How statement and proof provenance work

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✓ 10 results · all verified · 8 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: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passverified 2026-08-13 (gpt-5.6-sol-codex-subscription)↗ rests on later material (inherited)Open item page →

The rational function field $R (t)$ ordered by the eventual sign is an ordered field, worked out

Example

Let $R (t)$ be the field of fractions (For a field $F$ , $F (t) = Frac (F [t])$ is its rational function field; in particular $R (t) = Frac (R [t])$ ) of the polynomial ring $R [t]$ , and let

$P = {f \in R (t) : f \neq 0 and f (x) > 0 for all sufficiently large real x} .$

Not every ordered field is Archimedean proves that $(R (t), P)$ is an ordered field and that it is not Archimedean. This example works the order out in usable form. Three things are established below:

A computation rule. For $f = p / q$ with $p, q \in R [t]$ nonzero, $f \in P$ exactly when $lc (p) lc (q) > 0$ , where $lc$ is the leading coefficient. So comparing two rational functions is comparing one product of two real numbers.
That the rule is independent of the representative $p / q$ chosen, which is what makes it a definition of a function on $R (t)$ and not merely on pairs.
The two elements that make the field interesting: $t$ , which exceeds every canonical natural, and $1 / t$ , which is positive and lies below every positive rational. An element of the second kind is called an infinitesimal, and its existence is exactly the failure of the Archimedean property (Archimedean ordered field).

Facts & Assumptions

Given: The field $R (t)$ of fractions of $R [t]$ , whose elements are written $p / q$ with $p, q \in R [t]$ and $q \neq 0$ , with $p / q = p^{'} / q^{'}$ exactly when $p q^{'} = p^{'} q$ ; and the set $P$ above. For a nonzero $p \in R [t]$ , $lc (p)$ denotes its leading coefficient.

[L1]

$(R (t), P)$ is an ordered field, and $n \cdot 1 < t$ for every natural $n$ , so it is not Archimedean (Not every ordered field is Archimedean, Ordered field, Archimedean ordered field).

[L2]

For a commutative ring $R$ and a nonzero $f \in R [x]$ , $\deg f$ is the largest index carrying a nonzero coefficient and $lc (f)$ is that coefficient (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree). For a field $F$ the polynomial ring $F [t]$ is an integral domain (For a field $F$ , $F (t) = Frac (F [t])$ is its rational function field; in particular $R (t) = Frac (R [t])$ ), and $R$ is a field (The reals form a totally ordered field, Field). If $R$ is an integral domain and $f, g \in R [x]$ are nonzero, then $f g \neq 0$ , $\deg (f g) = \deg f + \deg g$ and $lc (f g) = lc (f) lc (g)$ (Over an integral domain, degrees add under multiplication of nonzero polynomials).

[L3]

In $R$ , a nonzero square is positive (Squares of nonzero elements are positive); a product of two nonzero reals is positive exactly when both are positive or both are negative (Sign rules for products and monotonicity of multiplication).

[L4]

In an ordered field, $f < g$ means $g - f \in P$ ; a positive element has a positive inverse (Inverses of positives are positive, and reciprocation reverses order, Ordered field).

[L5]

The canonical embedding of $Q$ into an ordered field is an order embedding, so a rational $q > 0$ names a positive element $q \cdot 1$ of $R (t)$ (The unique embedding of ℚ into an ordered field).

[L6]

In an ordered field $∣ u ∣ = u$ when $u \geq 0$ and $∣ u ∣ = - u$ when $u < 0$ (Absolute value in an ordered field), and $- ∣ u ∣ \leq u \leq ∣ u ∣$ for every $u$ (Basic properties of the absolute value); $R$ is a totally ordered field (The reals form a totally ordered field).

Verification

technique · direct

1.1

Let $p \in R [t]$ be nonzero, with $m = \deg p$ and $a = lc (p) \neq 0$ , so that $p (x) = a x^{m} + \sum_{i < m} a_{i} x^{i}$ for every real $x$ . If $m = 0$ then $p (x) = a$ for every $x$ . If $m \geq 1$ , put $C = \sum_{i < m} ∣ a_{i} ∣ \geq 0$ ; for $x \geq 1$ and $i < m$ one has $0 < x^{i} \leq x^{m - 1}$ , and $- ∣ a_{i} ∣ \leq a_{i} \leq ∣ a_{i} ∣$ , so $- ∣ a_{i} ∣ x^{m - 1} \leq a_{i} x^{i} \leq ∣ a_{i} ∣ x^{m - 1}$ , and adding these $m$ inequalities gives $- C x^{m - 1} \leq \sum_{i < m} a_{i} x^{i} \leq C x^{m - 1}$ . Hence if $a > 0$ then $p (x) \geq a x^{m} - C x^{m - 1} = x^{m - 1} (a x - C) > 0$ for every $x > \max (1, C / a)$ ; and if $a < 0$ , the same bound applied to $- p$ , whose leading coefficient is $- a > 0$ , gives $p (x) < 0$ for every $x > \max (1, C / (- a))$ . In every case there is a real $X_{p}$ such that $p (x) \neq 0$ and $p (x)$ has the sign of $lc (p)$ for every $x > X_{p}$ .

givenL2L3L6algebra

1.2

If $p / q = p^{'} / q^{'}$ then $p q^{'} = p^{'} q$ , so $lc (p) lc (q^{'}) = lc (p^{'}) lc (q)$ ; multiplying both sides by $lc (q) lc (q^{'})$ gives $lc (p) lc (q) \cdot lc (q^{'})^{2} = lc (p^{'}) lc (q^{'}) \cdot lc (q)^{2}$ , and both squares are positive, so $lc (p) lc (q)$ and $lc (p^{'}) lc (q^{'})$ have the same sign.

L2L3

2.1

For nonzero $p, q \in R [t]$ put $X = \max (X_{p}, X_{q})$ with $X_{p}, X_{q}$ as in step 1.1: for every $x > X$ neither $p$ nor $q$ vanishes, so $f = p / q$ has a value $f (x) = p (x) / q (x)$ there, and since $p (x)$ has the sign of $lc (p)$ and $q (x)$ the sign of $lc (q)$ , the sign of that value is the sign of $lc (p) lc (q)$ ; hence $f \in P$ exactly when $lc (p) lc (q) > 0$ .

step 1.1L3

3.1

The rule of step 2.1 is therefore independent of the representative and computes membership in $P$ ; combined with [L1] it computes the order: $p / q < p^{'} / q^{'}$ exactly when the numerator and denominator of $p^{'} / q^{'} - p / q$ , written in any representative, have leading coefficients of positive product.

step 1.2step 2.1L1L4

3.2

$1 / t \in P$ , since $lc (1) lc (t) = 1 > 0$ ; equivalently, $t \in P$ and inverses of positives are positive.

step 2.1L3L4

4.1

For every rational $q > 0$ : $q \cdot 1 - 1 / t = (q t - 1) / t$ , whose leading coefficients have product $q \cdot 1 = q > 0$ , so $1 / t < q \cdot 1$ . Together with step 3.2, $0 < 1 / t < q \cdot 1$ for every positive rational $q$ .

step 2.1step 3.1step 3.2L3L5

4.2

For every natural $n$ : $t - n \cdot 1$ has leading coefficients with product $1 > 0$ , so $n \cdot 1 < t$ ; and $t^{2} - t = t (t - 1)$ likewise gives $t < t^{2}$ .

step 2.1step 3.1L2L3

5.1

So $R (t)$ is an ordered field, computed by a single product of leading coefficients, in which $t$ is larger than every canonical natural and $1 / t$ is a positive infinitesimal.

step 3.1step 4.1step 4.2L1∎

Remarks

Why the eventual sign, and not the sign at a point. Evaluating at a fixed real $x_{0}$ is not even a function on all of $R (t)$ , since a rational function may have a pole at $x_{0}$ ; and even where evaluation is defined, its sign cannot give a positive cone on the field, because the nonzero rational function $t - x_{0}$ evaluates to $0$ , so trichotomy fails. The behaviour at $+ \infty$ is one representative-independent, multiplicative choice, and step 1.1 is exactly that statement.
What this field is and is not good for. It is the library's cheapest witness that an ordered field need not be Archimedean, and In $R (t)$ the rationals are not dense: no rational lies strictly between $0$ and $1 / t$ uses the infinitesimal $1 / t$ found above to show $Q$ need not be dense. It is not a witness for the completeness failures of FALSE: the nested interval property alone implies the least-upper-bound property or FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property: nothing in this library proves that $R (t)$ is Cauchy complete or that it has any nested interval property, and in fact it is neither. Those two need the larger field $R ((t^{- 1}))$ , which is why that field was built ( $R ((t^{- 1}))$ , the formal Laurent series field, is Cauchy complete, non-Archimedean, and lacks the least-upper-bound property).
The order is the one induced from $R ((t^{- 1}))$ in spirit but not by any embedding proved here. Both fields order an element by its behaviour at infinity, and in both the comparison looks at a single coefficient. This library constructs no embedding of one into the other and never uses one.

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

$Q (\sqrt{2})$ carries exactly two distinct field orders, exchanged by the conjugation $\sqrt{2} \mapsto - \sqrt{2}$

Example

Let $u : = \sqrt{2} \in R$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ) and

$F : = Q (\sqrt{2}) = {a + b u : a, b \in Q} \subseteq R .$

Then $F$ is a field, every element of it is $a + b u$ for exactly one pair $(a, b)$ of rationals, and the conjugation $σ (a + b u) : = a - b u$ is a field automorphism of $F$ .

$F$ carries exactly two positive cones (Ordered field):

$P_{1} = {x \in F : x > 0 in R}, P_{2} = {x \in F : σ (x) \in P_{1}},$

and $σ$ exchanges them. They differ: $u \in P_{1}$ and $u \notin P_{2}$ . In the second order $\sqrt{2}$ is negative, and indeed lies below every positive rational, while $- \sqrt{2}$ is positive; the rationals themselves are ordered the same way in both.

The point of the example is that an order is extra structure on a field, not a property of it: the same field is an ordered field in two inequivalent ways, and no algebraic property of $F$ can distinguish $u$ from $- u$ .

Facts & Assumptions

Given: $R$ with its order, $u = \sqrt{2}$ , the set $F$ above, and the map $σ (a + b u) = a - b u$ .

[L1]

$R$ is a complete ordered field and every $a \geq 0$ in it has a unique $s \geq 0$ with $s^{2} = a$ ; in particular $u > 0$ and $u^{2} = 2$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Complete ordered field (least-upper-bound property), The reals form a totally ordered field).

[L2]

No rational squares to $2$ (FALSE: some rational number squares to 2, The rationals as equivalence classes of pairs of integers); in particular $u \notin Q$ .

[L3]

Field axioms and arithmetic (Field); a positive cone is a subset $P$ satisfying trichotomy, exactly one of $x \in P$ , $x = 0$ , $- x \in P$ , and closure under addition and multiplication, and $x < y$ means $y - x \in P$ (Ordered field).

[L4]

In any ordered field: $0 < 1$ (The multiplicative identity is positive); $n \cdot 1 > 0$ for $n \geq 1$ (Canonical naturals are positive and strictly increasing); a nonzero square is positive (Squares of nonzero elements are positive); a positive element has a positive inverse (Inverses of positives are positive, and reciprocation reverses order); a product of two positives or of two negatives is positive and a product of a positive and a negative is negative (Sign rules for products and monotonicity of multiplication); sums of positives are positive and adding a constant preserves the order (Order is preserved by adding a constant and by adding inequalities). 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).

[L5]

$Q$ embeds in any ordered field as ${(p \cdot 1) (q \cdot 1)^{- 1}}$ , compatibly with the field operations (The unique embedding of ℚ into an ordered field).

Verification

technique · direct

1.1

$u \in R$ satisfies $u > 0$ and $u^{2} = 2$ , and $u \notin Q$ .

L1L2

1.2

In every ordered field the positivity of a rational is forced: $n \cdot 1 > 0$ for $n \geq 1$ , so for positive integers $p, q$ the element $(p \cdot 1) (q \cdot 1)^{- 1}$ is positive, and hence a rational is in the positive cone exactly when it is positive in the usual sense.

L4L5

2.1

$F$ is a subfield of $R$ : it contains $0$ and $1$ , is closed under subtraction, and $(a + b u) (c + d u) = (a c + 2 b d) + (a d + b c) u$ gives closure under multiplication; for $a + b u \neq 0$ one has $a^{2} - 2 b^{2} \neq 0$ , since $b \neq 0$ would otherwise give $(a / b)^{2} = 2$ against [L2] while $b = 0$ forces $a \neq 0$ , and then $(a + b u)^{- 1} = (a - b u) (a^{2} - 2 b^{2})^{- 1} \in F$ .

step 1.1L2L3

2.2

The representation is unique: $a + b u = a^{'} + b^{'} u$ with $b \neq b^{'}$ would give $u = (a - a^{'}) (b^{'} - b)^{- 1} \in Q$ , against step 1.1; so $b = b^{'}$ and then $a = a^{'}$ .

step 1.1L2L3

3.1

$σ$ is therefore a well-defined map $F \to F$ , and it is a field automorphism: it is additive by inspection, $σ (1) = 1$ , and $σ (a + b u) σ (c + d u) = (a c + 2 b d) - (a d + b c) u = σ ((a + b u) (c + d u))$ ; moreover $σ \circ σ$ is the identity, so $σ$ is a bijection.

step 2.1step 2.2L3

3.2

Let $Q$ be any positive cone on $F$ . Since $u \neq 0$ , exactly one of $u \in Q$ , $- u \in Q$ holds.

step 2.1L3

4.1

$Q$ is determined by that choice. Suppose $u \in Q$ (the other case is the same with $u$ replaced by $- u$ , which also squares to $2$ ). Let $x = a + b u \neq 0$ . If $b = 0$ then $x$ is a nonzero rational and step 1.2 decides it. If $b \neq 0$ then $x = b (u + c)$ with $c : = a / b \in Q$ , and by [L4] the membership of $x$ is decided by those of $b$ and of $u + c$ ; for $c \geq 0$ one has $u + c \in Q$ , while for $c < 0$ , writing $e : = - c > 0$ , the identity $(u - e) (u + e) = 2 - e^{2}$ with $u + e \in Q$ and $2 - e^{2} \neq 0$ shows that $u - e \in Q$ exactly when $2 - e^{2} > 0$ , a condition on a rational decided by step 1.2. So $Q$ is uniquely determined, and there are at most two positive cones on $F$ .

step 1.2step 2.1step 3.2L2L3L4

4.2

Both occur. $P_{1}$ is a positive cone on $F$ , being the restriction to the subfield $F$ of the positive cone of $R$ ; and $P_{2} = σ^{- 1} (P_{1})$ is one because $σ$ is a field automorphism, so trichotomy and closure transfer along it. They are distinct: $u \in P_{1}$ by step 1.1, whereas $σ (u) = - u \notin P_{1}$ , so $u \notin P_{2}$ .

step 1.1step 2.1step 3.1L3

5.1

Hence $F$ carries exactly two positive cones, $P_{1}$ and $P_{2}$ , and since $σ$ is an involution, $P_{2} = σ (P_{1})$ and $P_{1} = σ (P_{2})$ : the conjugation exchanges the two orders.

step 3.1step 4.1step 4.2∎

Remarks

Two orders, one field, and no way to tell them apart algebraically. The automorphism $σ$ carries $(F, P_{1})$ isomorphically onto $(F, P_{2})$ as an ordered field, so the two ordered fields are isomorphic even though the two orders on the underlying $F$ are different subsets. That is the precise sense in which an order is not determined by the field: what is determined here is the order up to isomorphism, not the order itself.
Contrast with $Q$ and with $R$ , each of which carries exactly one order. For $Q$ this is step 1.2: every rational is a quotient of canonical naturals, so its sign is forced. For $R$ it is Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ : the positives are exactly the nonzero squares, and the squares are fixed by the field structure alone. $Q (\sqrt{2})$ sits between the two and has room for exactly two, because $2$ acquires a square root while $F$ still has elements that are not squares.
What decides an order on $F$ is a single bit, the sign of $u$ , after which every other comparison reduces to a comparison of rationals. That is also why there are exactly two and not more: the sign of $u$ is the only free choice, and both of its values are realised.

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

$R ((t^{- 1}))$ , the formal Laurent series field, is Cauchy complete, non-Archimedean, and lacks the least-upper-bound property

Example

Let $K = R ((t^{- 1}))$ be the field of formal Laurent series in $t^{- 1}$ over $R$ (The formal Laurent series $R ((t^{- 1}))$ : support bounded below, valuation, leading coefficient), ordered by the sign of the leading coefficient ( $R ((t^{- 1}))$ is an ordered field, ordered by the sign of the leading coefficient). This example assembles, in one place and against the five properties of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness, what the field does and does not satisfy:

property	holds in $K$	reference
ordered field	yes	$R ((t^{- 1}))$ is an ordered field, ordered by the sign of the leading coefficient
Archimedean	no	$R ((t^{- 1}))$ is non-Archimedean, and the monomials $t^{- k}$ are cofinal below its positive elements
(CC) Cauchy completeness	yes	Every Cauchy sequence in $R ((t^{- 1}))$ converges: $K$ is sequentially Cauchy complete
(NIP) nested intervals, shrinking	yes	$R ((t^{- 1}))$ has the nested interval property for lengths tending to $0$
(LUB) least upper bound	no	$R ((t^{- 1}))$ does not have the least-upper-bound property; its canonical naturals have no supremum
(BW) Bolzano-Weierstrass	no	below
(MCT) monotone convergence	no	below

$K$ is therefore the witness for FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property, and a worked illustration of how far apart the two things called "completeness" can be. A concrete convergent Cauchy sequence is exhibited at the end.

Facts & Assumptions

Given: $K = R ((t^{- 1}))$ , whose elements are the functions $Z \to R$ with support bounded below, with $t^{- a}$ the function taking the value $1$ at $a$ and $0$ elsewhere.

[L1]

$K$ is an ordered field, in which $f > 0$ exactly when $f \neq 0$ and its lowest-index nonzero coefficient is a positive real ( $R ((t^{- 1}))$ is an ordered field, ordered by the sign of the leading coefficient, The formal Laurent series $R ((t^{- 1}))$ : support bounded below, valuation, leading coefficient); and $(g - h) (j) = g (j) - h (j)$ ( $R ((t^{- 1}))$ is a commutative ring: the product is a finite sum and both operations preserve support bounded below).

[L2]

$n \cdot 1_{K} < t$ for every natural $n$ , so $K$ is not Archimedean; $0_{K} < t^{- (k + 1)} < t^{- k}$ ; for every $ε > 0$ in $K$ there is $k \in N$ with $t^{- k} < ε$ ; and if $h (j) = 0$ for every $j \leq k$ then $∣ h ∣ < t^{- k}$ ( $R ((t^{- 1}))$ is non-Archimedean, and the monomials $t^{- k}$ are cofinal below its positive elements, Archimedean ordered field).

[L3]

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

[L4]

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

[L5]

Every nested sequence of closed intervals of $K$ whose lengths tend to $0$ in $K$ has exactly one common point, so (NIP) holds ( $R ((t^{- 1}))$ has the nested interval property for lengths tending to $0$ ).

[L6]

(BW) implies the Archimedean property (Bolzano-Weierstrass alone forces the Archimedean property, so it needs no separate Archimedean hypothesis) and so does (MCT) (The monotone convergence property alone forces the Archimedean property, so it carries no separate Archimedean hypothesis); and the five properties are equivalent once the Archimedean property is supplied where needed (For an ordered field the five completeness properties are equivalent, provided the Archimedean property is assumed alongside nested intervals and Cauchy completeness).

Verification

technique · direct

1.1

$K$ is an ordered field.

1.2

$K$ is not Archimedean: $t$ exceeds every canonical natural.

1.3

$K$ has (CC).

1.4

$K$ does not have (LUB): the canonical naturals are nonempty and bounded above and have no supremum in $K$ .

1.5

$K$ 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.

2.1

So $K$ is a Cauchy complete, non-Archimedean ordered field without the least-upper-bound property, which is what this example asserts, and it is the witness used in FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property and in FALSE: the nested interval property alone implies the least-upper-bound property.

step 1.1step 1.2step 1.3step 1.4step 1.5

2.2

$K$ has neither (BW) nor (MCT), since either would force $K$ to be Archimedean, which step 1.2 denies.

step 1.2L6

2.3

A concrete convergent Cauchy sequence: let $f^{(n)} : = \sum_{k = 0}^{n} t^{- k}$ , the function taking the value $1$ at each index $0 \leq j \leq n$ and $0$ elsewhere. For $n > m$ the difference $f^{(n)} - f^{(m)}$ vanishes at every index $j \leq m$ , so $∣ f^{(n)} - f^{(m)} ∣ < t^{- m}$ ; since the monomials $t^{- m}$ get below every positive element of $K$ , the sequence is Cauchy in $K$ . Its limit is the element $L$ with $L (j) = 1$ for $j \geq 0$ and $L (j) = 0$ for $j < 0$ , which lies in $K$ because its support is bounded below, and $f^{(n)} - L$ vanishes at every index $j \leq n$ , so $∣ f^{(n)} - L ∣ < t^{- n}$ and $f^{(n)} \to L$ in $K$ .

step 1.1step 1.3L1L2

3.1

The table of the Example is therefore established in every row, and $K$ separates Cauchy completeness from the least-upper-bound property.

step 2.1step 2.2step 2.3∎

Remarks

The one-line reason. Comparison in $K$ looks only at the first coefficient at which two elements differ, so $t$ is bigger than every real constant and $t^{- 1}$ is smaller than every positive real constant. The naturals are therefore bounded, which kills (LUB), (BW) and (MCT) at a stroke. Meanwhile a Cauchy sequence in $K$ must have each of its coefficients eventually constant, and reading off those eventual values builds the limit; nothing about the naturals being cofinal is needed for that.
Why the limit above is not a sum. The notation $\sum_{k \geq 0} t^{- k}$ for $L$ is a name for a function, not an infinite sum (The formal Laurent series $R ((t^{- 1}))$ : support bounded below, valuation, leading coefficient). What step 2.3 proves is a genuine limit in the order of $K$ , and it happens to agree with that notation; no notion of convergence is presupposed by the notation itself.
What this example does not give. It says nothing about $R (t)$ , the other non-Archimedean field in this library (The rational function field $R (t)$ ordered by the eventual sign is an ordered field, worked out), which is neither Cauchy complete nor nested-interval complete and cannot replace $K$ in any of these roles.
Uniqueness of the complete ordered field is untouched. $K$ is not a complete ordered field, so it is no counterexample to that uniqueness; it is a counterexample only to the habit of calling (CC) completeness.

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

The Cesaro means of $(- 1)^{k}$ converge to $0$ although the sequence diverges

Example

Let $(s_{k})$ be the alternating sequence 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 $s_{σ (k)} = - s_{k}$ , usually written $s_{k} = (- 1)^{k}$ . Its Cesaro means (The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability) are

$σ_{n} = \frac{s_{0} + \dots + s_{n}}{n + 1} = {\begin{cases} \frac{1}{n + 1} & n even, \\ 0 & n odd, \end{cases}$

so the first few values are

$σ_{0} = 1, σ_{1} = 0, σ_{2} = \frac{1}{3}, σ_{3} = 0, σ_{4} = \frac{1}{5}, σ_{5} = 0, \dots$

and $\lim_{n} σ_{n} = 0$ , while $(s_{k})$ does not converge at all. So $(s_{k})$ is $(C, 1)$ -summable to $0$ and divergent: it is the standard witness that $(C, 1)$ -summability is strictly weaker than convergence, and the one used in FALSE: if the Cesaro means of a sequence converge then the sequence converges.

The value $0$ is the one an average ought to give, since the sequence spends half its indices at $1$ and half at $- 1$ ; the classical way to say this is that the series $1 - 1 + 1 - 1 + \dots$ has Cesaro sum $\frac{1}{2}$ , that being the Cesaro limit of its partial sums rather than of its terms.

Facts & Assumptions

Given: The alternating sequence $(s_{k})$ with $s_{0} = 1$ and $s_{σ (k)} = - s_{k}$ , its partial sums $S_{n} = \sum_{k < n} s_{k}$ , and its Cesaro means $σ_{n} = (n + 1)^{- 1} S_{n + 1}$ .

[L1]

The alternating sequence and its index maps $e$ (even indices) and $o$ (odd indices), with $N$ the disjoint union of their ranges and $s_{e_{j}} = 1$ , $s_{o_{j}} = - 1$ (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]

Its partial sums satisfy $S_{e_{j}} = 0$ and $S_{o_{j}} = 1$ , and consequently $∣ σ_{n} ∣ \leq (n + 1)^{- 1}$ and $σ_{n} \to 0$ ; this is proved in FALSE: if the Cesaro means of a sequence converge then the sequence converges, steps 2.1, 3.1, 4.1 and 5.1 there.

[L3]

$(s_{k})$ is bounded and does not converge (FALSE: every bounded sequence converges).

[L4]

The Cesaro means and $(C, 1)$ -summability (The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability, Finite sums and finite products, by recursion, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L5]

Convergence of a real sequence (Limits and Cauchy sequences of reals); the reciprocal Archimedean property (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L6]

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 (Complete ordered field (least-upper-bound property), Ordered field).

Verification

technique · direct

1.1

$S_{m} = 0$ when $m$ is even and $S_{m} = 1$ when $m$ is odd, since $N$ is the disjoint union of the ranges of $e$ and $o$ and $S_{e_{j}} = 0$ , $S_{o_{j}} = 1$ .

L1L2

1.2

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

2.1

Hence $σ_{n} = (n + 1)^{- 1} S_{n + 1}$ equals $(n + 1)^{- 1}$ when $n$ is even, because $n + 1$ is then odd, and equals $0$ when $n$ is odd; in particular $σ_{0} = 1$ , $σ_{1} = 0$ , $σ_{2} = 1 / 3$ , $σ_{3} = 0$ , $σ_{4} = 1 / 5$ and $σ_{5} = 0$ .

step 1.1L4L6

2.2

$∣ σ_{n} ∣ \leq (n + 1)^{- 1}$ for every $n$ , and given a real $ε > 0$ a natural $m \geq 1$ with $1 / m < ε$ gives $∣ σ_{n} ∣ \leq (n + 1)^{- 1} < ε$ for all $n \geq m$ ; so $\lim_{n} σ_{n} = 0$ .

step 1.1L2L5L6

3.1

$(s_{k})$ is therefore $(C, 1)$ -summable to $0$ and divergent.

step 1.2step 2.1step 2.2L4∎

Remarks

The means converge but are not monotone, and they are not even eventually of one shape: they alternate between $0$ and a positive value shrinking like $1 / (n + 1)$ . Convergence of a Cesaro transform therefore carries no monotonicity information, which is another way of seeing that the transform loses the oscillation rather than damping it.
Where the $1 / 2$ comes from. The classical assertion " $1 - 1 + 1 - 1 + \dots = 1 / 2$ " is about the partial sums $S_{m}$ , which are $0, 1, 0, 1, \dots$ ; their Cesaro means tend to $1 / 2$ . This library has no theory of series yet, so nothing above asserts it; the sequence averaged here is $(s_{k})$ itself, whose means tend to $0$ .
This is not a failure of the Cesaro matrix. That matrix is regular (The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem): it never changes a limit that exists. What it does here is assign a value where no limit exists, which is exactly what a summability method is for.

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

Stolz-Cesaro gives $(1 + 2 + \dots + n) / n^{2} \to 1 / 2$ and $(1^{p} + \dots + n^{p}) / n^{p + 1} \to 1 / (p + 1)$ for natural $p$

Example

Fix a natural number $p \geq 1$ and put

$a_{n} : = \sum_{k = 0}^{n} k^{p} = 1^{p} + 2^{p} + \dots + n^{p}, b_{n} : = n^{p + 1},$

the first equality because $0^{p} = 0$ for $p \geq 1$ (Integer powers $a^{m}$ ). Then $(b_{n})$ is strictly increasing and unbounded with $b_{0} = 0$ and $b_{n} > 0$ for $n \geq 1$ , so 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 $a_{k} / b_{k} \to L$ applies with $K_{0} = 1$ , and

$\lim_{n} \frac{1^{p} + 2^{p} + \dots + n^{p}}{n^{p + 1}} = \frac{1}{p + 1},$

the limit being taken over the indices $n \geq 1$ , where the quotient is defined. For $p = 1$ this is

$\lim_{n} \frac{1 + 2 + \dots + n}{n^{2}} = \frac{1}{2} .$

No closed form for $a_{n}$ is used. That is the point of the example: the difference quotient of Stolz-Cesaro replaces a summation formula by a single algebraic identity, the factorisation of $(n + 1)^{p + 1} - n^{p + 1}$ .

Facts & Assumptions

Given: A natural $p \geq 1$ , the sequences $a_{n} = \sum_{k = 0}^{n} k^{p}$ and $b_{n} = n^{p + 1}$ , and their difference quotients $d_{n} = (a_{n + 1} - a_{n}) / (b_{n + 1} - b_{n})$ .

[L1]

Stolz-Cesaro in the $\infty / \infty$ form: for $(b_{n})$ strictly increasing with range not bounded above and $(d_{n})$ convergent, the tail of $(a_{n} / b_{n})$ beyond an index where $b$ becomes positive converges to $\lim_{n} d_{n}$ (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 $a_{k} / b_{k} \to L$ ); convergence depends only on a tail (Convergence depends only on the tail); limits are unique (A sequence has at most one limit).

[L2]

Powers: $a^{0} = 1$ , $a^{n + 1} = a^{n} a$ , so $0^{p} = 0$ for $p \geq 1$ (Integer powers $a^{m}$ ); $a^{m + n} = a^{m} a^{n}$ and $a^{- m} = (a^{m})^{- 1}$ for $a \neq 0$ (Laws of integer exponents); for $0 \leq a < b$ and $n \geq 1$ , $a^{n} < b^{n}$ , and $a \geq 1$ with $n \geq 1$ gives $a^{n} \geq a$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L3]

Factorisation: $b^{m} - a^{m} = (b - a) \sum_{k = 0}^{m - 1} a^{k} b^{m - 1 - k}$ for $m \geq 1$ (Factorisation of $b^{n} - a^{n}$ , and the resulting Lipschitz estimate).

[L4]

Finite sums and their laws (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L5]

Algebra of limits for sums, products, scalar multiples and quotients with nonvanishing denominators (Algebra of limits: sums, scalar multiples, products and quotients); convergence of real sequences (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L6]

The Archimedean property of $R$ and its reciprocal form (Every complete ordered field is Archimedean, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ); strict monotonicity and boundedness of real sequences (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, Lower bound, bounded below, bounded set).

[L7]

Induction principle (The principle of mathematical induction).

[L8]

Order arithmetic: canonical naturals are positive and strictly increasing (Canonical naturals are positive and strictly increasing); a positive element has a positive inverse and reciprocation reverses the order (Inverses of positives are positive, and reciprocation reverses order); 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).

Verification

technique · direct

1.1

$(b_{n})$ is strictly increasing: for naturals $m < n$ one has $0 \leq m < n$ and $p + 1 \geq 1$ , so $m^{p + 1} < n^{p + 1}$ .

L2L6L8

1.2

The range of $(b_{n})$ is not bounded above: $n \geq 1$ gives $n^{p + 1} \geq n$ , and no real bounds every canonical natural.

L2L6L8

1.3

$b_{0} = 0^{p + 1} = 0$ and $b_{n} > 0$ for $n \geq 1$ , so $K_{0} = 1$ is an index beyond which $b$ is positive.

L2L8

1.4

$a_{n + 1} - a_{n} = (n + 1)^{p}$ and $b_{n + 1} - b_{n} = (n + 1)^{p + 1} - n^{p + 1}$ , so $d_{n} = (n + 1)^{p} ((n + 1)^{p + 1} - n^{p + 1})^{- 1}$ .

L2L4

1.5

Put $τ_{n} : = n (n + 1)^{- 1}$ ; then $1 - τ_{n} = (n + 1)^{- 1}$ , so given a real $ε > 0$ and a natural $m \geq 1$ with $1 / m < ε$ one has $∣ τ_{n} - 1 ∣ = (n + 1)^{- 1} < ε$ for all $n \geq m$ , that is $\lim_{n} τ_{n} = 1$ .

L5L6L8

2.1

By the factorisation at $m = p + 1$ , $b = n + 1$ and $a = n$ : $(n + 1)^{p + 1} - n^{p + 1} = ((n + 1) - n) \sum_{k = 0}^{p} n^{k} (n + 1)^{p - k} = \sum_{k = 0}^{p} n^{k} (n + 1)^{p - k}$ .

step 1.4L3L4

2.2

By induction on $k$ , using the product rule for limits, $\lim_{n} τ_{n}^{k} = 1$ for every $k$ ; by induction on $p$ , using the sum rule, $\lim_{n} \sum_{k = 0}^{p} τ_{n}^{k} = p + 1$ .

step 1.5L5L7

3.1

Dividing numerator and denominator of $d_{n}$ by $(n + 1)^{p} > 0$ and using $n^{k} (n + 1)^{p - k} (n + 1)^{- p} = τ_{n}^{k}$ gives $d_{n} = (\sum_{k = 0}^{p} τ_{n}^{k})^{- 1}$ , and $\sum_{k = 0}^{p} τ_{n}^{k} \geq 1 > 0$ for every $n$ , the term at $k = 0$ being $1$ and all terms being $\geq 0$ .

step 1.4step 1.5step 2.1L2L4L8

4.1

Since the denominators are nonzero and their limit $p + 1$ is nonzero, the quotient rule gives $\lim_{n} d_{n} = (p + 1)^{- 1}$ .

step 2.2step 3.1L5L8

5.1

Steps 1.1, 1.2 and 4.1 are the hypotheses of Stolz-Cesaro, so the tail $j \mapsto a_{j + 1} / b_{j + 1}$ converges to $(p + 1)^{- 1}$ ; that is, $\lim_{n} (1^{p} + \dots + n^{p}) n^{- (p + 1)} = (p + 1)^{- 1}$ over the indices $n \geq 1$ .

step 1.1step 1.2step 1.3step 4.1L1

6.1

At $p = 1$ this reads $\lim_{n} (1 + 2 + \dots + n) n^{- 2} = 1 / 2$ .

step 5.1∎

Remarks

Why the limit is taken from $n = 1$ . $b_{0} = 0$ , so $a_{0} / b_{0}$ does not denote anything, and 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 $a_{k} / b_{k} \to L$ is stated for the tail exactly for this reason. Nothing is lost: convergence is a property of a tail (Convergence depends only on the tail).
The closed form is available and is not needed. For $p = 1$ one has $a_{n} = n (n + 1) / 2$ , and dividing by $n^{2}$ gives the limit $1 / 2$ directly. For general $p$ the closed form is Faulhaber's formula, which this library does not prove; the difference quotient sidesteps it entirely, and that is the practical content of Stolz-Cesaro.
A sanity check on the answer. The quotient compares a sum of $n$ terms, the largest of which is $n^{p}$ , with $n^{p + 1} = n \cdot n^{p}$ , so the limit must lie in $[0, 1]$ ; and the terms $k^{p}$ grow, so the sum should be a definite fraction of the largest term times $n$ . The fraction is $1 / (p + 1)$ , which is what an integral comparison would also predict. No such comparison is used above.

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27↗ rests on later material (inherited)Open item page →

In $R (t)$ the rationals are not dense: no rational lies strictly between $0$ and $1 / t$

Statement refuted

Refuted claim: in every ordered field $F$ the image of $Q$ is dense, that is, for all $x < y$ in $F$ there is a rational $q$ with $x < q \cdot 1_{F} < y$ .

The witness is $R (t)$ with the eventual-sign order (Not every ordered field is Archimedean, The rational function field $R (t)$ ordered by the eventual sign is an ordered field, worked out), and the pair $x = 0$ , $y = 1 / t$ : the interval between them contains no rational at all.

The true statement requires the Archimedean property and is ℚ is dense in every Archimedean ordered field; $R (t)$ is not Archimedean, and this counterexample is exactly the failure that the Archimedean hypothesis rules out.

Facts & Assumptions

Given: The ordered field $R (t)$ with positive cone $P = {f \neq 0 : f (x) > 0 for all sufficiently large real x}$ , and its element $1 / t$ .

[L1]

$R (t)$ is an ordered field and is not Archimedean (Not every ordered field is Archimedean, Archimedean ordered field).

[L2]

$0 < 1 / t$ , and $1 / t < q \cdot 1$ for every rational $q > 0$ (The rational function field $R (t)$ ordered by the eventual sign is an ordered field, worked out).

[L3]

The canonical embedding of $Q$ into an ordered field is an embedding of ordered fields, so $q \cdot 1 > 0$ if and only if $q > 0$ , and $q \cdot 1 \leq 0$ when $q \leq 0$ (The unique embedding of ℚ into an ordered field).

[L4]

$Q$ is dense in every Archimedean ordered field (ℚ is dense in every Archimedean ordered field).

[L5]

In an ordered field the order is total and transitive, exactly one of $u < v$ , $u = v$ , $v < u$ holds, and a positive element has a positive inverse (Ordered field, Inverses of positives are positive, and reciprocation reverses order).

Counterexample

technique · direct

1.1

$R (t)$ is an ordered field, it is not Archimedean, and $0 < 1 / t$ in it.

L1L2L5

1.2

For every rational $q > 0$ one has $1 / t < q \cdot 1$ .

2.1

No rational $q$ satisfies $0 < q \cdot 1 < 1 / t$ : if $q \leq 0$ then $q \cdot 1 \leq 0$ and the left inequality fails, while if $q > 0$ then $1 / t < q \cdot 1$ by step 1.2, so $q \cdot 1 < 1 / t$ fails by trichotomy.

step 1.1step 1.2L3L5

3.1

So $0 < 1 / t$ in $R (t)$ with no rational strictly between them: the image of $Q$ is not dense in $R (t)$ , and the claim is false.

step 1.1step 2.1

4.1

The hypothesis the claim omitted is the Archimedean property, which $R (t)$ lacks and under which the conclusion does hold.

step 1.1L1L4∎

Remarks

What density really needs. Given $0 < x < y$ in an Archimedean field one finds $n$ with $1 / n < y - x$ and then a multiple of $1 / n$ in the gap; the Archimedean property is used precisely to make the mesh $1 / n$ finer than the gap. In $R (t)$ the gap $1 / t - 0$ is smaller than every $1 / n$ , so no mesh built from rationals is ever fine enough.
An element like $1 / t$ is called an infinitesimal: positive, and below every positive rational. A non-Archimedean ordered field always has one, since if $x$ exceeds every canonical natural then $1 / x$ is below every $1 / n$ (Inverses of positives are positive, and reciprocation reverses order). So the failure of density is not special to this field; it happens in every non-Archimedean ordered field, including $R ((t^{- 1}))$ .
Density is not the same as completeness. $Q$ is dense in itself and in $R$ , and $Q$ is not complete. What this counterexample shows is only that density of $Q$ needs the Archimedean property, which is also the hypothesis missing from 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.

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

On a closed interval of $Q$ there is a continuous unbounded function, a bounded one with no maximum, and one without the intermediate value property

Statement refuted

The notion of continuity used here is stated in full, and is not imported. Let $F$ be an ordered field, $D \subseteq F$ and $φ : D \to F$ . Say $φ$ is continuous at $c \in D$ when

$for every ε > 0 in F there is δ > 0 in F with ∣ φ (x) - φ (c) ∣ < ε whenever x \in D and ∣ x - c ∣ < δ,$

and continuous on $D$ when it is continuous at every point of $D$ . This is the ordinary $ε$ - $δ$ condition, read entirely inside $F$ . Nothing below cites a definition of continuity from elsewhere in this library, because there is none yet.

Refuted claim: over every ordered field $F$ , a function $φ : [a, b]_{F} \to F$ that is continuous on the closed interval $[a, b]_{F}$ (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field) is bounded there, attains a maximum there, and takes every value between $φ (a)$ and $φ (b)$ . In other words, the extreme value theorem and the intermediate value theorem hold over an arbitrary ordered field.

The witness is $F = Q$ and $D = [0, 2]_{Q}$ , with three functions, one for each clause:

$f (x) = \frac{1}{x^{2} - 2}, g (x) = - (x^{2} - 2)^{2}, h (x) = x^{2} - 2.$

All three are continuous on $D$ in the sense above. $f$ is unbounded; $g$ is bounded and has no maximum; $h$ satisfies $h (0) = - 2 < 0 < 2 = h (2)$ and never takes the value $0$ . What $Q$ lacks is 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, and each of the three clauses fails because of that single omission.

Facts & Assumptions

Given: The ordered field $Q$ ; the set $D : = [0, 2]_{Q} = {x \in Q : 0 \leq x \leq 2}$ ; the functions $f, g, h$ above; and the map $T (x) : = (2 x + 2) (x + 2)^{- 1}$ .

[L1]

$Q$ is an ordered field (The rationals form a totally ordered field, The rationals as equivalence classes of pairs of integers, Field, Ordered field) and is Archimedean (The rationals are Archimedean).

[L2]

No rational squares to $2$ (FALSE: some rational number squares to 2).

[L3]

In a complete ordered field every element $\geq 0$ has a square root (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Complete ordered field (least-upper-bound property)).

[L4]

Closed intervals $[a, b]_{F}$ of an ordered field (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field); the properties (LUB) and the rest (The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness).

[L5]

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

[L6]

Powers: $a^{0} = 1$ , $a^{n + 1} = a^{n} a$ (Integer powers $a^{m}$ ); for $0 \leq a < b$ and $n \geq 1$ , $a^{n} < b^{n}$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ); $(1 + x)^{n} \geq 1 + n x$ for $x \geq - 1$ (Bernoulli's inequality $(1 + x)^{n} \geq 1 + n x$ ).

[L7]

Recursion theorem (The recursion theorem) and induction principle (The principle of mathematical induction).

[L8]

Order arithmetic: a positive element is invertible with positive inverse and reciprocation reverses the order (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); canonical naturals are positive (Canonical naturals are positive and strictly increasing); the order is total and transitive (Ordered field).

Counterexample

technique · direct

1.1

For every $x \in D$ one has $x^{2} \neq 2$ , so $x^{2} - 2 \neq 0$ and $∣ x^{2} - 2 ∣ > 0$ ; and $0 \leq x \leq 2$ gives $0 \leq x^{2} \leq 4$ , so $∣ x^{2} - 2 ∣ \leq 2$ . Hence $f$ , $g$ and $h$ are defined on all of $D$ .

L2L5L6L8

1.2

$Q$ is an ordered field that is not complete: a complete ordered field has a square root of $2$ , and no rational squares to $2$ .

L1L2L3L4

1.3

For $x \in D$ one has $x + 2 \geq 2 > 0$ , so $T (x) = 2 - 2 (x + 2)^{- 1}$ is defined, and $(x + 2)^{- 1}$ lies between $(4)^{- 1}$ and $(2)^{- 1}$ , so $1 \leq T (x) \leq 3 / 2$ and $T (x) \in D$ ; moreover $T (x)^{2} - 2 = 2 (x^{2} - 2) (x + 2)^{- 2}$ and $(x + 2)^{2} \geq 4$ , so $∣ T (x)^{2} - 2 ∣ \leq ∣ x^{2} - 2 ∣ (2)^{- 1}$ .

L5L6L8

1.4

For all $x, c \in D$ : $∣ x^{2} - c^{2} ∣ = ∣ x - c ∣ ∣ x + c ∣ \leq 4 ∣ x - c ∣$ , since $0 \leq x + c \leq 4$ .

L5L8

2.1

$h$ is continuous on $D$ : given $ε > 0$ take $δ : = ε (4)^{- 1}$ , and $∣ x - c ∣ < δ$ gives $∣ h (x) - h (c) ∣ = ∣ x^{2} - c^{2} ∣ \leq 4 ∣ x - c ∣ < ε$ .

step 1.4L5L8

2.2

$g$ is continuous on $D$ : $∣ g (x) - g (c) ∣ = ∣ x^{2} - c^{2} ∣ \cdot ∣ (x^{2} - 2) + (c^{2} - 2) ∣ \leq 4 ∣ x - c ∣ \cdot 4 = 16 ∣ x - c ∣$ , using step 1.1 for the second factor, so $δ : = ε (16)^{- 1}$ works.

step 1.1step 1.4L5L8

2.3

$f$ is continuous on $D$ : fix $c \in D$ and put $A : = ∣ c^{2} - 2 ∣ > 0$ ; for $x \in D$ with $∣ x - c ∣ < A (8)^{- 1}$ one gets $∣ x^{2} - c^{2} ∣ < A (2)^{- 1}$ and hence $∣ x^{2} - 2 ∣ \geq A - ∣ x^{2} - c^{2} ∣ > A (2)^{- 1}$ , so $∣ f (x) - f (c) ∣ = ∣ c^{2} - x^{2} ∣ (∣ x^{2} - 2 ∣ ∣ c^{2} - 2 ∣)^{- 1} < 8 ∣ x - c ∣ A^{- 2}$ ; taking $δ$ to be the smaller of $A (8)^{- 1}$ and $ε A^{2} (8)^{- 1}$ gives $∣ f (x) - f (c) ∣ < ε$ .

step 1.1step 1.4L5L8

2.4

By the recursion theorem applied to $D$ , the element $1 \in D$ and the map $T$ , there is a sequence $(x_{n})$ in $D$ with $x_{0} = 1$ and $x_{n + 1} = T (x_{n})$ ; and by induction $∣ x_{n}^{2} - 2 ∣ \leq (2^{n})^{- 1}$ , the base case being $∣ 1 - 2 ∣ = 1$ and the step being step 1.3.

step 1.3L6L7

2.5

$g$ is bounded on $D$ , with $- 4 \leq g (x) \leq 0$ , and has no maximum: for every $c \in D$ the point $T (c)$ lies in $D$ and satisfies $∣ T (c)^{2} - 2 ∣ \leq ∣ c^{2} - 2 ∣ (2)^{- 1} < ∣ c^{2} - 2 ∣$ , so $(T (c)^{2} - 2)^{2} < (c^{2} - 2)^{2}$ and $g (T (c)) > g (c)$ .

step 1.1step 1.3L6L8

3.1

$f$ is unbounded on $D$ : $∣ f (x_{n}) ∣ = ∣ x_{n}^{2} - 2 ∣^{- 1} \geq 2^{n}$ , and given any $M \in Q$ the Archimedean property supplies $n$ with $M < n \cdot 1$ , whence $2^{n} \geq 1 + n \cdot 1 > M$ by Bernoulli.

step 2.4L1L6L8

3.2

$h$ is continuous on $D$ with $h (0) = - 2$ and $h (2) = 2$ , so $0$ lies strictly between $h (0)$ and $h (2)$ , and yet $h (x) = 0$ has no solution in $D$ , since that would be a rational squaring to $2$ .

step 1.1step 2.1L2

4.1

Over the ordered field $Q$ , on the closed interval $D = [0, 2]_{Q}$ : $f$ is continuous and unbounded, $g$ is continuous and bounded with no maximum, and $h$ is continuous and omits a value strictly between its values at the endpoints. All three clauses of the claim are therefore false, and the field involved is exactly one failing (LUB).

step 1.2step 2.2step 2.3step 2.5step 3.1step 3.2∎

Remarks

One mechanism, three failures. All three functions are built from $x^{2} - 2$ , whose zero is missing from $Q$ . The map $T$ is a contraction towards that missing zero: it halves $∣ x^{2} - 2 ∣$ at every step while staying inside $D$ . So $∣ x^{2} - 2 ∣$ has infimum $0$ on $D$ and does not attain it, and the three failures are three ways of reading that one sentence.
Nothing here is peculiar to $Q$ . The same construction runs in any ordered subfield of $R$ that omits $\sqrt{2}$ , since every step above uses only the field operations, the order, and the absence of a square root of $2$ . This item exhibits the cheapest witness; no claim is made here about ordered fields in general.
This item does not use, and does not need, a general theory of continuous functions. The $ε$ - $δ$ condition is stated in the Statement refuted and every use of it above is a direct verification, so the item is self-contained and nothing here waits on a later page. That is deliberate and not a placeholder: the claim refuted here is a claim about an arbitrary ordered field, and it is refuted over $Q$ , so a definition of continuity written for real functions on subsets of $R$ would not apply to it. This library has no notion of continuity over a general ordered field and needs none elsewhere, and inventing an id for one would put an unused definition on a page about completeness properties. The condition above is the ordinary one read inside $F$ , and it specialises to the real-variable definition at $F = R$ .
What is true over $Q$ . Continuity, sums and products of continuous functions, and composition all behave normally; what fails is every statement whose proof needs a supremum. That is the content of the page this one belongs to.

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

Over $Q$ there is a nonconstant differentiable function with identically zero derivative, so Rolle and the mean value theorem both fail

Statement refuted

The notion of derivative used here is stated in full, and is not imported. Let $F$ be an ordered field, $D \subseteq F$ , $φ : D \to F$ and $c \in D$ a point that is not isolated in $D$ , meaning that for every $δ > 0$ in $F$ there is $x \in D$ with $0 < ∣ x - c ∣ < δ$ . Say $φ$ is differentiable at $c$ with derivative $L \in F$ when

$for every ε > 0 in F there is δ > 0 in F with ∣ \frac{φ (x) - φ (c)}{x - c} - L ∣ < ε for all x \in D with 0 < ∣ x - c ∣ < δ,$

and write $φ^{'} (c) = L$ . This is the ordinary difference-quotient condition, read entirely inside $F$ . Nothing below cites a definition of the derivative from elsewhere in this library, because there is none yet.

Refuted claim: over every ordered field $F$ , if $φ : [a, b]_{F} \to F$ with $a < b$ is differentiable at every point of $[a, b]_{F}$ (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field), then

(Rolle) $φ (a) = φ (b)$ implies $φ^{'} (c) = 0$ for some $c \in [a, b]_{F}$ , and
(Mean value) $φ (b) - φ (a) = φ^{'} (c) (b - a)$ for some $c \in [a, b]_{F}$ .

The witness is $F = Q$ and $D = [0, 2]_{Q}$ with

$φ (x) = {\begin{cases} 0 & x^{2} < 2, \\ 1 & x^{2} > 2, \end{cases} and ρ (x) = φ (x) - \frac{x}{2} .$

$φ$ is well defined on $D$ because no rational squares to $2$ (FALSE: some rational number squares to 2). It is locally constant, hence differentiable everywhere on $D$ with $φ^{'} \equiv 0$ , and it is not constant, since $φ (0) = 0$ and $φ (2) = 1$ ; that refutes clause 2. And $ρ$ satisfies $ρ (0) = ρ (2) = 0$ while $ρ^{'} (c) = - 1 / 2$ for every $c$ ; that refutes clause 1.

Facts & Assumptions

Given: The ordered field $Q$ ; $D : = [0, 2]_{Q}$ ; the functions $φ$ and $ρ$ above.

[L1]

$Q$ is an ordered field (The rationals form a totally ordered field, The rationals as equivalence classes of pairs of integers, Ordered field); closed intervals $[a, b]_{F}$ are as in Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field.

[L2]

No rational squares to $2$ (FALSE: some rational number squares to 2); consequently $x^{2} \neq 2$ and $∣ x^{2} - 2 ∣ > 0$ for every $x \in D$ , and $Q$ is not a complete ordered field, since a complete one would contain a square root of $2$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Complete ordered field (least-upper-bound property), The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness). This is step 1.1 and step 1.2 of On a closed interval of $Q$ there is a continuous unbounded function, a bounded one with no maximum, and one without the intermediate value property.

[L3]

For all $x, c \in D$ : $∣ x^{2} - c^{2} ∣ \leq 4 ∣ x - c ∣$ , which is step 1.4 of On a closed interval of $Q$ there is a continuous unbounded function, a bounded one with no maximum, and one without the intermediate value property.

[L4]

Absolute value: $∣ u ∣ \geq 0$ , $∣ u ∣ = u$ for $u \geq 0$ , $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ , and $∣ u ∣ < c$ exactly when $- c < u < c$ (Basic properties of the absolute value); powers (Integer powers $a^{m}$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L5]

Order arithmetic: a positive element is invertible with positive inverse (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); $2 \cdot 1 > 0$ and $8 \cdot 1 > 0$ (Canonical naturals are positive and strictly increasing); the order is total and transitive (Ordered field).

Counterexample

technique · direct

1.1

Every $x \in D$ has $x^{2} \neq 2$ , so exactly one of $x^{2} < 2$ , $x^{2} > 2$ holds and $φ$ is well defined on $D$ ; moreover $0^{2} = 0 < 2$ and $1^{2} = 1 < 2$ and $2^{2} = 4 > 2$ , so $φ (0) = φ (1) = 0$ and $φ (2) = 1$ , and $φ$ is not constant on $D$ .

L1L2L5

1.2

For all $x, c \in D$ one has $∣ x^{2} - c^{2} ∣ \leq 4 ∣ x - c ∣$ .

1.3

No point of $D$ is isolated in $D$ : given $c \in D$ and $δ > 0$ , let $η$ be the smaller of $δ$ and $1$ , and take $x : = c + η / 2$ if $c \leq 1$ and $x : = c - η / 2$ otherwise; then $x \in D$ and $0 < ∣ x - c ∣ < δ$ .

L1L5

2.1

$φ$ is differentiable at every $c \in D$ with $φ^{'} (c) = 0$ . Put $A : = ∣ c^{2} - 2 ∣ > 0$ and $δ_{0} : = A (8)^{- 1} > 0$ . For $x \in D$ with $∣ x - c ∣ < δ_{0}$ step 1.2 gives $∣ x^{2} - c^{2} ∣ < A (2)^{- 1}$ ; so if $c^{2} < 2$ , that is $A = 2 - c^{2}$ , then $x^{2} < c^{2} + A (2)^{- 1} = (c^{2} + 2) (2)^{- 1} < 2$ , while if $c^{2} > 2$ , that is $A = c^{2} - 2$ , then $x^{2} > c^{2} - A (2)^{- 1} = (c^{2} + 2) (2)^{- 1} > 2$ . In either case $φ (x) = φ (c)$ , so the difference quotient is $0$ for every such $x$ with $x \neq c$ , and $∣ 0 - 0 ∣ < ε$ for every $ε > 0$ ; the same $δ_{0}$ serves for every $ε$ .

step 1.1step 1.2step 1.3L4L5

3.1

$ρ$ is differentiable at every $c \in D$ with $ρ^{'} (c) = - 1 / 2$ : with $δ_{0}$ as in step 2.1, every $x \in D$ with $0 < ∣ x - c ∣ < δ_{0}$ has $(ρ (x) - ρ (c)) (x - c)^{- 1} = (φ (x) - φ (c)) (x - c)^{- 1} - (2)^{- 1} = - (2)^{- 1}$ , so the quotient is constantly $- 1 / 2$ near $c$ ; and $ρ (0) = 0$ , $ρ (2) = 1 - 1 = 0$ , $ρ (1) = - 1 / 2$ .

step 1.1step 2.1L5

3.2

The mean value clause fails for $φ$ on $D$ : $φ (2) - φ (0) = 1$ while $φ^{'} (c) (2 - 0) = 0$ for every $c \in D$ , and $1 \neq 0$ .

step 1.1step 2.1L5

4.1

The Rolle clause fails for $ρ$ on $D$ : $ρ$ is differentiable at every point of $D$ , $ρ (0) = ρ (2)$ , and yet $ρ^{'} (c) = - 1 / 2 \neq 0$ for every $c \in D$ .

step 3.1L5

5.1

So over the ordered field $Q$ , on the closed interval $D = [0, 2]_{Q}$ , both clauses of the claim are false, and $Q$ is an ordered field without the least-upper-bound property.

step 1.1step 3.2step 4.1L2∎

Remarks

Where the classical proof breaks. Rolle's theorem is proved by taking a point where the function attains its maximum and showing the derivative vanishes there. Over $Q$ the maximum need not exist: that is On a closed interval of $Q$ there is a continuous unbounded function, a bounded one with no maximum, and one without the intermediate value property, proved on the same interval and by the same missing $\sqrt{2}$ . So this counterexample is not independent of that one, it is its consequence for the differential calculus.
A locally constant function need not be constant when the domain is disconnected, and $D$ is disconnected in exactly the way $Q$ is: the sets ${x \in D : x^{2} < 2}$ and ${x \in D : x^{2} > 2}$ are disjoint, nonempty, cover $D$ , and each is open in the $ε$ - $δ$ sense. Over $R$ no such split of an interval exists, and that is the connectedness that the mean value theorem really rests on.
The derivative here is genuinely a derivative, not a degenerate reading: the difference quotient is not merely small near $c$ , it is exactly $0$ for $φ$ and exactly $- 1 / 2$ for $ρ$ on a whole punctured neighbourhood, so the limit exists in the strongest possible sense.
This item does not use, and does not need, a general theory of differentiation. The difference-quotient condition is stated in the Statement refuted and every use of it above is a direct verification, so the item is self-contained and nothing here waits on a later page. As with On a closed interval of $Q$ there is a continuous unbounded function, a bounded one with no maximum, and one without the intermediate value property, that is deliberate: the claim refuted is a claim about an arbitrary ordered field and is refuted over $Q$ , so a derivative defined for real functions on subsets of $R$ would not apply to it. The condition above is the ordinary difference-quotient one read inside $F$ , and it specialises to the real-variable definition at $F = R$ .

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

$a_{k} = (- 1)^{k}$ , $b_{k} = k$ have $a_{k} / b_{k} \to 0$ while the difference quotient oscillates, so Stolz-Cesaro has no converse

Statement refuted

Refuted claim: the converse 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 $a_{k} / b_{k} \to L$ . That is: if $(b_{k})$ is strictly increasing with range not bounded above and the quotients $a_{k} / b_{k}$ converge, then the difference quotients $(a_{k + 1} - a_{k}) / (b_{k + 1} - b_{k})$ converge too.

The witness is $a_{k} = s_{k}$ , the alternating sequence 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 $s \circ o \equiv - 1$ usually written $(- 1)^{k}$ , and $b_{k} = k$ . Then $b_{0} = 0$ , so the quotient $a_{k} / b_{k}$ is formed for $k \geq 1$ only, exactly as 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 $a_{k} / b_{k} \to L$ is stated; and there

$\frac{a_{k}}{b_{k}} = \frac{s_{k}}{k}, ∣ \frac{a_{k}}{b_{k}} ∣ = \frac{1}{k} ⟶ 0,$

while the difference quotient is

$\frac{a_{k + 1} - a_{k}}{b_{k + 1} - b_{k}} = s_{k + 1} - s_{k} = - 2 s_{k},$

which takes the value $- 2$ at even $k$ and $2$ at odd $k$ and does not converge.

Facts & Assumptions

Given: The alternating sequence $(s_{k})$ with $s_{0} = 1$ and $s_{σ (k)} = - s_{k}$ ; the sequences $a_{k} : = s_{k}$ and $b_{k} : = k \cdot 1_{R}$ ; and the difference quotients $d_{k} : = (a_{k + 1} - a_{k}) (b_{k + 1} - b_{k})^{- 1}$ .

[L1]

The alternating sequence: $∣ s_{k} ∣ = 1$ for every $k$ , $s_{k + 1} = - s_{k}$ , $s_{e_{j}} = 1$ and $s_{o_{j}} = - 1$ along the even and odd index maps (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]

$(s_{k})$ is bounded and does not converge (FALSE: every bounded sequence converges).

[L3]

The canonical naturals of $R$ are positive for $k \geq 1$ and strictly increasing (Canonical naturals are positive and strictly increasing), and no real bounds them all (Every complete ordered field is Archimedean); strict monotonicity and boundedness of real sequences (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, Lower bound, bounded below, bounded set, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L4]

Convergence of real sequences and uniqueness of limits (Limits and Cauchy sequences of reals, A sequence has at most one limit); convergence depends only on a tail (Convergence depends only on the tail); the reciprocal Archimedean property (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L5]

Algebra of limits, in particular that a scalar multiple of a convergent sequence converges (Algebra of limits: sums, scalar multiples, products and quotients).

[L6]

Order arithmetic: $∣ u ∣ \geq 0$ and $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ (Basic properties of the absolute value); a positive element is invertible with positive inverse and reciprocation reverses the order (Inverses of positives are positive, and reciprocation reverses order); the order is total and transitive (Complete ordered field (least-upper-bound property), Ordered field).

[L7]

The hypotheses and conclusion of Stolz-Cesaro in the $\infty / \infty$ form (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 $a_{k} / b_{k} \to L$ ).

Counterexample

technique · direct

1.1

$(b_{k})$ is strictly increasing, its range is not bounded above, $b_{0} = 0$ and $b_{k} > 0$ for $k \geq 1$ ; so the hypotheses 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 $a_{k} / b_{k} \to L$ on $(b_{k})$ hold with $K_{0} = 1$ .

L3L7

1.2

$∣ s_{k} ∣ = 1$ for every $k$ , and $(s_{k})$ does not converge.

L1L2

2.1

The tail quotients $q_{j} : = a_{j + 1} / b_{j + 1} = s_{j + 1} (j + 1)^{- 1}$ satisfy $∣ q_{j} ∣ = (j + 1)^{- 1}$ ; given a real $ε > 0$ and a natural $m \geq 1$ with $1 / m < ε$ , every $j \geq m$ has $∣ q_{j} - 0 ∣ = (j + 1)^{- 1} < ε$ , so $q_{j} \to 0$ .

step 1.1step 1.2L4L6

2.2

$b_{k + 1} - b_{k} = 1$ for every $k$ , so $d_{k} = a_{k + 1} - a_{k} = s_{k + 1} - s_{k} = - s_{k} - s_{k} = - 2 s_{k}$ , which is $- 2$ when $k$ is even and $2$ when $k$ is odd.

step 1.1step 1.2L1L3

3.1

$(d_{k})$ does not converge: were $d_{k} \to M$ , then $s_{k} = (- 1 / 2) d_{k}$ would converge to $- M / 2$ by the scalar-multiple rule, contradicting step 1.2.

step 1.2step 2.2L5

4.1

So $(b_{k})$ is strictly increasing and unbounded, the quotients $a_{k} / b_{k}$ converge to $0$ over the indices $k \geq 1$ , and the difference quotients do not converge: the converse of Stolz-Cesaro is false.

step 2.1step 3.1L7∎

Remarks

The implication is genuinely one-way, and the reason is averaging. The conclusion 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 $a_{k} / b_{k} \to L$ is obtained by summing the difference quotients against the weights $b_{k + 1} - b_{k}$ and dividing by $b_{n}$ , which is a weighted average. An average can converge while what is averaged oscillates, and here it does: the quotients $a_{k} / b_{k}$ are damped by the growing denominator, and no information about $d_{k}$ survives.
The same phenomenon, in the summability language. With $b_{k} = k$ the weights are equal, so this is the Cesaro situation of FALSE: if the Cesaro means of a sequence converge then the sequence converges and The Cesaro means of $(- 1)^{k}$ converge to $0$ although the sequence diverges in another costume; the divergent object is the same alternating sequence.
The index range is not a technicality. $b_{0} = 0$ , so $a_{0} / b_{0}$ does not denote anything, and the quotient sequence exists only from $k = 1$ . That is why 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 $a_{k} / b_{k} \to L$ is stated for the tail, and why the convergence asserted above is asserted over $k \geq 1$ .

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

A summability matrix failing exactly one Silverman-Toeplitz condition and transforming a convergent sequence to a divergent one

Statement refuted

Refuted claim: a summability matrix whose columns tend to $0$ and whose row sums tend to $1$ is regular; equivalently, the uniform bound on the row absolute sums in 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 redundant.

The witness is the matrix with exactly two nonzero entries in each row,

$c_{n, n} : = - (n + 1), c_{n, n + 1} : = n + 2, c_{n, k} : = 0 for k \notin {n, n + 1},$

together with the null sequence $x_{k} : = s_{k} / (k + 1)$ , where $(s_{k})$ is the alternating sequence 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 $s \circ o \equiv - 1$ . Every column of $c$ is eventually $0$ ; every row sum is exactly $1$ ; the row absolute sums are $2 n + 3$ and are unbounded. The transform of $(x_{k})$ is

$y_{n} = - (n + 1) \frac{s_{n}}{n + 1} + (n + 2) \frac{s_{n + 1}}{n + 2} = - s_{n} + s_{n + 1} = - 2 s_{n},$

which does not converge although $x_{k} \to 0$ . So $c$ is not regular, and the third condition 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 is not redundant.

Facts & Assumptions

Given: The matrix $c$ above, the alternating sequence $(s_{k})$ with $s_{0} = 1$ and $s_{σ (k)} = - s_{k}$ , and the sequence $x_{k} : = s_{k} ((k + 1) \cdot 1_{R})^{- 1}$ .

[L1]

Summability matrices, the transform, row sums, row absolute sums 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); finite sums and their laws (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L2]

The Silverman-Toeplitz conditions and the theorem that they characterise 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).

[L3]

The alternating sequence: $∣ s_{k} ∣ = 1$ , $s_{k + 1} = - s_{k}$ (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 $s \circ o \equiv - 1$ ); it does not converge (FALSE: every bounded sequence converges).

[L4]

Convergence of real sequences (Limits and Cauchy sequences of reals); a sequence that is eventually $0$ converges to $0$ ; the reciprocal Archimedean property (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ); no real bounds every canonical natural (Every complete ordered field is Archimedean, Lower bound, bounded below, bounded set).

[L5]

Algebra of limits, in particular the scalar-multiple rule (Algebra of limits: sums, scalar multiples, products and quotients).

[L6]

Order arithmetic: $(k + 1) \cdot 1_{R} > 0$ (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); $∣ u ∣ \geq 0$ and $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ (Basic properties of the absolute value); the order is total and transitive (Complete ordered field (least-upper-bound property), Ordered field).

Counterexample

technique · direct

1.1

$c$ is a summability matrix: row $n$ vanishes at every $k > n + 1$ , so $n + 1$ is an admissible bound for row $n$ .

1.2

Every column of $c$ converges to $0$ : for fixed $k$ , the entry $c_{n, k}$ is nonzero only when $n = k$ or $n + 1 = k$ , so $c_{n, k} = 0$ for every $n \geq k + 1$ and the column is eventually $0$ .

L1L4

1.3

Every row sum is $1$ : $\sum_{k} c_{n, k} = - (n + 1) + (n + 2) = 1$ , so the row sums form the constant sequence $1$ and converge to $1$ .

1.4

The row absolute sums are not bounded above: $\sum_{k} ∣ c_{n, k} ∣ = (n + 1) + (n + 2) = 2 n + 3$ , and no real exceeds every canonical natural.

L1L4L6

1.5

$(x_{k})$ converges to $0$ : $∣ x_{k} ∣ = ((k + 1) \cdot 1_{R})^{- 1}$ , and given a real $ε > 0$ and a natural $m \geq 1$ with $1 / m < ε$ , every $k \geq m$ has $∣ x_{k} - 0 ∣ < ε$ .

L3L4L6

2.1

The transform of $(x_{k})$ by $c$ is $y_{n} = c_{n, n} x_{n} + c_{n, n + 1} x_{n + 1} = - (n + 1) s_{n} ((n + 1) \cdot 1)^{- 1} + (n + 2) s_{n + 1} ((n + 2) \cdot 1)^{- 1} = - s_{n} + s_{n + 1} = - 2 s_{n}$ .

step 1.1step 1.5L1L3L6

3.1

$(y_{n})$ does not converge: were $y_{n} \to M$ , then $s_{n} = (- 1 / 2) y_{n}$ would converge to $- M / 2$ by the scalar-multiple rule, contradicting [L3].

step 2.1L3L5

4.1

So $c$ has null columns and row sums tending to $1$ , yet transforms the convergent sequence $(x_{k})$ into a divergent one and is therefore not regular; the claim is false, and by 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 what fails is exactly the uniform bound on the row absolute sums, as step 1.4 confirms.

step 1.2step 1.3step 1.4step 1.5step 3.1L1L2∎

Remarks

Exactly one condition fails, and the counterexample is arranged so. The columns are eventually $0$ and the row sums are constantly $1$ , so conditions 1 and 2 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 hold outright; only the uniform bound fails, and the failure is visible in a single line, $2 n + 3$ being unbounded.
How the failure is exploited. The two entries of row $n$ are large and of opposite sign, so they nearly cancel on a slowly varying input and do not cancel at all on an alternating one. The input $x_{k} = s_{k} / (k + 1)$ is chosen so that the large factors $n + 1$ and $n + 2$ exactly cancel the small factors $1 / (n + 1)$ and $1 / (n + 2)$ , leaving the undamped oscillation $- 2 s_{n}$ . That is the gliding hump of the necessity proof in 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, in its simplest possible instance.
The failure is the stronger of the two possible ones. A matrix can be irregular by changing a limit, for instance $c_{n, n} = 2$ and all other entries $0$ , whose row sums tend to $2$ rather than $1$ and which sends $(x_{k})$ to $(2 x_{k})$ . The matrix above destroys convergence altogether.
Contrast with the Cesaro matrix, whose rows are nonnegative and sum to $1$ , so its row absolute sums are constantly $1$ and it is regular (The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem). Uniform boundedness of the row absolute sums is what stops a weighting from amplifying, and it is the only one of the three conditions that is not tested by a single fixed input.

Equivalent Forms of Completeness: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The rational function field R(t)R(t) ordered by the eventual sign is an ordered field, worked out

Example

Facts & Assumptions

Verification

Remarks

Q(2)Q(2​) carries exactly two distinct field orders, exchanged by the conjugation 2↦−22​↦−2​

Example

Facts & Assumptions

Verification

Remarks

R((t−1))R((t−1)), the formal Laurent series field, is Cauchy complete, non-Archimedean, and lacks the least-upper-bound property

Example

Facts & Assumptions

Verification

Remarks

The Cesaro means of (−1)k(−1)k converge to 00 although the sequence diverges

Example

Facts & Assumptions

Verification

Remarks

Stolz-Cesaro gives (1+2+⋯+n)/n2→1/2(1+2+⋯+n)/n2→1/2 and (1p+⋯+np)/np+1→1/(p+1)(1p+⋯+np)/np+1→1/(p+1) for natural pp

Example

Facts & Assumptions

Verification

Remarks

In R(t)R(t) the rationals are not dense: no rational lies strictly between 00 and 1/t1/t

Statement refuted

Facts & Assumptions

Counterexample

Remarks

On a closed interval of QQ there is a continuous unbounded function, a bounded one with no maximum, and one without the intermediate value property

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Over QQ there is a nonconstant differentiable function with identically zero derivative, so Rolle and the mean value theorem both fail

Statement refuted

Facts & Assumptions

Counterexample

Remarks

ak=(−1)kak​=(−1)k, bk=kbk​=k have ak/bk→0ak​/bk​→0 while the difference quotient oscillates, so Stolz-Cesaro has no converse

Statement refuted

Facts & Assumptions

Counterexample

Remarks

A summability matrix failing exactly one Silverman-Toeplitz condition and transforming a convergent sequence to a divergent one

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Sources

The rational function field $R (t)$ ordered by the eventual sign is an ordered field, worked out

$Q (\sqrt{2})$ carries exactly two distinct field orders, exchanged by the conjugation $\sqrt{2} \mapsto - \sqrt{2}$

$R ((t^{- 1}))$ , the formal Laurent series field, is Cauchy complete, non-Archimedean, and lacks the least-upper-bound property

The Cesaro means of $(- 1)^{k}$ converge to $0$ although the sequence diverges

Stolz-Cesaro gives $(1 + 2 + \dots + n) / n^{2} \to 1 / 2$ and $(1^{p} + \dots + n^{p}) / n^{p + 1} \to 1 / (p + 1)$ for natural $p$

In $R (t)$ the rationals are not dense: no rational lies strictly between $0$ and $1 / t$

On a closed interval of $Q$ there is a continuous unbounded function, a bounded one with no maximum, and one without the intermediate value property

Over $Q$ there is a nonconstant differentiable function with identically zero derivative, so Rolle and the mean value theorem both fail

$a_{k} = (- 1)^{k}$ , $b_{k} = k$ have $a_{k} / b_{k} \to 0$ while the difference quotient oscillates, so Stolz-Cesaro has no converse