimedean property, so it needs no separate Archimedean hypothesis), but that fact is not used below.

Facts & Assumptions

Given: An ordered field $F$ with (BW), and a Cauchy sequence $(x_{k})$ in $F$ .

[L1]

The properties (BW) and (CC) of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness.

[L2]

Sequences in an ordered field: boundedness, subsequences, convergence in $F$ and Cauchyness in $F$ (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field).

[L3]

In any ordered field, a Cauchy sequence is bounded (clause 4 of Sequence basics in an arbitrary ordered field: limits are unique, limits preserve non-strict inequalities, convergent sequences are Cauchy, Cauchy sequences are bounded, and a Cauchy sequence with a convergent subsequence converges), and a Cauchy sequence with a subsequence converging to $L$ converges to $L$ (clause 5 of the same lemma).

Proof

technique · direct

1.1

Being Cauchy in $F$ , the sequence $(x_{k})$ is bounded.

L2L3

2.1

By (BW) there is a strictly increasing $n : N \to N$ and an $L \in F$ with $x_{n_{j}} \to L$ in $F$ .

step 1.1L1L2

3.1

A Cauchy sequence with a convergent subsequence converges to the same limit, so $x_{k} \to L$ in $F$ .

step 2.1L2L3

4.1

An arbitrary Cauchy sequence in $F$ therefore converges in $F$ , which is (CC).

step 3.1L1∎

Remarks

This is the only implication on the page that is pure bookkeeping, and that is precisely because both of its ingredients were extracted into Sequence basics in an arbitrary ordered field: limits are unique, limits preserve non-strict inequalities, convergent sequences are Cauchy, Cauchy sequences are bounded, and a Cauchy sequence with a convergent subsequence converges and proved there for an arbitrary ordered field. Written out inline it would repeat the boundedness induction and the three-term triangle estimate of that lemma.
The converse fails: $R ((t^{- 1}))$ has (CC) (Every Cauchy sequence in $R ((t^{- 1}))$ converges: $K$ is sequentially Cauchy complete) and, being non-Archimedean ( $R ((t^{- 1}))$ is non-Archimedean, and the monomials $t^{- k}$ are cofinal below its positive elements), fails (BW) by Bolzano-Weierstrass alone forces the Archimedean property, so it needs no separate Archimedean hypothesis.

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Cauchy completeness plus the Archimedean property imply the monotone convergence property

Statement

Let $F$ be an Archimedean ordered field (Archimedean ordered field) with Cauchy completeness (CC). Then $F$ has the monotone convergence property (MCT) of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness: every nondecreasing sequence in $F$ that is bounded above converges in $F$ .

The Archimedean hypothesis is not decoration. Without it the implication is false: $R ((t^{- 1}))$ has (CC) (Every Cauchy sequence in $R ((t^{- 1}))$ converges: $K$ is sequentially Cauchy complete) and fails (MCT), since (MCT) would force it to be Archimedean (The monotone convergence property alone forces the Archimedean property, so it carries no separate Archimedean hypothesis) and it is not ( $R ((t^{- 1}))$ is non-Archimedean, and the monomials $t^{- k}$ are cofinal below its positive elements).

Facts & Assumptions

Given: An Archimedean ordered field $F$ with (CC), and a nondecreasing sequence $(x_{k})$ in $F$ with $x_{k} \leq B$ for every $k$ and some $B \in F$ .

[L1]

The properties (MCT) and (CC) of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness.

[L2]

Sequences in an ordered field: $(x_{k})$ is nondecreasing when $x_{j} \leq x_{k}$ for all $j \leq k$ ; it is Cauchy in $F$ when for every $ε > 0$ in $F$ there is $N$ with $∣ x_{k} - x_{l} ∣ < ε$ for all $k, l \geq N$ ; convergence in $F$ is as fixed there (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field).

[L3]

Archimedean property: for every $z \in F$ there is a natural $n \geq 1$ with $z < n \cdot 1_{F}$ (Archimedean ordered field); the canonical naturals satisfy $n \cdot 1_{F} > 0$ for $n \geq 1$ and $(m + n) \cdot 1_{F} = m \cdot 1_{F} + n \cdot 1_{F}$ , with $0 \cdot 1_{F} = 0$ (Canonical naturals are positive and strictly increasing).

[L4]

Recursion theorem (The recursion theorem); well-ordering principle, every nonempty subset of $N$ has a least element (The well-ordering principle); induction principle (The principle of mathematical induction); the order on $N$ is total ( $\leq$ is a linear order on $N$ ).

[L5]

Absolute value: $∣ u ∣ = u$ whenever $u \geq 0$ (Basic properties of the absolute value).

[L6]

Order arithmetic: adding a constant preserves the strict order and strict inequalities add (Order is preserved by adding a constant and by adding inequalities), the nonstrict forms following with the equality cases; for $c > 0$ one has $a < b$ if and only if $a c < b c$ (Sign rules for products and monotonicity of multiplication); a positive element is invertible with positive inverse (Inverses of positives are positive, and reciprocation reverses order); the order is total and transitive (Ordered field).

Proof

technique · contradiction

1.1

Suppose $(x_{k})$ is not Cauchy in $F$ : there is $ε > 0$ in $F$ such that for every $N \in N$ there are $k, l \geq N$ with $∣ x_{k} - x_{l} ∣ \geq ε$ .

L2assume-contra

2.1

For every $n \in N$ there is $k > n$ with $x_{k} - x_{n} \geq ε$ : apply step 1.1 with $N : = n + 1$ to get $k, l \geq n + 1$ with $∣ x_{k} - x_{l} ∣ \geq ε$ , name them so that $l \leq k$ , note that monotonicity gives $x_{l} \leq x_{k}$ and hence $∣ x_{k} - x_{l} ∣ = x_{k} - x_{l} \geq ε$ , and note that $n < l$ gives $x_{n} \leq x_{l}$ , so $x_{k} - x_{n} \geq x_{k} - x_{l} \geq ε$ with $k \geq n + 1 > n$ .

step 1.1L2L4L5L6

3.1

For each $n$ the set ${k \in N : k > n and x_{k} - x_{n} \geq ε}$ is therefore nonempty and has a least element, so $f (n) : = \min {k : k > n, x_{k} - x_{n} \geq ε}$ is a total function $N \to N$ ; the recursion theorem applied to $N$ , the element $0$ and $f$ gives indices $k_{0} = 0$ and $k_{j + 1} = f (k_{j})$ , with $k_{j} < k_{j + 1}$ and $x_{k_{j + 1}} - x_{k_{j}} \geq ε$ for every $j$ .

step 2.1L4

4.1

By induction on $j$ , $x_{k_{j}} - x_{k_{0}} \geq (j \cdot 1_{F}) ε$ for every $j$ : at $j = 0$ both sides are $0$ , and adding $x_{k_{j + 1}} - x_{k_{j}} \geq ε$ to the inductive inequality gives $x_{k_{j + 1}} - x_{k_{0}} \geq (j \cdot 1_{F}) ε + ε = ((j + 1) \cdot 1_{F}) ε$ .

step 3.1L3L4L6

5.1

Since $ε > 0$ is invertible with $ε^{- 1} > 0$ , the Archimedean property supplies $j \geq 1$ with $(B - x_{k_{0}}) ε^{- 1} < j \cdot 1_{F}$ , hence $B - x_{k_{0}} < (j \cdot 1_{F}) ε$ and $x_{k_{j}} \geq x_{k_{0}} + (j \cdot 1_{F}) ε > B$ , contradicting the hypothesis that $B$ bounds every term of $(x_{k})$ .

step 4.1L3L6

6.1

The assumption of step 1.1 is therefore untenable, so $(x_{k})$ is Cauchy in $F$ and (CC) makes it converge in $F$ ; as $(x_{k})$ was an arbitrary nondecreasing sequence bounded above, $F$ has (MCT).

step 5.1L1L2discharge-contradiction∎

Remarks

What the Archimedean property does here. It is used exactly once, in the final estimate, to say that a fixed positive $ε$ added to itself often enough exceeds a given element. In a non-Archimedean field the increments $ε$ of the recursion can be infinitesimal relative to $B - x_{k_{0}}$ , and the sequence $(x_{k_{j}})$ climbs forever without ever passing $B$ ; that is exactly how (CC) survives while (MCT) fails.
No choice is used: the recursion takes the least admissible index, supplied by The well-ordering principle, and it is The recursion theorem applied to a function defined outright.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

The monotone convergence property alone forces the Archimedean property, so it carries no separate Archimedean hypothesis

Statement

Let $F$ be an ordered field with the monotone convergence property (MCT) of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness. Then $F$ is Archimedean (Archimedean ordered field).

So (MCT), like (BW) and like (LUB), carries the Archimedean property on its own, and the hypothesis attached to (CC) in Cauchy completeness plus the Archimedean property imply the monotone convergence property need not be attached here. This is what lets Which of the five completeness properties carry the Archimedean property on their own, and which must be handed it sort the five properties into those that do and those that do not.

Facts & Assumptions

Given: An ordered field $F$ with (MCT).

[L1]

The property (MCT): every nondecreasing sequence in $F$ that is bounded above converges in $F$ (The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness).

[L2]

Sequences in an ordered field: a sequence is a function $N \to F$ ; it is nondecreasing when $x_{j} \leq x_{k}$ for all $j \leq k$ ; convergence and Cauchyness in $F$ are as fixed there (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field).

[L3]

Archimedean property: $F$ is Archimedean when for every $x \in F$ there is a natural $n$ with $x < n \cdot 1_{F}$ , where $0 \cdot 1_{F} = 0$ and $(n + 1) \cdot 1_{F} = n \cdot 1_{F} + 1_{F}$ (Archimedean ordered field).

[L4]

Canonical naturals: $n \cdot 1_{F} > 0$ for $n \geq 1$ and $n \mapsto n \cdot 1_{F}$ is strictly increasing on ${1, 2, 3, \dots}$ (Canonical naturals are positive and strictly increasing).

[L5]

[L6]

Order arithmetic: $0 < 1_{F}$ (The multiplicative identity is positive); the order is total, so the failure of $x < y$ is $y \leq x$ ; adding a constant preserves the order (Order is preserved by adding a constant and by adding inequalities, Ordered field); and $∣ u ∣ = u$ whenever $u \geq 0$ (Basic properties of the absolute value). Here Order is preserved by adding a constant and by adding inequalities states the STRICT forms and only those; the nonstrict forms used below are those together with the equality cases, which trichotomy settles, the order being total (Ordered field).

Proof

technique · contradiction

1.1

Suppose $F$ has (MCT) and is not Archimedean; then there is $x \in F$ with $n \cdot 1_{F} \leq x$ for every $n \in N$ .

L3L6assume-contra

1.2

Let $(y_{k})$ be the sequence $y_{k} : = k \cdot 1_{F}$ in $F$ , so $y_{0} = 0$ and $y_{k + 1} = y_{k} + 1_{F}$ ; it is nondecreasing, since $0 = y_{0} < y_{n}$ for $n \geq 1$ and $j \mapsto j \cdot 1_{F}$ is strictly increasing on the positive naturals.

L2L3L4

2.1

$(y_{k})$ is bounded above by $x$ , so (MCT) makes it converge in $F$ to some $L$ .

step 1.1step 1.2L1L2

3.1

Being convergent, $(y_{k})$ is Cauchy in $F$ , so, $1_{F}$ being positive, there is $N \in N$ with $∣ y_{k} - y_{l} ∣ < 1_{F}$ for all $k, l \geq N$ .

step 2.1L2L5L6

4.1

But $y_{N + 1} - y_{N} = 1_{F} > 0$ , so $∣ y_{N + 1} - y_{N} ∣ = 1_{F}$ , which is not $< 1_{F}$ ; this contradicts step 3.1.

step 1.2step 3.1L6

5.1

The assumption of step 1.1 is therefore untenable, and an ordered field with (MCT) is Archimedean.

step 4.1discharge-contradiction∎

Remarks

Why this item exists. Without it, the natural reading of the equivalence theorem would attach an Archimedean hypothesis to (MCT) as well, and Which of the five completeness properties carry the Archimedean property on their own, and which must be handed it would answer its own question wrongly. With it the answer is clean: (LUB), (BW) and (MCT) each imply the Archimedean property, while (NIP) and (CC) do not (FALSE: the nested interval property alone implies the least-upper-bound property, FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property).
The witness is the same sequence as in Bolzano-Weierstrass alone forces the Archimedean property, so it needs no separate Archimedean hypothesis, the canonical naturals, but the two arguments use different failures of it. There the sequence is bounded and has no convergent subsequence; here it is nondecreasing and bounded above and has no limit. Neither argument implies the other, because neither (BW) nor (MCT) is assumed in the other's proof.
The gap of exactly $1_{F}$ between consecutive terms is what does the work, and it is available in every ordered field: $1_{F} > 0$ by The multiplicative identity is positive, and no smallness of $1_{F}$ relative to $x$ is possible, since the Cauchy condition is tested at the threshold $1_{F}$ itself.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

The monotone convergence property plus the Archimedean property imply the least-upper-bound property

Statement

Let $F$ be an Archimedean ordered field (Archimedean ordered field) with the monotone convergence property (MCT) of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness. Then $F$ has the least-upper-bound property (LUB), that is, $F$ is a complete ordered field (Complete ordered field (least-upper-bound property)).

The Archimedean hypothesis is stated for symmetry with the other implications on this page and is in fact redundant here: (MCT) implies it on its own (The monotone convergence property alone forces the Archimedean property, so it carries no separate Archimedean hypothesis).

The supremum is produced by bisection between an upper bound and a non-upper bound, and it is identified as a limit of both bracketing sequences.

Facts & Assumptions

Given: An Archimedean ordered field $F$ with (MCT), a nonempty $S \subseteq F$ bounded above by some $B \in F$ , and an element $s_{0} \in S$ .

[L1]

The properties (MCT) and (LUB), and least upper bounds: $u$ is an upper bound of $S$ when $s \leq u$ for all $s \in S$ , and a least upper bound when moreover $u \leq v$ for every upper bound $v$ ; (LUB) says every nonempty subset bounded above has one (The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness, Complete ordered field (least-upper-bound property), Upper bound, least upper bound, and strict upper bound).

[L2]

Sequences in an ordered field: nondecreasing, nonincreasing, bounded above, and convergence in $F$ (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field).

[L3]

Archimedean property: for every $z \in F$ there is a natural $n \geq 1$ with $z < n \cdot 1_{F}$ (Archimedean ordered field); the canonical naturals are positive for $n \geq 1$ and satisfy $n \cdot 1_{F} \leq m \cdot 1_{F}$ for $n \leq m$ (Canonical naturals are positive and strictly increasing).

[L4]

Recursion theorem (The recursion theorem), induction principle (The principle of mathematical induction), and totality of the order on $N$ ( $\leq$ is a linear order on $N$ ).

[L5]

Powers and Bernoulli: $a^{0} = 1$ , $a^{n + 1} = a^{n} a$ (Integer powers $a^{m}$ ); $(1_{F} + x)^{n} \geq 1_{F} + n \cdot x$ for $x \geq - 1_{F}$ (Bernoulli's inequality $(1 + x)^{n} \geq 1 + n x$ ).

[L6]

Order arithmetic: $0 < 1_{F}$ (The multiplicative identity is positive); adding a constant preserves the strict order and strict inequalities add (Order is preserved by adding a constant and by adding inequalities), the nonstrict forms following with the equality cases; $a > 0$ gives $a^{- 1} > 0$ and $0 < a < b$ gives $0 < b^{- 1} < a^{- 1}$ (Inverses of positives are positive, and reciprocation reverses order); the order is total and transitive (Ordered field).

[L7]

Absolute value: $∣ u ∣ \geq 0$ , $∣ u ∣ = ∣ - u ∣$ , $∣ u ∣ = u$ for $u \geq 0$ (Basic properties of the absolute value); and $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ (The triangle inequality).

[L8]

Limits in an ordered field preserve non-strict inequalities (clause 2 of Sequence basics in an arbitrary ordered field: limits are unique, limits preserve non-strict inequalities, convergent sequences are Cauchy, Cauchy sequences are bounded, and a Cauchy sequence with a convergent subsequence converges).

Proof

technique · constructive

1.1

Put $u_{0} : = B$ and $l_{0} : = s_{0} - 1_{F}$ ; then $u_{0}$ is an upper bound of $S$ , $l_{0}$ is not one because $s_{0} \in S$ and $l_{0} < s_{0}$ , and $l_{0} < s_{0} \leq u_{0}$ .

L1L6construct

1.2

Writing $m (l, u) : = (l + u) (2 \cdot 1_{F})^{- 1}$ , define $f : F \times F \to F \times F$ by $f (l, u) : = (l, m (l, u))$ when $m (l, u)$ is an upper bound of $S$ and $f (l, u) : = (m (l, u), u)$ otherwise; the recursion theorem applied to $F \times F$ , the element $(l_{0}, u_{0})$ and $f$ gives a unique $g : N \to F \times F$ with $g (0) = (l_{0}, u_{0})$ and $g (n + 1) = f (g (n))$ , and we write $g (n) = (l_{n}, u_{n})$ .

L4L6construct

1.3

A constant sequence in $F$ converges to its value, since $∣ a - a ∣ = 0 < ε$ for every $ε > 0$ .

L2L7

2.1

By induction on $n$ : $u_{n}$ is an upper bound of $S$ ; $l_{n}$ is not an upper bound of $S$ ; $l_{n} \leq l_{n + 1} \leq u_{n + 1} \leq u_{n}$ ; and $u_{n} - l_{n} = (u_{0} - l_{0}) ((2 \cdot 1_{F})^{n})^{- 1}$ . The base case is step 1.1 together with $(2 \cdot 1_{F})^{0} = 1_{F}$ ; for the step, $m : = m (l_{n}, u_{n})$ satisfies $l_{n} \leq m \leq u_{n}$ and $m - l_{n} = u_{n} - m = (u_{n} - l_{n}) (2 \cdot 1_{F})^{- 1}$ , and whichever of the two clauses of $f$ applies, the retained pair again brackets $S$ in the stated sense with half the previous length.

step 1.1step 1.2L1L5L6

3.1

The lengths tend to $0$ in $F$ : given $ε > 0$ , the element $(u_{0} - l_{0}) ε^{- 1}$ is positive, so [L3] supplies $n \geq 1$ with $(u_{0} - l_{0}) ε^{- 1} < n \cdot 1_{F}$ , and for every $p \geq n$ Bernoulli at $x = 1_{F}$ gives $(2 \cdot 1_{F})^{p} \geq 1_{F} + p \cdot 1_{F} > p \cdot 1_{F} \geq n \cdot 1_{F} > (u_{0} - l_{0}) ε^{- 1} > 0$ , whence $u_{p} - l_{p} = (u_{0} - l_{0}) ((2 \cdot 1_{F})^{p})^{- 1} < ε$ .

step 2.1L3L5L6

3.2

The sequence $(- u_{n})$ is nondecreasing and is bounded above by $- l_{0}$ , since $l_{0} \leq l_{n} \leq u_{n}$ for every $n$ ; so (MCT) gives $w \in F$ with $- u_{n} \to w$ , and putting $c : = - w$ one has $∣ u_{n} - c ∣ = ∣ - ((- u_{n}) - w) ∣ = ∣ (- u_{n}) - w ∣$ , so $u_{n} \to c$ in $F$ .

step 2.1L1L2L6L7

4.1

$l_{n} \to c$ in $F$ : given $ε > 0$ , step 3.1 supplies $N_{1}$ with $u_{n} - l_{n} < ε / 2$ for $n \geq N_{1}$ and step 3.2 supplies $N_{2}$ with $∣ u_{n} - c ∣ < ε / 2$ for $n \geq N_{2}$ , and for $n$ beyond both, $∣ l_{n} - c ∣ \leq ∣ l_{n} - u_{n} ∣ + ∣ u_{n} - c ∣ = (u_{n} - l_{n}) + ∣ u_{n} - c ∣ < ε$ .

step 3.1step 3.2L2L4L6L7

4.2

$c$ is an upper bound of $S$ : for $s \in S$ one has $s \leq u_{n}$ for every $n$ by step 2.1, and the constant sequence with value $s$ converges to $s$ while $u_{n} \to c$ , so $s \leq c$ by [L8].

step 1.3step 2.1step 3.2L1L8

5.1

$c$ is the least upper bound: let $v$ be any upper bound of $S$ ; for each $n$ the element $l_{n}$ is not an upper bound, so some $s \in S$ has $l_{n} < s \leq v$ and hence $l_{n} \leq v$ ; since $l_{n} \to c$ and the constant sequence with value $v$ converges to $v$ , [L8] gives $c \leq v$ .

step 1.3step 2.1step 4.1L1L6L8

6.1

So $c = \sup S$ exists in $F$ ; as $S$ was an arbitrary nonempty subset bounded above, $F$ has (LUB) and is a complete ordered field.

step 4.2step 5.1L1discharge-construct∎

Remarks

Both bracketing sequences are needed. The upper endpoints give the upper-bound half of the conclusion and the lower endpoints give minimality; the shrinking lengths are what force the two to have the same limit, and that is the only place the Archimedean property is used.
Nonincreasing sequences are handled by reflection, as announced in The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness: (MCT) is stated only for nondecreasing sequences, and step 3.2 applies it to $(- u_{n})$ rather than assuming a second form of the property.
No choice is used. The bisection rule keeps the left half exactly when the midpoint is an upper bound of $S$ , which is a definite condition, so $f$ is a function and The recursion theorem applies.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)Open item page →

For an ordered field the five completeness properties are equivalent, provided the Archimedean property is assumed alongside nested intervals and Cauchy completeness

Statement

Let $F$ be an ordered field, with the five properties (LUB), (MCT), (NIP), (BW), (CC) and the Archimedean property (ARCH) as in The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness and Archimedean ordered field. The following five statements about $F$ are equivalent:

(LUB);
(ARCH) and (NIP);
(BW);
(ARCH) and (CC);
(MCT).

Moreover each of (LUB), (BW) and (MCT) implies (ARCH) on its own, so in statements 1, 3 and 5 the Archimedean property is a consequence rather than a hypothesis.

The Archimedean hypothesis in statements 2 and 4 may not be dropped. It is not an artefact of the proof: the nested interval property without it does not imply (LUB) (FALSE: the nested interval property alone implies the least-upper-bound property), and neither does Cauchy completeness without it (FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property). Both are refuted by the same witness, the formal Laurent series field $R ((t^{- 1}))$ .

The equivalence is proved as a single cycle $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 1$ , each arrow being one lemma of this page.

Facts & Assumptions

Given: An ordered field $F$ .

[L1]

(LUB) implies (ARCH) and (NIP) (An ordered field with the least-upper-bound property has the nested interval property and is Archimedean).

[L2]

(ARCH) together with (NIP) implies (BW) (Nested intervals plus the Archimedean property imply Bolzano-Weierstrass, by repeated bisection).

[L3]

(BW) implies (ARCH) (Bolzano-Weierstrass alone forces the Archimedean property, so it needs no separate Archimedean hypothesis).

[L4]

(BW) implies (CC) (Bolzano-Weierstrass implies Cauchy completeness in any ordered field).

[L5]

(ARCH) together with (CC) implies (MCT) (Cauchy completeness plus the Archimedean property imply the monotone convergence property).

[L6]

(MCT) implies (ARCH) (The monotone convergence property alone forces the Archimedean property, so it carries no separate Archimedean hypothesis).

[L7]

(ARCH) together with (MCT) implies (LUB) (The monotone convergence property plus the Archimedean property imply the least-upper-bound property).

Proof

technique · direct

1.1

Statement 1 implies statement 2: (LUB) gives both (ARCH) and (NIP).

1.2

Statement 2 implies statement 3: (ARCH) with (NIP) gives (BW).

1.3

Statement 3 implies statement 4: (BW) gives (ARCH), and (BW) gives (CC), so it gives their conjunction.

L3L4

1.4

Statement 4 implies statement 5: (ARCH) with (CC) gives (MCT).

1.5

Statement 5 implies statement 1: (MCT) gives (ARCH) by [L6], and (ARCH) with (MCT) gives (LUB) by [L7].

L6L7

2.1

Steps 1.1 to 1.5 form a cycle passing through all five statements, so for any two of them there is a chain of implications from the first to the second; the five are therefore equivalent.

step 1.1step 1.2step 1.3step 1.4step 1.5

2.2

Each of (LUB), (BW) and (MCT) implies (ARCH): the first by [L1], the second by [L3], the third by [L6].

step 1.1step 1.3step 1.5L1L3L6

3.1

Both assertions of the statement are established, by steps 2.1 and 2.2.

step 2.1step 2.2∎

Remarks

What the cycle costs. Seven lemmas suffice for the whole equivalence, because a single cycle through all five statements yields every implication between them, and the arrangement is chosen so that no lemma has to carry an Archimedean hypothesis it cannot discharge. Statement 3 is deliberately the hinge: (BW) is the one property that both implies (ARCH) and is implied by a nested interval argument, so the cycle can enter and leave it without an extra hypothesis.
Read as a statement about $R$ , the theorem says that the five familiar theorems of a first analysis course are not five theorems but one, and that the least-upper-bound axiom could have been replaced by any of the other four (with (ARCH) alongside, where required). This library takes (LUB) as the axiom (Complete ordered field (least-upper-bound property)) and proves the others from it on earlier pages; nothing here re-proves them for $R$ , and nothing here may be cited as a proof about $R$ that is not already there.
The two failures are genuinely different from the three successes. (NIP) and (CC) are both statements about sequences whose data are already close together, and neither of them ever produces a new element far away; that is why an infinitesimal layer can be added to a field without disturbing them, and why the naturals can stay bounded. (LUB), (BW) and (MCT) each quantify over an object that is only assumed bounded, so each of them can be tested against the canonical naturals themselves, and each fails at once when those are bounded. Which of the five completeness properties carry the Archimedean property on their own, and which must be handed it develops this.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability

Definition

Let $(x_{k})$ be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences), with finite sums as in Finite sums and finite products, by recursion. For $n \in N$ the $n$ -th Cesaro mean of $(x_{k})$ is

$σ_{n} : = \frac{1}{n + 1} \sum_{k = 0}^{n} x_{k} = \frac{x_{0} + x_{1} + \dots + x_{n}}{n + 1},$

where $n + 1$ denotes the canonical natural $(n + 1) \cdot 1_{R}$ .

This is well defined. The only thing that could fail is the division: since $n + 1 \geq 1$ , the canonical natural $(n + 1) \cdot 1_{R}$ is strictly positive (Canonical naturals are positive and strictly increasing), hence nonzero by trichotomy (Complete ordered field (least-upper-bound property)), hence invertible. The sum $\sum_{k = 0}^{n} x_{k}$ is the finite sum $\sum_{k < n + 1} x_{k}$ of Finite sums and finite products, by recursion, a single well-determined real for each $n$ . So $(σ_{n})_{n \in N}$ is again a sequence of reals.

The sequence $(x_{k})$ is $(C, 1)$ -summable to $L \in R$ , or Cesaro summable to $L$ , when the sequence of Cesaro means converges to $L$ (Limits and Cauchy sequences of reals). Limits of real sequences are unique (A sequence has at most one limit), so such an $L$ is unique when it exists, and we write it $\lim_{n} σ_{n}$ .

Remarks

The indexing starts at $0$ and the denominator is $n + 1$ , not $n$ . Sequences here are functions on $N$ and $N$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), so $σ_{n}$ averages the $n + 1$ terms $x_{0}, \dots, x_{n}$ . Writing $σ_{n} = (x_{1} + \dots + x_{n}) / n$ , as texts indexing from $1$ do, would leave $σ_{0}$ undefined and would not be a sequence on $N$ at all. The convention chosen here is also what makes the Cesaro matrix $c_{n, k} = 1 / (n + 1)$ for $k \leq n$ a genuine $N \times N$ summability matrix (A summability (Toeplitz) matrix, the transformed sequence $y_{n} = \sum_{k} c_{n, k} x_{k}$ , and regularity, The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem).
$(C, 1)$ -summability is strictly weaker than convergence. Every convergent sequence is $(C, 1)$ -summable to its limit (If $x_{k} \to L$ then $σ_{n} \to L$ : convergence implies $(C, 1)$ -summability to the same value), and the converse fails (FALSE: if the Cesaro means of a sequence converge then the sequence converges): the Cesaro means of the alternating sequence converge to $0$ while the sequence itself diverges (The Cesaro means of $(- 1)^{k}$ converge to $0$ although the sequence diverges ↗). That is the entire point of the notion: it assigns a value to some divergent sequences, consistently with the ordinary limit wherever the ordinary limit exists.
Nothing here sums a series. The object averaged is the sequence $(x_{k})$ itself, not its partial sums, and $\sum_{k = 0}^{n} x_{k}$ is a finite sum in the sense of Finite sums and finite products, by recursion. Applied instead to the partial sums of a series, the same definition gives the classical Cesaro summation of divergent series; series are not available until the next page of this track and nothing above presupposes them.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

If $x_{k} \to L$ then $σ_{n} \to L$ : convergence implies $(C, 1)$ -summability to the same value

Statement

Let $(x_{k})$ be a sequence of reals that converges (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals), and let $(σ_{n})$ be its sequence of Cesaro means (The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability). Then $(σ_{n})$ converges as well, and

$\lim_{n} σ_{n} = \lim_{k} x_{k} .$

Both limits are asserted to exist: the right-hand one by hypothesis, the left-hand one as part of the conclusion. Equivalently: a convergent sequence is $(C, 1)$ -summable, to its own limit. The notation is licensed by uniqueness of limits of real sequences (A sequence has at most one limit).

The converse is false (FALSE: if the Cesaro means of a sequence converge then the sequence converges).

Facts & Assumptions

Given: A sequence $(x_{k})$ of reals converging to $L : = \lim_{k} x_{k}$ , and its Cesaro means $σ_{n} = (n + 1)^{- 1} \sum_{k = 0}^{n} x_{k}$ .

[L1]

The Cesaro means and $(C, 1)$ -summability (The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability); $(n + 1) \cdot 1_{R} > 0$ for every $n \in N$ (Canonical naturals are positive and strictly increasing).

[L2]

Finite sums (Finite sums and finite products, by recursion) and their laws: additivity, scaling with $\sum_{k < d} λ = d λ$ , splitting $\sum_{k < n} = \sum_{k < m} + \sum_{k = m}^{n - 1}$ for $m \leq n$ , and monotonicity of $\sum$ in its terms (Laws of finite sums and finite products).

[L3]

Triangle inequality for finite sums: $∣ \sum_{k < n} a_{k} ∣ \leq \sum_{k < n} ∣ a_{k} ∣$ (Triangle inequality for finite sums).

[L4]

Convergence: for every rational $ε > 0$ there is $K$ with $∣ x_{k} - L ∣ < ε$ for all $k \geq K$ , and equivalently for every real $ε > 0$ , since below every positive real lies a positive rational (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences, The rationals embed densely in the reals).

[L5]

Reciprocal Archimedean property: for every real $ε > 0$ there is a natural $m \geq 1$ with $1 / m < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L6]

Order arithmetic: $a > 0$ gives $a^{- 1} > 0$ and $0 < a < b$ gives $0 < b^{- 1} < a^{- 1}$ (Inverses of positives are positive, and reciprocation reverses order); for $c > 0$ , $a \leq b$ gives $a c \leq b c$ (Sign rules for products and monotonicity of multiplication); adding a constant preserves the order and inequalities add (Order is preserved by adding a constant and by adding inequalities); the order is total and transitive (Complete ordered field (least-upper-bound property), Ordered field); $∣ u ∣ \geq 0$ (Basic properties of the absolute value); and $d \cdot 1_{R} \leq e \cdot 1_{R}$ whenever $d \leq e$ in $N$ (Canonical naturals are positive and strictly increasing). In each clause above, Sign rules for products and monotonicity of multiplication and Order is preserved by adding a constant and by adding inequalities state the STRICT forms and only those; the nonstrict forms used below are those together with the equality cases, which trichotomy settles, the order being total (Ordered field).

[L7]

The order on $N$ is total, so two indices have a larger one ( $\leq$ is a linear order on $N$ ).

Proof

technique · direct

1.1

Let $ε > 0$ be an arbitrary real; choose $K \in N$ with $∣ x_{k} - L ∣ < ε / 2$ for every $k \geq K$ .

L4L6choose

1.2

Put $C : = \sum_{k < K} ∣ x_{k} - L ∣$ , a real with $C \geq 0$ because every summand is $\geq 0$ .

L2L6

2.1

For every $n \geq K$ : since $\sum_{k = 0}^{n} L = (n + 1) L$ , one has $σ_{n} - L = (n + 1)^{- 1} \sum_{k = 0}^{n} (x_{k} - L)$ , hence $∣ σ_{n} - L ∣ \leq (n + 1)^{- 1} \sum_{k = 0}^{n} ∣ x_{k} - L ∣ = (n + 1)^{- 1} (C + \sum_{k = K}^{n} ∣ x_{k} - L ∣) \leq (n + 1)^{- 1} (C + (n + 1 - K) (ε / 2)) \leq C (n + 1)^{- 1} + ε / 2$ , the last two steps using $∣ x_{k} - L ∣ < ε / 2$ for $K \leq k \leq n$ and $n + 1 - K \leq n + 1$ .

step 1.1step 1.2L1L2L3L6

2.2

Since $ε (2 (C + 1))^{- 1} > 0$ , choose a natural $m \geq 1$ with $1 / m < ε (2 (C + 1))^{- 1}$ ; then for every $n \geq m$ one has $n + 1 > m$ , so $C (n + 1)^{- 1} \leq C / m \leq (C + 1) / m < ε / 2$ .

step 1.2L5L6choose

3.1

Let $N$ be the larger of $K$ and $m$ ; for every $n \geq N$ both estimates apply and $∣ σ_{n} - L ∣ < ε / 2 + ε / 2 = ε$ .

step 2.1step 2.2L6L7

4.1

As $ε > 0$ was arbitrary, $(σ_{n})$ converges to $L$ , so $\lim_{n} σ_{n}$ exists and equals $L = \lim_{k} x_{k}$ .

step 3.1L1L4∎

Remarks

The shape of the argument is the shape of every regularity proof. The terms split into a fixed head, whose contribution is a constant divided by $n + 1$ and therefore eventually negligible, and a tail, all of whose terms are already within $ε / 2$ of $L$ and whose weights sum to at most $1$ . A summability matrix with only finitely many nonzero entries per row is regular iff each column tends to $0$ , the row sums tend to $1$ , and the row absolute sums are uniformly bounded is exactly this argument carried out for an arbitrary weighting, and The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem recovers the theorem above from it.
The bound $C + 1$ rather than $C$ in the second estimate is there only so that the divisor is positive when $C = 0$ , which happens whenever the first $K$ terms already equal $L$ .
Nothing here needs boundedness of $(x_{k})$ as a separate hypothesis: it follows from convergence, and in any case only the fixed head $x_{0}, \dots, x_{K - 1}$ is estimated crudely, and it is finite.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Stolz-Cesaro, $\infty / \infty$ form: if $b_{k}$ is strictly increasing and unbounded and $(a_{k + 1} - a_{k}) / (b_{k + 1} - b_{k}) \to L$ then $a_{k} / b_{k} \to L$

Statement

Let $(a_{k})$ and $(b_{k})$ be sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) such that $(b_{k})$ is strictly increasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) and its range is not bounded above (Lower bound, bounded below, bounded set). Then $b_{k + 1} - b_{k} > 0$ for every $k$ , so the difference quotient

$d_{k} : = \frac{a_{k + 1} - a_{k}}{b_{k + 1} - b_{k}}$

is defined for every $k \in N$ . Suppose $(d_{k})$ converges.

Then there is $K_{0} \in N$ with $b_{k} > 0$ for every $k \geq K_{0}$ , so that $q_{j} : = a_{j + K_{0}} / b_{j + K_{0}}$ is a sequence of reals; that sequence converges, and

$\lim_{j} q_{j} = \lim_{k} d_{k} .$

Both limits are asserted to exist: the right-hand one by hypothesis, the left-hand one as part of the conclusion. If moreover $b_{k} \neq 0$ for every $k$ , so that $(a_{k} / b_{k})$ is itself a sequence of reals, then it converges and $\lim_{k} a_{k} / b_{k} = \lim_{k} d_{k}$ , because convergence depends only on a tail (Convergence depends only on the tail).

Why the statement is about a tail. Nothing forbids $b_{0} = 0$ , and the two standard applications on the companion examples page have exactly that, $b_{k} = k$ in one and $b_{n} = n^{p + 1}$ in the other. Writing the conclusion for the whole sequence would be writing a quotient that does not denote at $k = 0$ .

The notation is licensed by uniqueness of limits of real sequences (A sequence has at most one limit).

Facts & Assumptions

Given: Sequences $(a_{k})$ , $(b_{k})$ of reals with $(b_{k})$ strictly increasing and with range not bounded above, and the difference quotients $d_{k} = (a_{k + 1} - a_{k}) / (b_{k + 1} - b_{k})$ , assumed to converge; write $L : = \lim_{k} d_{k}$ .

[L1]

Sequences, tails and strict monotonicity: $(b_{k})$ strictly increasing means $b_{j} < b_{k}$ for $j < k$ , so $b_{k + 1} - b_{k} > 0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L2]

Not bounded above: no real is $\geq$ every element of the range (Lower bound, bounded below, bounded set).

[L3]

A nondecreasing sequence whose range is not bounded above diverges to $+ \infty$ , that is, for every real $M$ there is $K$ with $b_{k} > M$ for all $k \geq K$ (A nondecreasing sequence that is not bounded above diverges to $+ \infty$ , Divergence to $+ \infty$ and to $- \infty$ ).

[L4]

Finite sums (Finite sums and finite products, by recursion) and their laws: telescoping $\sum_{k < n} (c_{k + 1} - c_{k}) = c_{n} - c_{0}$ , splitting, additivity, scaling and monotonicity in the terms (Laws of finite sums and finite products).

[L5]

Triangle inequalities: $∣ \sum_{k < n} u_{k} ∣ \leq \sum_{k < n} ∣ u_{k} ∣$ (Triangle inequality for finite sums) and $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ (The triangle inequality); also $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ and $∣ u ∣ = u$ for $u \geq 0$ (Basic properties of the absolute value).

[L6]

Convergence: for every real $ε > 0$ there is $K$ beyond which the terms are within $ε$ of the limit, the rational and real formulations agreeing (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences, The rationals embed densely in the reals); limits are unique (A sequence has at most one limit).

[L7]

For a sequence of strictly positive reals, converging to $0$ is equivalent to the reciprocal sequence diverging to $+ \infty$ (For positive terms, null and divergence to $+ \infty$ are reciprocal).

[L8]

Convergence depends only on the tail (Convergence depends only on the tail).

[L9]

Algebra of limits: a scalar multiple of a convergent sequence converges to the scalar multiple of the limit (Algebra of limits: sums, scalar multiples, products and quotients).

[L10]

Order arithmetic: $a > 0$ gives $a^{- 1} > 0$ and $0 < a < b$ gives $0 < b^{- 1} < a^{- 1}$ (Inverses of positives are positive, and reciprocation reverses order); for $c > 0$ , $a < b$ if and only if $a c < b c$ (Sign rules for products and monotonicity of multiplication); adding a constant preserves the order and inequalities add (Order is preserved by adding a constant and by adding inequalities); the order is total and transitive (Complete ordered field (least-upper-bound property), Ordered field); and of two indices one is the larger ( $\leq$ is a linear order on $N$ ).

Proof

technique · direct

1.1

$(b_{k})$ is nondecreasing and its range is not bounded above, so it diverges to $+ \infty$ ; taking $M = 0$ gives $K_{0} \in N$ with $b_{k} > 0$ for every $k \geq K_{0}$ , and then $q_{j} : = a_{j + K_{0}} / b_{j + K_{0}}$ is a well-defined sequence of reals.

L1L2L3L10

1.2

For every $k$ one has $b_{k + 1} - b_{k} > 0$ , so each $d_{k}$ is defined and $a_{k + 1} - a_{k} = d_{k} (b_{k + 1} - b_{k})$ .

L1L10

2.1

For all $K < n$ : telescoping gives $a_{n} - a_{K} = \sum_{k = K}^{n - 1} (a_{k + 1} - a_{k}) = \sum_{k = K}^{n - 1} d_{k} (b_{k + 1} - b_{k})$ and $b_{n} - b_{K} = \sum_{k = K}^{n - 1} (b_{k + 1} - b_{k})$ , hence $a_{n} - a_{K} - L (b_{n} - b_{K}) = \sum_{k = K}^{n - 1} (d_{k} - L) (b_{k + 1} - b_{k})$ .

step 1.2L4

2.2

Let $ε > 0$ be an arbitrary real; choose $K \geq K_{0}$ with $∣ d_{k} - L ∣ < ε / 2$ for every $k \geq K$ .

step 1.1L6L10choose

3.1

For every $n > K$ : $∣ a_{n} - a_{K} - L (b_{n} - b_{K}) ∣ \leq \sum_{k = K}^{n - 1} ∣ d_{k} - L ∣ (b_{k + 1} - b_{k}) \leq (ε / 2) (b_{n} - b_{K})$ , and $b_{K} > 0$ with $b_{n} > b_{K}$ gives $0 < b_{n} - b_{K} < b_{n}$ , so dividing by $b_{n} > 0$ yields $∣ a_{n} / b_{n} - L - (a_{K} - L b_{K}) / b_{n} ∣ < ε / 2$ and therefore $∣ a_{n} / b_{n} - L ∣ < ε / 2 + ∣ a_{K} - L b_{K} ∣ / b_{n}$ .

step 2.1step 2.2L4L5L10

3.2

The sequence $j \mapsto 1 / b_{j + K_{0}}$ has strictly positive terms and its reciprocal $j \mapsto b_{j + K_{0}}$ diverges to $+ \infty$ , so it converges to $0$ ; multiplying by the constant $∣ a_{K} - L b_{K} ∣$ , the sequence $j \mapsto ∣ a_{K} - L b_{K} ∣ / b_{j + K_{0}}$ converges to $0$ .

step 1.1step 2.2L3L7L9

4.1

So there is $N \geq K$ with $∣ a_{K} - L b_{K} ∣ / b_{n} < ε / 2$ for every $n \geq N$ , and then $∣ a_{n} / b_{n} - L ∣ < ε / 2 + ε / 2 = ε$ for every $n \geq N$ ; equivalently $∣ q_{j} - L ∣ < ε$ for every $j \geq N - K_{0}$ .

step 3.1step 3.2L6L10

5.1

As $ε > 0$ was arbitrary, $(q_{j})$ converges to $L$ , so $\lim_{j} q_{j}$ exists and equals $L = \lim_{k} d_{k}$ ; and if $b_{k} \neq 0$ for every $k$ then $(a_{k} / b_{k})$ is a sequence of reals whose $K_{0}$ -th tail is $(q_{j})$ , so it too converges to $L$ .

step 4.1L6L8∎

Remarks

This is a discrete l'Hopital rule, and the shape of the proof says why: the increments of $a$ are compared with the increments of $b$ , the comparison is summed by telescoping, and the fixed head $a_{K} - L b_{K}$ is washed out by the divergence of $b_{n}$ . The head is exactly the constant of integration.
Strict increase is used twice, once so that $b_{k + 1} - b_{k} \neq 0$ and the quotients $d_{k}$ exist, and once so that $b_{n} - b_{K} > 0$ and the sum estimate keeps its sign. Unboundedness is used twice as well, once to reach positive $b_{k}$ and once to kill the head.
There is no converse, not even under the same hypotheses: $a_{k} = (- 1)^{k}$ , $b_{k} = k$ have $a_{k} / b_{k} \to 0$ while the difference quotient oscillates, so Stolz-Cesaro has no converse ↗ exhibits $a_{k}$ and $b_{k}$ for which $a_{k} / b_{k}$ converges while $(d_{k})$ oscillates between two values.
The Cesaro mean theorem is the special case $a_{n} = \sum_{k = 0}^{n} x_{k}$ , $b_{n} = n + 1$ : then $d_{n} = x_{n + 1}$ , and the conclusion reads $σ_{n} \to \lim_{k} x_{k}$ . That deduction is not carried out here, because If $x_{k} \to L$ then $σ_{n} \to L$ : convergence implies $(C, 1)$ -summability to the same value is proved directly and earlier; the observation is recorded so the reader can see the two results are one.

CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Stolz-Cesaro, $0 / 0$ form: if $b_{k}$ is strictly decreasing to $0$ , $a_{k} \to 0$ , and the difference quotient converges, then $a_{k} / b_{k}$ converges to the same value

Statement

Let $(a_{k})$ and $(b_{k})$ be sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) with $(b_{k})$ strictly decreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences), $\lim_{k} b_{k} = 0$ and $\lim_{k} a_{k} = 0$ . Then $b_{k + 1} - b_{k} < 0$ for every $k$ , so the difference quotient

$d_{k} : = \frac{a_{k + 1} - a_{k}}{b_{k + 1} - b_{k}}$

is defined for every $k$ , exactly as in Stolz-Cesaro, $\infty / \infty$ form: if $b_{k}$ is strictly increasing and unbounded and $(a_{k + 1} - a_{k}) / (b_{k + 1} - b_{k}) \to L$ then $a_{k} / b_{k} \to L$ . Suppose $(d_{k})$ converges.

Then $b_{k} > 0$ for every $k$ , so $(a_{k} / b_{k})$ is a sequence of reals; that sequence converges, and

$\lim_{k} \frac{a_{k}}{b_{k}} = \lim_{k} d_{k} .$

Both limits are asserted to exist: the right-hand one by hypothesis, the left-hand one as part of the conclusion. The notation is licensed by uniqueness of limits of real sequences (A sequence has at most one limit).

Facts & Assumptions

Given: Sequences $(a_{k})$ , $(b_{k})$ of reals with $(b_{k})$ strictly decreasing, $\lim_{k} b_{k} = 0$ , $\lim_{k} a_{k} = 0$ , and difference quotients $d_{k} = (a_{k + 1} - a_{k}) / (b_{k + 1} - b_{k})$ assumed to converge; write $L : = \lim_{k} d_{k}$ .

[L1]

Sequences and strict decrease: $b_{j} > b_{k}$ for $j < k$ , so $b_{k} - b_{k + 1} > 0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L2]

Convergence: for every real $ε > 0$ there is $K$ beyond which the terms are within $ε$ of the limit, the rational and real formulations agreeing (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences, The rationals embed densely in the reals); a constant sequence converges to its value; limits are unique (A sequence has at most one limit).

[L3]

Limits preserve non-strict inequalities, hypotheses needed only eventually (Limits preserve non-strict inequalities).

[L4]

Algebra of limits for sums, differences and scalar multiples (Algebra of limits: sums, scalar multiples, products and quotients).

[L5]

Finite sums (Finite sums and finite products, by recursion) and their laws: telescoping, splitting, additivity, scaling and monotonicity in the terms (Laws of finite sums and finite products).

[L6]

Triangle inequality for finite sums (Triangle inequality for finite sums); $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ , $∣ u ∣ = u$ for $u \geq 0$ , and $∣ u ∣ \leq c$ exactly when $- c \leq u \leq c$ (Basic properties of the absolute value).

[L7]

Order arithmetic: $a > 0$ gives $a^{- 1} > 0$ (Inverses of positives are positive, and reciprocation reverses order); for $c > 0$ , $a \leq b$ if and only if $a c \leq b c$ (Sign rules for products and monotonicity of multiplication); adding a constant preserves the order and inequalities add (Order is preserved by adding a constant and by adding inequalities); the order is total and transitive (Complete ordered field (least-upper-bound property), Ordered field); of two indices one is the larger ( $\leq$ is a linear order on $N$ ). In each clause above, Sign rules for products and monotonicity of multiplication and Order is preserved by adding a constant and by adding inequalities state the STRICT forms and only those; the nonstrict forms used below are those together with the equality cases, which trichotomy settles, the order being total (Ordered field).

Proof

technique · direct

1.1

For every $k$ one has $b_{k} - b_{k + 1} > 0$ , so $b_{k + 1} - b_{k} \neq 0$ , each $d_{k}$ is defined, and $a_{k} - a_{k + 1} = d_{k} (b_{k} - b_{k + 1})$ .

L1L7

1.2

$b_{m} > 0$ for every $m$ : for all $k \geq m + 1$ one has $b_{k} \leq b_{m + 1}$ , so comparing $(b_{k})$ with the constant sequence $b_{m + 1}$ gives $0 = \lim_{k} b_{k} \leq b_{m + 1}$ , and $b_{m + 1} < b_{m}$ then gives $b_{m} > 0$ .

L1L2L3

1.3

Let $ε > 0$ be an arbitrary real; choose $K$ with $∣ d_{k} - L ∣ < ε / 2$ for every $k \geq K$ .

L2L7choose

2.1

For all $n > m \geq K$ : telescoping gives $a_{m} - a_{n} = \sum_{k = m}^{n - 1} (a_{k} - a_{k + 1}) = \sum_{k = m}^{n - 1} d_{k} (b_{k} - b_{k + 1})$ and $b_{m} - b_{n} = \sum_{k = m}^{n - 1} (b_{k} - b_{k + 1})$ , so $∣ a_{m} - a_{n} - L (b_{m} - b_{n}) ∣ \leq \sum_{k = m}^{n - 1} ∣ d_{k} - L ∣ (b_{k} - b_{k + 1}) \leq (ε / 2) (b_{m} - b_{n})$ , that is $- (ε / 2) (b_{m} - b_{n}) \leq a_{m} - a_{n} - L (b_{m} - b_{n}) \leq (ε / 2) (b_{m} - b_{n})$ .

step 1.1step 1.3L5L6L7

3.1

Fix $m \geq K$ and let $n$ vary over the indices $> m$ : since $a_{n} \to 0$ and $b_{n} \to 0$ , the middle quantity converges in $n$ to $a_{m} - L b_{m}$ , the right-hand one to $(ε / 2) b_{m}$ and the left-hand one to $- (ε / 2) b_{m}$ ; the two eventual inequalities of step 2.1 therefore pass to the limit and give $- (ε / 2) b_{m} \leq a_{m} - L b_{m} \leq (ε / 2) b_{m}$ .

step 2.1L2L3L4

4.1

Since $b_{m} > 0$ , dividing by $b_{m}$ gives $∣ a_{m} / b_{m} - L ∣ \leq ε / 2 < ε$ for every $m \geq K$ .

step 1.2step 3.1L6L7

5.1

As $ε > 0$ was arbitrary, $(a_{k} / b_{k})$ converges to $L$ , so $\lim_{k} a_{k} / b_{k}$ exists and equals $L = \lim_{k} d_{k}$ .

step 4.1L2∎

Remarks

This is a companion of Stolz-Cesaro, $\infty / \infty$ form: if $b_{k}$ is strictly increasing and unbounded and $(a_{k + 1} - a_{k}) / (b_{k + 1} - b_{k}) \to L$ then $a_{k} / b_{k} \to L$ rather than a formal consequence of it, and the cor- prefix should be read that way. The two statements share hypotheses of the same shape and the same telescoping device, but neither is obtained by substituting data into the other: there the fixed head $a_{K} - L b_{K}$ is washed out because $b_{n}$ grows without bound, here the head is removed by letting the far end of the telescope go to $0$ . Nothing in the proof above cites the theorem.
Where the limit is taken matters. In step 3.1 the index $m$ is frozen and $n$ runs; the estimate of step 2.1 is uniform in $n$ for that fixed $m$ , which is why the passage to the limit is legitimate. Taking both indices to infinity at once would prove nothing.
The conclusion is non-strict at the level of $ε / 2$ and is turned into the strict inequality demanded by the definition of a limit only at the last division, where $ε / 2 < ε$ . That is the usual price of passing an inequality to a limit (Limits preserve non-strict inequalities): strictness is not preserved, so it has to be recovered by halving.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

A summability (Toeplitz) matrix, the transformed sequence $y_{n} = \sum_{k} c_{n, k} x_{k}$ , and regularity

Definition

A summability matrix, also called a Toeplitz matrix, is a function

$c : N \times N \to R, (n, k) \mapsto c_{n, k},$

with finite row support: for every $n \in N$ there is $R \in N$ such that $c_{n, k} = 0$ for every $k > R$ . Such an $R$ is called an admissible bound for row $n$ . Rows are indexed by $n$ and columns by $k$ ; the $k$ -th column of $c$ is the sequence $n \mapsto c_{n, k}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

The transform. Let $(x_{k})$ be a sequence of reals. The transform of $(x_{k})$ by $c$ is the sequence $(y_{n})$ given by

$y_{n} : = \sum_{k = 0}^{R} c_{n, k} x_{k},$

where $R$ is any admissible bound for row $n$ and the sum is the finite sum of Finite sums and finite products, by recursion. We write $\sum_{k} c_{n, k} x_{k}$ for this value.

This is well defined, and the check is the reason finite row support is part of the definition. Suppose $R \leq R^{'}$ are both admissible bounds for row $n$ . Splitting the longer sum (Laws of finite sums and finite products) gives

$\sum_{k = 0}^{R^{'}} c_{n, k} x_{k} = \sum_{k = 0}^{R} c_{n, k} x_{k} + \sum_{k = R + 1}^{R^{'}} c_{n, k} x_{k},$

and every term of the second sum has $k > R$ , hence $c_{n, k} = 0$ and $c_{n, k} x_{k} = 0$ ; a finite sum all of whose terms are $0$ is $0$ , by the scaling law of Laws of finite sums and finite products with $λ = 0$ . So the two agree. For two arbitrary admissible bounds $R_{1}, R_{2}$ , the order on $N$ is total ( $\leq$ is a linear order on $N$ ), so each may be compared with the larger of the two, and the three values agree. Hence $y_{n}$ is a single well-determined real for each $n$ , and $(y_{n})$ is a sequence of reals.

Two instances of the transform have their own names. The row sum of row $n$ is $\sum_{k} c_{n, k}$ , the transform of the constant sequence $1$ ; the row absolute sum is $\sum_{k} ∣ c_{n, k} ∣$ , the transform of the constant sequence $1$ by the matrix $(n, k) \mapsto ∣ c_{n, k} ∣$ , which again has finite row support with the same admissible bounds.

Regularity. The summability matrix $c$ is regular when for every convergent sequence $(x_{k})$ of reals the transform $(y_{n})$ converges and

$\lim_{n} y_{n} = \lim_{k} x_{k} .$

Both limits are asserted to exist there: the right-hand one by hypothesis on $(x_{k})$ , the left-hand one as part of the condition. Limits of real sequences are unique (A sequence has at most one limit, Limits and Cauchy sequences of reals), so the condition is unambiguous.

Remarks

Finite row support is not a technical convenience, it is what makes the transform mean anything at all at this point of the library. The classical definition allows every row to be an infinite series $\sum_{k = 0}^{\infty} c_{n, k} x_{k}$ , and asks that each such series converge. Series are not defined anywhere in this library yet; they arrive on the next page of this track. Every sum above is therefore a finite sum in the sense of Finite sums and finite products, by recursion, and no convergence question arises inside a row.
What is lost, and what is not. The restriction excludes matrices such as the Abel and Borel means, whose rows are genuine series. It does not exclude anything needed here: the Cesaro matrix has $c_{n, k} = 1 / (n + 1)$ for $k \leq n$ and $0$ beyond (The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem), so row $n$ has admissible bound $n$ ; and the counterexample of A summability matrix failing exactly one Silverman-Toeplitz condition and transforming a convergent sequence to a divergent one ↗ has two nonzero entries per row. A summability matrix with only finitely many nonzero entries per row is regular iff each column tends to $0$ , the row sums tend to $1$ , and the row absolute sums are uniformly bounded characterises regularity within this class.
Regularity says the transform is a genuine generalisation of the limit. It is exactly the condition that $c$ never changes the value of a limit that already exists. It says nothing at all about sequences that do not converge, and the interest of such matrices is precisely that a regular one may still assign a value to a divergent sequence: the Cesaro matrix does so for the alternating sequence (The Cesaro means of $(- 1)^{k}$ converge to $0$ although the sequence diverges ↗).
A matrix that fails regularity can fail it in two different ways, by changing a limit or by destroying convergence outright. A summability matrix failing exactly one Silverman-Toeplitz condition and transforming a convergent sequence to a divergent one ↗ does the second, which is the stronger failure.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

A summability matrix with only finitely many nonzero entries per row is regular iff each column tends to $0$ , the row sums tend to $1$ , and the row absolute sums are uniformly bounded

Statement

Let $c$ be a summability matrix (A summability (Toeplitz) matrix, the transformed sequence $y_{n} = \sum_{k} c_{n, k} x_{k}$ , and regularity), so that every row has only finitely many nonzero entries. Then $c$ is regular if and only if all three of the following hold:

(Columns are null.) For every $k \in N$ the $k$ -th column converges with $\lim_{n} c_{n, k} = 0$ .
(Row sums tend to $1$ .) The sequence of row sums converges with $\lim_{n} \sum_{k} c_{n, k} = 1$ .
(Row absolute sums are uniformly bounded.) There is $M \in R$ with $\sum_{k} ∣ c_{n, k} ∣ \leq M$ for every $n \in N$ .

In 1 and 2 the existence of the limit is part of the assertion. The notation is licensed by uniqueness of limits of real sequences (A sequence has at most one limit).

Condition 3 is the one that cannot be seen on any single sequence: 1 and 2 are read off two particular convergent inputs, while the necessity of 3 needs a sequence built against the matrix, by a gliding hump.

Facts & Assumptions

Given: A summability matrix $c : N \times N \to R$ with finite row support. For a sequence $(x_{k})$ of reals we write $(y_{n})$ for its transform, $y_{n} = \sum_{k} c_{n, k} x_{k}$ , and $r_{n} : = \sum_{k} ∣ c_{n, k} ∣$ for the row absolute sums.

[L1]

Summability matrices: finite row support, the transform and its independence of the admissible row bound used, the row sum, the row absolute sum, and regularity (A summability (Toeplitz) matrix, the transformed sequence $y_{n} = \sum_{k} c_{n, k} x_{k}$ , and regularity, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L2]

Finite sums (Finite sums and finite products, by recursion) and their laws: additivity, scaling with $\sum_{k < d} λ = d λ$ , splitting, and monotonicity in the terms (Laws of finite sums and finite products).

[L3]

Triangle inequality for finite sums (Triangle inequality for finite sums); $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ , $∣ u ∣ \geq 0$ , $∣ ∣ u ∣ ∣ = ∣ u ∣$ and $∣ u ∣ = u$ for $u \geq 0$ (Basic properties of the absolute value).

[L4]

Convergence: for every real $ε > 0$ there is $N$ beyond which the terms are within $ε$ of the limit, the rational and real formulations agreeing (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences, The rationals embed densely in the reals); limits are unique (A sequence has at most one limit); a sequence that is eventually $0$ converges to $0$ .

[L5]

Every convergent sequence of reals is bounded (Every convergent sequence is bounded).

[L6]

Algebra of limits for sums and scalar multiples (Algebra of limits: sums, scalar multiples, products and quotients).

[L7]

Archimedean property of $R$ : for every real $z$ there is a natural $n \geq 1$ with $z < n \cdot 1_{R}$ (Every complete ordered field is Archimedean); equivalently, for every real $ε > 0$ there is a natural $J \geq 1$ with $1 / J < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L8]

Least upper bounds: a nonempty subset of $R$ bounded above has a supremum, which dominates every element of the set (Complete ordered field (least-upper-bound property), Upper bound, least upper bound, and strict upper bound, Lower bound, bounded below, bounded set).

[L9]

Every nonempty finite set of reals has a maximum, which lies in the set and dominates it (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L10]

Recursion theorem (The recursion theorem); well-ordering principle (The well-ordering principle); induction principle (The principle of mathematical induction); totality of the order on $N$ ( $\leq$ is a linear order on $N$ ); and consecutive comparisons suffice for strict increase, with $k_{j} \geq j$ for a strictly increasing index map (A strictly increasing index map satisfies $n_{k} \geq k$ ).

[L11]

Order arithmetic: $a > 0$ gives $a^{- 1} > 0$ and $0 < a < b$ gives $0 < b^{- 1} < a^{- 1}$ (Inverses of positives are positive, and reciprocation reverses order); for $c > 0$ , $a \leq b$ if and only if $a c \leq b c$ (Sign rules for products and monotonicity of multiplication); adding a constant preserves the order and inequalities add (Order is preserved by adding a constant and by adding inequalities); canonical naturals are positive and increasing (Canonical naturals are positive and strictly increasing); the order is total and transitive (Complete ordered field (least-upper-bound property), Ordered field). In each clause above, Sign rules for products and monotonicity of multiplication and Order is preserved by adding a constant and by adding inequalities state the STRICT forms and only those; the nonstrict forms used below are those together with the equality cases, which trichotomy settles, the order being total (Ordered field).

Proof

technique · direct

1.1

Sufficiency. Assume conditions 1, 2 and 3, let $(x_{k})$ converge to $L$ , and let $ε > 0$ be an arbitrary real; fix $M \geq 0$ as in condition 3, and fix $D \geq 0$ with $∣ x_{k} - L ∣ \leq D$ for every $k$ , which exists because a convergent sequence is bounded.

L1L3L4L5L11choose

1.2

Choose $K \in N$ with $∣ x_{k} - L ∣ < ε (3 (M + 1))^{- 1}$ for every $k \geq K$ .

L4L11choose

1.3

For every $n$ , choosing an admissible bound $R \geq K$ for row $n$ , one has $y_{n} - L = \sum_{k = 0}^{R} c_{n, k} (x_{k} - L) + (\sum_{k} c_{n, k} - 1) L$ , since $\sum_{k = 0}^{R} c_{n, k} L = L \sum_{k} c_{n, k}$ .

L1L2

1.4

Necessity. The remaining steps, apart from 2.1, 2.2, 2.3 and 3.1 which finish the sufficiency argument above, assume instead that $c$ is regular.

1.5

Suppose, towards a contradiction, that the row absolute sums $(r_{n})$ are not bounded above, that is, for every $T \in R$ there is $n$ with $r_{n} > T$ .

L1L11assume-contra

1.6

For each $n$ the set of admissible bounds for row $n$ is a nonempty subset of $N$ , so it has a least element $ρ (n)$ ; thus $c_{n, k} = 0$ for every $k > ρ (n)$ , and every $R \geq ρ (n)$ is admissible for row $n$ .

L1L10

2.1

For every $n$ : $∣ \sum_{k = 0}^{R} c_{n, k} (x_{k} - L) ∣ \leq \sum_{k < K} ∣ c_{n, k} ∣ ∣ x_{k} - L ∣ + \sum_{k = K}^{R} ∣ c_{n, k} ∣ ∣ x_{k} - L ∣ \leq D \sum_{k < K} ∣ c_{n, k} ∣ + ε (3 (M + 1))^{- 1} r_{n} \leq D \sum_{k < K} ∣ c_{n, k} ∣ + ε / 3$ , the last step because $r_{n} \leq M < M + 1$ .

step 1.1step 1.2step 1.3L2L3L11

2.2

By condition 1 each of the finitely many columns $k < K$ satisfies $c_{n, k} \to 0$ , hence $∣ c_{n, k} ∣ \to 0$ since $∣ ∣ c_{n, k} ∣ - 0 ∣ = ∣ c_{n, k} - 0 ∣$ ; a sum of finitely many null sequences is null, by induction on the number of summands, so $\sum_{k < K} ∣ c_{n, k} ∣ \to 0$ in $n$ and there is $N_{1}$ with $\sum_{k < K} ∣ c_{n, k} ∣ < ε (3 (D + 1))^{- 1}$ for every $n \geq N_{1}$ .

step 1.1step 1.2L3L4L6L10L11choose

2.3

By condition 2 there is $N_{2}$ with $∣ \sum_{k} c_{n, k} - 1 ∣ < ε (3 (∣ L ∣ + 1))^{- 1}$ for every $n \geq N_{2}$ , so that $∣ (\sum_{k} c_{n, k} - 1) L ∣ < ε / 3$ for such $n$ .

step 1.1L3L4L11choose

2.4

Condition 1 holds. Fix $k$ and let $e$ be the sequence with $e_{k} = 1$ and $e_{j} = 0$ for $j \neq k$ ; it is eventually $0$ , so it converges to $0$ , and its transform at row $n$ is $c_{n, k}$ because every other term of the row sum vanishes. Regularity gives $\lim_{n} c_{n, k} = 0$ .

step 1.4L1L2L4

2.5

Condition 2 holds. The constant sequence with value $1$ converges to $1$ and its transform at row $n$ is the row sum $\sum_{k} c_{n, k}$ , so regularity gives $\lim_{n} \sum_{k} c_{n, k} = 1$ .

step 1.4L1L4

3.1

For every $n$ beyond both $N_{1}$ and $N_{2}$ : $∣ y_{n} - L ∣ \leq D \sum_{k < K} ∣ c_{n, k} ∣ + ε / 3 + ε / 3 < ε / 3 + ε / 3 + ε / 3 = ε$ ; as $ε$ was arbitrary, $y_{n} \to L$ , and as $(x_{k})$ was an arbitrary convergent sequence, $c$ is regular.

step 1.3step 2.1step 2.2step 2.3L1L4L10L11

3.2

Each column converges, hence is bounded, so $M_{k} : = \sup {∣ c_{n, k} ∣ : n \in N}$ exists in $R$ for every $k$ ; putting $B_{m} : = \sum_{k = 0}^{m} M_{k}$ one has $\sum_{k = 0}^{m} ∣ c_{n, k} ∣ \leq B_{m}$ for every $n$ and every $m$ , and $B_{m} \geq 0$ .

step 2.4L2L3L5L8L11

4.1

Define by recursion $k_{0} : = 0$ , $n_{0} : = 0$ and, for $j \geq 1$ , first $T_{j} : =$ the larger of $\max {r_{n} : n \leq n_{j - 1}}$ and $B_{k_{j - 1}} + j (j + B_{k_{j - 1}})$ , then $n_{j} : = \min {n : r_{n} > T_{j}}$ , which exists by step 1.5 and well-ordering, and then $k_{j} : =$ the larger of $k_{j - 1} + 1$ and $ρ (n_{j})$ ; then $n_{j} > n_{j - 1}$ , because $r_{n_{j}}$ exceeds every $r_{n}$ with $n \leq n_{j - 1}$ , and $k_{j - 1} < k_{j}$ with $c_{n_{j}, k} = 0$ for every $k > k_{j}$ , and $r_{n_{j}} > B_{k_{j - 1}} + j (j + B_{k_{j - 1}})$ .

step 1.5step 1.6step 3.2L9L10L11construct

5.1

Since $(k_{j})$ is strictly increasing with $k_{0} = 0$ , every $k \geq 1$ lies in exactly one block $k_{j - 1} < k \leq k_{j}$ with $j \geq 1$ ; define $x_{0} : = 0$ and, for $k$ in the $j$ -th block, $x_{k} : = sgn (c_{n_{j}, k}) j^{- 1}$ , where $sgn (t) : = 1$ for $t \geq 0$ and $sgn (t) : = - 1$ for $t < 0$ . Then $∣ x_{k} ∣ \leq 1$ for every $k$ , and $x_{k} \to 0$ : given $ε > 0$ , take $J \geq 1$ with $1 / J < ε$ , and every $k > k_{J}$ lies in a block with index $j > J$ , so $∣ x_{k} ∣ = 1 / j < 1 / J < ε$ .

step 4.1L3L7L10L11construct

6.1

For every $j \geq 1$ : the terms of $y_{n_{j}}$ with $k > k_{j}$ vanish, so $y_{n_{j}} = \sum_{k \leq k_{j - 1}} c_{n_{j}, k} x_{k} + \sum_{k_{j - 1} < k \leq k_{j}} c_{n_{j}, k} x_{k}$ ; the second sum equals $j^{- 1} \sum_{k_{j - 1} < k \leq k_{j}} ∣ c_{n_{j}, k} ∣ = j^{- 1} (r_{n_{j}} - \sum_{k \leq k_{j - 1}} ∣ c_{n_{j}, k} ∣) \geq j^{- 1} (r_{n_{j}} - B_{k_{j - 1}})$ , while the first has absolute value at most $\sum_{k \leq k_{j - 1}} ∣ c_{n_{j}, k} ∣ \leq B_{k_{j - 1}}$ ; hence $y_{n_{j}} \geq j^{- 1} (r_{n_{j}} - B_{k_{j - 1}}) - B_{k_{j - 1}} > j^{- 1} j (j + B_{k_{j - 1}}) - B_{k_{j - 1}} = j$ .

step 3.2step 4.1step 5.1L2L3L11

7.1

But $(x_{k})$ converges to $0$ , so regularity makes $(y_{n})$ converge, hence bounded, so some $S \in R$ has $∣ y_{n} ∣ \leq S$ for every $n$ ; taking $j$ with $j \cdot 1_{R} > S$ , available by the Archimedean property, step 6.1 gives $y_{n_{j}} > j > S$ , a contradiction.

step 1.4step 5.1step 6.1L1L5L7L11

8.1

The assumption of step 1.5 is therefore untenable and condition 3 holds; with steps 2.4 and 2.5, regularity implies all three conditions.

step 2.4step 2.5step 7.1discharge-contradiction

9.1

Sufficiency is step 3.1 and necessity is step 8.1, so $c$ is regular exactly when conditions 1, 2 and 3 all hold.

step 3.1step 8.1∎

Remarks

No one of the three conditions follows from the other two, and each is tested by a different input. Condition 1 is what a single nonzero coordinate detects, condition 2 what the constant sequence detects, and condition 3 is invisible to any fixed sequence: for each individual bounded input a matrix with unbounded row absolute sums may behave perfectly well, and the failure only appears against a sequence whose signs are chosen row by row. The substantial case is 3, and it is A summability matrix failing exactly one Silverman-Toeplitz condition and transforming a convergent sequence to a divergent one ↗, which exhibits a matrix satisfying 1 and 2 and failing 3, together with a null sequence whose transform diverges. The other two are settled in a line each and are recorded here rather than given items of their own: the matrix with $c_{n, 0} = 1$ and every other entry $0$ has row sums and row absolute sums constantly $1$ , so it satisfies 2 and 3, while its $0$ -th column is constantly $1$ and fails 1; and the zero matrix has null columns and row absolute sums $0$ , so it satisfies 1 and 3, while its row sums are constantly $0$ and fail 2.
The gliding hump. The witness of the necessity argument is built in blocks: on the $j$ -th block its terms have modulus $1 / j$ , so the sequence tends to $0$ , and their signs are chosen to align with the entries of one row $n_{j}$ , so that on that row the transform picks up almost the whole row absolute sum, divided by $j$ . Choosing $r_{n_{j}}$ larger than $j$ times its own head bound makes the transform exceed $j$ there. The bound $B_{k_{j - 1}}$ on the head is available before $n_{j}$ is chosen, because it depends only on the earlier block boundary, and that is what keeps the construction from circling.
No choice is used. Every stage of the recursion takes a least element or a maximum of a finite set; the row bound $ρ (n)$ is the least admissible one; and $M_{k}$ is a supremum, that is, a definite element of $R$ rather than a selected bound.

CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem

Statement

Define $c_{n, k} : = (n + 1)^{- 1}$ for $k \leq n$ and $c_{n, k} : = 0$ for $k > n$ . Then:

$c$ is a summability matrix (A summability (Toeplitz) matrix, the transformed sequence $y_{n} = \sum_{k} c_{n, k} x_{k}$ , and regularity), with $n$ an admissible bound for row $n$ ;
the transform of a sequence $(x_{k})$ by $c$ is exactly its sequence of Cesaro means (The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability), $y_{n} = σ_{n}$ ;
$c$ satisfies the three conditions of A summability matrix with only finitely many nonzero entries per row is regular iff each column tends to $0$ , the row sums tend to $1$ , and the row absolute sums are uniformly bounded, so $c$ is regular.

Consequently every convergent sequence has $σ_{n} \to \lim_{k} x_{k}$ , which is a second proof of If $x_{k} \to L$ then $σ_{n} \to L$ : convergence implies $(C, 1)$ -summability to the same value, obtained from the general characterisation rather than from a direct estimate.

Facts & Assumptions

Given: The matrix $c$ with $c_{n, k} = (n + 1)^{- 1}$ for $k \leq n$ and $c_{n, k} = 0$ for $k > n$ .

[L1]

Summability matrices: finite row support, the transform, the row sum, the row absolute sum and regularity (A summability (Toeplitz) matrix, the transformed sequence $y_{n} = \sum_{k} c_{n, k} x_{k}$ , and regularity, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L2]

The Cesaro means $σ_{n} = (n + 1)^{- 1} \sum_{k = 0}^{n} x_{k}$ (The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability).

[L3]

Silverman-Toeplitz: a summability matrix is regular exactly when every column tends to $0$ , the row sums tend to $1$ , and the row absolute sums are uniformly bounded (A summability matrix with only finitely many nonzero entries per row is regular iff each column tends to $0$ , the row sums tend to $1$ , and the row absolute sums are uniformly bounded).

[L4]

Finite sums and their laws, in particular $\sum_{k < d} λ = d λ$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L5]

Convergence, and the fact that a constant sequence converges to its value (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L6]

[L7]

Order arithmetic: $(n + 1) \cdot 1_{R} > 0$ for every $n \in N$ (Canonical naturals are positive and strictly increasing); a positive element is invertible with positive inverse, and reciprocation reverses the order (Inverses of positives are positive, and reciprocation reverses order); $∣ u ∣ = u$ for $u \geq 0$ (Basic properties of the absolute value); the order is total and transitive (Complete ordered field (least-upper-bound property), Ordered field).

Proof

technique · direct

1.1

Row $n$ of $c$ vanishes at every $k > n$ , so $n$ is an admissible bound for row $n$ and $c$ is a summability matrix; its transform is $y_{n} = \sum_{k = 0}^{n} (n + 1)^{- 1} x_{k} = (n + 1)^{- 1} \sum_{k = 0}^{n} x_{k} = σ_{n}$ .

L1L2L4

1.2

For every $n$ the canonical natural $(n + 1) \cdot 1_{R}$ is positive, hence invertible with $(n + 1)^{- 1} > 0$ ; so $∣ c_{n, k} ∣ = c_{n, k}$ for all $n, k$ .

2.1

Columns are null. Fix $k$ and let $ε > 0$ ; choose $m \geq 1$ with $1 / m < ε$ . For $n \geq m$ one has $n + 1 > m$ , so $∣ c_{n, k} - 0 ∣ \leq (n + 1)^{- 1} < 1 / m < ε$ , the case $n < k$ giving $c_{n, k} = 0$ outright. Hence $\lim_{n} c_{n, k} = 0$ .

step 1.2L5L6L7

2.2

Row sums tend to $1$ . For every $n$ , $\sum_{k} c_{n, k} = \sum_{k = 0}^{n} (n + 1)^{- 1} = (n + 1) (n + 1)^{- 1} = 1$ , a constant sequence, which converges to $1$ .

step 1.1step 1.2L1L4L5

2.3

Row absolute sums are uniformly bounded. For every $n$ , $\sum_{k} ∣ c_{n, k} ∣ = \sum_{k} c_{n, k} = 1 \leq 1$ .

step 1.1step 1.2L1L4

3.1

All three conditions hold, so $c$ is regular.

step 2.1step 2.2step 2.3L3

4.1

Therefore, for every convergent sequence $(x_{k})$ , the transform $(σ_{n})$ converges with $\lim_{n} σ_{n} = \lim_{k} x_{k}$ .

step 1.1step 3.1L1∎

Remarks

Which condition is doing what. For the Cesaro matrix the row sums and the row absolute sums are not merely convergent and bounded, they are constantly $1$ ; the whole content is that the columns are null, that is, that each single term contributes a weight $1 / (n + 1)$ which fades away. That is the precise sense in which averaging forgets any finite head, and it is why FALSE: if the Cesaro means of a sequence converge then the sequence converges is false: forgetting the head is not the same as recovering the sequence.
Two proofs, two costs. If $x_{k} \to L$ then $σ_{n} \to L$ : convergence implies $(C, 1)$ -summability to the same value is proved directly by a head-and-tail estimate, with no machinery at all; A summability matrix with only finitely many nonzero entries per row is regular iff each column tends to $0$ , the row sums tend to $1$ , and the row absolute sums are uniformly bounded proves the same estimate once for every weighting and then reads the Cesaro case off three trivial verifications. Both are kept, because the direct proof is what a reader should see first and the general one is what generalises.
A weighting with unbounded row absolute sums need not be regular, and the Cesaro matrix is as far from that as possible, its rows being nonnegative and summing to $1$ . See A summability matrix failing exactly one Silverman-Toeplitz condition and transforming a convergent sequence to a divergent one ↗ for the contrast.

RemarkRemark: AI-generatedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Which of the five completeness properties carry the Archimedean property on their own, and which must be handed it

The statement of For an ordered field the five completeness properties are equivalent, provided the Archimedean property is assumed alongside nested intervals and Cauchy completeness attaches the Archimedean property to two of its five clauses and not to the other three. This remark says exactly why, clause by clause, and records what is proved on this page rather than what is customary.

The three that carry it. Each of the following is proved here, with no Archimedean hypothesis anywhere in sight:

(LUB) implies the Archimedean property. This is claim 1 of An ordered field with the least-upper-bound property has the nested interval property and is Archimedean, which is Every complete ordered field is Archimedean applied to the field: a complete ordered field is Archimedean, because otherwise the canonical naturals would be a nonempty set $A$ bounded above, and then $\sup A - 1$ , being smaller than $\sup A$ , is not an upper bound of $A$ , so some $n \cdot 1_{F}$ exceeds it and $(n + 1) \cdot 1_{F}$ exceeds $\sup A$ .
(BW) implies the Archimedean property, by Bolzano-Weierstrass alone forces the Archimedean property, so it needs no separate Archimedean hypothesis. If the canonical naturals were bounded they would form a bounded sequence, and every subsequence of it has consecutive terms at distance at least $1$ , so no subsequence converges.
(MCT) implies the Archimedean property, by The monotone convergence property alone forces the Archimedean property, so it carries no separate Archimedean hypothesis. If the canonical naturals were bounded above they would be a nondecreasing bounded sequence, hence convergent, hence Cauchy, which the gap of $1$ between consecutive terms forbids.

The two that do not. Neither (NIP) nor (CC) implies the Archimedean property, and one field refutes both: the formal Laurent series field $K = R ((t^{- 1}))$ is not Archimedean ( $R ((t^{- 1}))$ is non-Archimedean, and the monomials $t^{- k}$ are cofinal below its positive elements), has (CC) (Every Cauchy sequence in $R ((t^{- 1}))$ converges: $K$ is sequentially Cauchy complete) and has (NIP) in the shrinking form of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness ( $R ((t^{- 1}))$ has the nested interval property for lengths tending to $0$ ). The consequences are the two false statements of this page, FALSE: the nested interval property alone implies the least-upper-bound property and FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property: without the Archimedean hypothesis neither clause 2 nor clause 4 of the equivalence theorem implies clause 1.

What distinguishes the two groups. (LUB), (BW) and (MCT) each quantify over an object that is assumed only to be bounded: a bounded set, a bounded sequence, a nondecreasing sequence bounded above. In a non-Archimedean field the canonical naturals are such an object, so each of the three can be tested against them directly, and each fails on them at once. (NIP) and (CC) quantify instead over data that are already forced together: nested intervals whose lengths tend to $0$ in the field, and sequences whose terms get arbitrarily close to each other in the field. In a non-Archimedean field that is a much stronger hypothesis than it looks, because "arbitrarily close" now means below every infinitesimal as well; so few sequences and few interval families qualify, and the ones that do converge for reasons that have nothing to do with the naturals being cofinal.

Two corollaries worth stating plainly.

An Archimedean hypothesis is never needed alongside (LUB), (BW) or (MCT), and writing one there is not merely redundant but misleading, since it suggests the property is weaker than it is.
The customary phrase "complete ordered field" is ambiguous in exactly one place, and that place is (CC). This library resolves it by reserving complete for the least-upper-bound property (Complete ordered field (least-upper-bound property)) and always writing Cauchy complete for the other, as Every Cauchy sequence in $R ((t^{- 1}))$ converges: $K$ is sequentially Cauchy complete does. A text that says "the reals are the unique complete ordered field" and means (CC) is stating something false, and $K$ is the counterexample.

A note on what is not claimed. Nothing above says that (NIP) and (CC) are equivalent to each other, or that either is equivalent to the Archimedean property's negation, or that $K$ is the only witness. What is proved is the implication pattern of For an ordered field the five completeness properties are equivalent, provided the Archimedean property is assumed alongside nested intervals and Cauchy completeness and the two failures just named.

5 · Examples, counterexamples and false statements

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)Open item page →

FALSE: the nested interval property alone implies the least-upper-bound property

Statement

False claim: every ordered field with the nested interval property (NIP) of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness has the least-upper-bound property (LUB).

This is clause 2 of For an ordered field the five completeness properties are equivalent, provided the Archimedean property is assumed alongside nested intervals and Cauchy completeness with its Archimedean hypothesis deleted, and the deletion is exactly what makes it false. The witness is the formal Laurent series field $K = R ((t^{- 1}))$ , which satisfies (NIP) and has no least upper bound for the set of its own canonical naturals.

Note that the false claim is being refuted in the shrinking form of (NIP), which is the weaker hypothesis and therefore makes the implication stronger.

Facts & Assumptions

Given: The formal Laurent series field $K = R ((t^{- 1}))$ .

[L1]

$K$ is an ordered field ( $R ((t^{- 1}))$ is an ordered field, ordered by the sign of the leading coefficient).

[L2]

Every nested sequence of closed intervals of $K$ whose lengths tend to $0$ in $K$ has exactly one point in its intersection ( $R ((t^{- 1}))$ has the nested interval property for lengths tending to $0$ ); intervals, nesting and lengths tending to $0$ in an ordered field are as in Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field, and (NIP) asks exactly that such an intersection be nonempty (The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness).

[L3]

$K$ is not a complete ordered field: the set $A = {n \cdot 1_{K} : n \in N}$ is nonempty and bounded above by $t$ and has no least upper bound in $K$ ( $R ((t^{- 1}))$ does not have the least-upper-bound property; its canonical naturals have no supremum, Complete ordered field (least-upper-bound property)).

[L4]

$K$ is not Archimedean, since $n \cdot 1_{K} < t$ for every natural $n$ ( $R ((t^{- 1}))$ is non-Archimedean, and the monomials $t^{- k}$ are cofinal below its positive elements, Archimedean ordered field).

[L5]

For an ordered field, the Archimedean property together with (NIP) does imply (LUB) (For an ordered field the five completeness properties are equivalent, provided the Archimedean property is assumed alongside nested intervals and Cauchy completeness, clause 2 implies clause 1).

Refutation

technique · direct

1.1

$K$ is an ordered field.

1.2

$K$ has (NIP): any nested sequence of closed intervals of $K$ whose lengths tend to $0$ in $K$ has a point in its intersection, indeed exactly one.

1.3

$K$ does not have (LUB), the set of its canonical naturals being nonempty, bounded above and without a least upper bound.

2.1

So $K$ is an ordered field with (NIP) and without (LUB), and the claim is false.

step 1.1step 1.2step 1.3

3.1

What fails in $K$ is precisely the hypothesis that the claim deleted: $K$ is not Archimedean, and with that hypothesis restored the implication is true.

step 1.1L4L5∎

Remarks

The failure is not an accident of one field. By An ordered field with the least-upper-bound property has the nested interval property and is Archimedean every field with (LUB) is Archimedean, so any witness at all must be non-Archimedean; and in a non-Archimedean field the shrinking hypothesis in (NIP) is a severe restriction, because a length that tends to $0$ in the order of the field must get below every infinitesimal. That is why checking shrinking (NIP) in $K$ is substantive, and why $K$ can satisfy (NIP) while failing (LUB) at all.
$R (t)$ will not do as a witness, although it is the library's other non-Archimedean ordered field (Not every ordered field is Archimedean). Nothing in this library establishes any nested interval property for it, and the page that built $K$ says why a new field was constructed rather than reusing that one.
The companion failure is FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property, refuted by the same field. Together they are the exact content of the Archimedean hypotheses in clauses 2 and 4 of For an ordered field the five completeness properties are equivalent, provided the Archimedean property is assumed alongside nested intervals and Cauchy completeness.

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property

Statement

False claim: every ordered field in which every Cauchy sequence converges, that is, every ordered field with (CC) as in The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness, has the least-upper-bound property (LUB).

This is clause 4 of For an ordered field the five completeness properties are equivalent, provided the Archimedean property is assumed alongside nested intervals and Cauchy completeness with its Archimedean hypothesis deleted. The witness is again the formal Laurent series field $K = R ((t^{- 1}))$ : every Cauchy sequence in $K$ converges in $K$ , and $K$ has no least upper bound for the set of its own canonical naturals.

This is the sharpest of the failures on this page, because "complete" is the word most often used loosely for both properties at once. In $R$ they coincide; in an ordered field they do not, and the difference is exactly the Archimedean property.

Facts & Assumptions

Given: The formal Laurent series field $K = R ((t^{- 1}))$ .

[L1]

$K$ is an ordered field ( $R ((t^{- 1}))$ is an ordered field, ordered by the sign of the leading coefficient).

[L2]

Every sequence in $K$ that is Cauchy in $K$ converges in $K$ (Every Cauchy sequence in $R ((t^{- 1}))$ converges: $K$ is sequentially Cauchy complete); Cauchyness and convergence in an ordered field are as in Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field, and that is exactly (CC) (The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness).

[L3]

[L4]

[L5]

For an ordered field, the Archimedean property together with (CC) does imply (LUB) (For an ordered field the five completeness properties are equivalent, provided the Archimedean property is assumed alongside nested intervals and Cauchy completeness, clause 4 implies clause 1).

Refutation

technique · direct

1.1

$K$ is an ordered field.

1.2

$K$ has (CC): every Cauchy sequence in $K$ converges in $K$ .

1.3

$K$ does not have (LUB), the set of its canonical naturals being nonempty, bounded above and without a least upper bound.

2.1

So $K$ is an ordered field with (CC) and without (LUB), and the claim is false.

step 1.1step 1.2step 1.3

3.1

What fails in $K$ is precisely the hypothesis that the claim deleted: $K$ is not Archimedean, and with that hypothesis restored the implication is true.

step 1.1L4L5∎

Remarks

Where the thresholds are read is what makes this possible. Cauchyness in $K$ is tested against every positive element of $K$ , including the infinitesimals (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field), so the condition is much stronger in $K$ than the same words read with rational thresholds. It is strong enough that only sequences whose coefficients freeze can satisfy it, and those all converge. Meanwhile the canonical naturals, which are what (LUB) fails on, are not Cauchy at all, so (CC) never gets a chance to see them.
The three properties $K$ has and the three it lacks. It has (CC) and (NIP) in the shrinking form ( $R ((t^{- 1}))$ has the nested interval property for lengths tending to $0$ ) and it is an ordered field; it lacks (LUB), and hence also (BW) and (MCT), each of which would force it to be Archimedean (Bolzano-Weierstrass alone forces the Archimedean property, so it needs no separate Archimedean hypothesis, The monotone convergence property alone forces the Archimedean property, so it carries no separate Archimedean hypothesis).
A reader who wants a single sentence: Cauchy completeness says the field has no holes that a sequence can point at; the least-upper-bound property says it has no holes at all. In a non-Archimedean field a sequence indexed by $N$ is too short to point at the holes.

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

FALSE: if the Cesaro means of a sequence converge then the sequence converges

Statement

False claim: if the Cesaro means $(σ_{n})$ of a sequence $(x_{k})$ of reals converge (The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability), then $(x_{k})$ converges.

The implication in the opposite direction is true and is If $x_{k} \to L$ then $σ_{n} \to L$ : convergence implies $(C, 1)$ -summability to the same value. The claim above asserts its converse, and it is refuted below by the alternating sequence $s_{k} = (- 1)^{k}$ , whose Cesaro means converge to $0$ while the sequence itself does not converge at all.

That is the whole reason Cesaro summability is worth defining: it is a strictly larger notion than convergence, consistent with it where both apply.

Facts & Assumptions

Given: The alternating sequence $(s_{k})$ of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ , the unique sequence of reals with $s_{0} = 1$ and $s_{σ (k)} = - s_{k}$ , together with the index maps $e$ and $o$ of that lemma; and its partial sums $S_{n} : = \sum_{k < n} s_{k}$ (Finite sums and finite products, by recursion), and its Cesaro means $σ_{n} = (n + 1)^{- 1} S_{n + 1}$ (The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability).

[L1]

The alternating sequence: $e$ and $o$ are the unique maps with $e_{0} = 0$ , $e_{σ (j)} = σ (σ (e_{j}))$ , $o_{0} = σ (0)$ , $o_{σ (j)} = σ (σ (o_{j}))$ ; both are strictly increasing; $N$ is the disjoint union of their ranges; $(s_{k})$ is the unique sequence with $s_{0} = 1$ and $s_{σ (k)} = - s_{k}$ ; $∣ s_{k} ∣ = 1$ , $s_{e_{j}} = 1$ and $s_{o_{j}} = - 1$ (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ).

[L2]

Finite sums: $\sum_{k < 0} s_{k} = 0$ and $\sum_{k < n + 1} s_{k} = \sum_{k < n} s_{k} + s_{n}$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L3]

Induction principle (The principle of mathematical induction).

[L4]

Cesaro means: $σ_{n} = (n + 1)^{- 1} \sum_{k = 0}^{n} x_{k}$ , and $\sum_{k = 0}^{n} = \sum_{k < n + 1}$ (The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability, Finite sums and finite products, by recursion).

[L5]

The alternating sequence does not converge: it is bounded and divergent, which is the refutation of FALSE: every bounded sequence converges, carried out there for the very same sequence, the one determined by $s_{0} = 1$ and $s_{σ (k)} = - s_{k}$ , which The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ shows is unique.

[L6]

Reciprocal Archimedean property (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ); convergence of a real sequence (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L7]

Order arithmetic: $(n + 1) \cdot 1_{R} > 0$ (Canonical naturals are positive and strictly increasing) hence invertible with positive inverse, and reciprocation reverses the order (Inverses of positives are positive, and reciprocation reverses order); $∣ u ∣ = u$ for $u \geq 0$ (Basic properties of the absolute value); the order is total and transitive (Complete ordered field (least-upper-bound property), Ordered field).

Refutation

technique · direct

1.1

By induction on $j$ , $o_{j} = σ (e_{j})$ and $e_{σ (j)} = σ (o_{j})$ : at $j = 0$ one has $o_{0} = σ (0) = σ (e_{0})$ , and $e_{σ (j)} = σ (σ (e_{j})) = σ (o_{j})$ follows from the first identity at $j$ , while $o_{σ (j)} = σ (σ (o_{j})) = σ (e_{σ (j)})$ carries the first identity to $σ (j)$ .

L1L3

1.2

The partial sums satisfy $S_{0} = 0$ and $S_{σ (n)} = S_{n} + s_{n}$ , and $σ_{n} = (n + 1)^{- 1} S_{σ (n)}$ .

L2L4L7

1.3

$(s_{k})$ does not converge.

2.1

By induction on $j$ : $S_{e_{j}} = 0$ and $S_{o_{j}} = 1$ . At $j = 0$ : $S_{e_{0}} = S_{0} = 0$ and $S_{o_{0}} = S_{σ (0)} = S_{0} + s_{0} = 1$ . For the step, $S_{e_{σ (j)}} = S_{σ (o_{j})} = S_{o_{j}} + s_{o_{j}} = 1 + (- 1) = 0$ and $S_{o_{σ (j)}} = S_{σ (e_{σ (j)})} = S_{e_{σ (j)}} + s_{e_{σ (j)}} = 0 + 1 = 1$ .

step 1.1step 1.2L1L3

3.1

Every natural number is $e_{j}$ for exactly one $j$ or $o_{j}$ for exactly one $j$ , so $S_{m} \in {0, 1}$ for every $m$ ; in particular $0 \leq S_{σ (n)} \leq 1$ for every $n$ .

step 1.1step 2.1L1

4.1

Hence $0 \leq σ_{n} = (n + 1)^{- 1} S_{σ (n)} \leq (n + 1)^{- 1}$ and so $∣ σ_{n} ∣ \leq (n + 1)^{- 1}$ for every $n$ .

step 1.2step 3.1L7

5.1

Given a real $ε > 0$ , choose $m \geq 1$ with $1 / m < ε$ ; for every $n \geq m$ one has $n + 1 > m$ and therefore $∣ σ_{n} - 0 ∣ \leq (n + 1)^{- 1} < 1 / m < ε$ . So $(σ_{n})$ converges to $0$ , that is, $(s_{k})$ is $(C, 1)$ -summable to $0$ .

step 4.1L4L6L7

6.1

So $(s_{k})$ has convergent Cesaro means and does not converge, and the claim is false.

step 1.3step 5.1∎

Remarks

What averaging destroys. The Cesaro mean of the first $n + 1$ terms of an alternating sequence is either $0$ or $1 / (n + 1)$ , because the terms cancel in pairs and at most one is left over. The oscillation is real and is not damped by any tail condition; it is simply invisible to the average. So the transform loses information, and no regular summability method can be expected to recover a limit that does not exist (The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem).
The worked computation of the means, with the values displayed, is The Cesaro means of $(- 1)^{k}$ converge to $0$ although the sequence diverges ↗.
A correct converse needs an extra hypothesis. The classical one is Tauberian: if the Cesaro means converge and in addition $k (x_{k} - x_{k - 1})$ is bounded, then $(x_{k})$ converges. No such theorem is proved in this library, and none may be cited from it; the statement is mentioned only to say what the repaired claim would look like.
The failure is not caused by unboundedness. The witness is bounded, with $∣ s_{k} ∣ = 1$ at every index. It is the same sequence that refutes the claim that bounded sequences converge (FALSE: every bounded sequence converges), and for the same underlying reason: boundedness forbids escaping, not oscillating.

Sources

Standard references

Recommended treatments; not extraction sources.

Equivalent Forms of Completeness

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness

Definition

Remarks

Sequence basics in an arbitrary ordered field: limits are unique, limits preserve non-strict inequalities, convergent sequences are Cauchy, Cauchy sequences are bounded, and a Cauchy sequence with a convergent subsequence converges

Statement

Facts & Assumptions

Proof

Remarks

An ordered field with the least-upper-bound property has the nested interval property and is Archimedean

Statement

Facts & Assumptions

Proof

Remarks

Nested intervals plus the Archimedean property imply Bolzano-Weierstrass, by repeated bisection

Statement

Facts & Assumptions

Proof

Remarks

Bolzano-Weierstrass alone forces the Archimedean property, so it needs no separate Archimedean hypothesis

Statement

Facts & Assumptions

Proof

Remarks

Bolzano-Weierstrass implies Cauchy completeness in any ordered field

Statement

Facts & Assumptions

Proof

Remarks

Cauchy completeness plus the Archimedean property imply the monotone convergence property

Statement

Facts & Assumptions

Proof

Remarks

The monotone convergence property alone forces the Archimedean property, so it carries no separate Archimedean hypothesis

Statement

Facts & Assumptions

Proof

Remarks

The monotone convergence property plus the Archimedean property imply the least-upper-bound property

Statement

Facts & Assumptions

Proof

Remarks

For an ordered field the five completeness properties are equivalent, provided the Archimedean property is assumed alongside nested intervals and Cauchy completeness

Statement

Facts & Assumptions

Proof

Remarks

The Cesaro means σn=(x0+⋯+xn)/(n+1)σn​=(x0​+⋯+xn​)/(n+1) and (C,1)(C,1)-summability

Definition

Remarks

If xk→Lxk​→L then σn→Lσn​→L: convergence implies (C,1)(C,1)-summability to the same value

Statement

Facts & Assumptions

Proof

Remarks

Stolz-Cesaro, ∞/∞∞/∞ form: if bkbk​ is strictly increasing and unbounded and (ak+1−ak)/(bk+1−bk)→L(ak+1​−ak​)/(bk+1​−bk​)→L then ak/bk→Lak​/bk​→L

Statement

Facts & Assumptions

Proof

Remarks

Stolz-Cesaro, 0/00/0 form: if bkbk​ is strictly decreasing to 00, ak→0ak​→0, and the difference quotient converges, then ak/bkak​/bk​ converges to the same value

Statement

Facts & Assumptions

Proof

Remarks

A summability (Toeplitz) matrix, the transformed sequence yn=∑kcn,kxkyn​=∑k​cn,k​xk​, and regularity

Definition

Remarks

A summability matrix with only finitely many nonzero entries per row is regular iff each column tends to 00, the row sums tend to 11, and the row absolute sums are uniformly bounded

Statement

Facts & Assumptions

Proof

Remarks

The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem

The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability

If $x_{k} \to L$ then $σ_{n} \to L$ : convergence implies $(C, 1)$ -summability to the same value

Stolz-Cesaro, $\infty / \infty$ form: if $b_{k}$ is strictly increasing and unbounded and $(a_{k + 1} - a_{k}) / (b_{k + 1} - b_{k}) \to L$ then $a_{k} / b_{k} \to L$

Stolz-Cesaro, $0 / 0$ form: if $b_{k}$ is strictly decreasing to $0$ , $a_{k} \to 0$ , and the difference quotient converges, then $a_{k} / b_{k}$ converges to the same value

A summability (Toeplitz) matrix, the transformed sequence $y_{n} = \sum_{k} c_{n, k} x_{k}$ , and regularity

A summability matrix with only finitely many nonzero entries per row is regular iff each column tends to $0$ , the row sums tend to $1$ , and the row absolute sums are uniformly bounded