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✓ 11 results · all verified · 6 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 5 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Monotone Sequences, Bolzano-Weierstrass, and Cauchy 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-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

The Babylonian sequence $x_{1} = 2$ , $x_{k + 1} = (x_{k} + 2 / x_{k}) / 2$ decreases to $\sqrt{2}$

Example

Let $(x_{k})_{k \geq 1}$ be given by

$x_{1} = 2, x_{k + 1} = \frac{1}{2} (x_{k} + \frac{2}{x_{k}}) (k \geq 1) .$

Then $(x_{k})$ is strictly decreasing, every term satisfies $x_{k} > \sqrt{2}$ , and

$\lim_{k} x_{k} = \sqrt{2} .$

This is the Babylonian, or Heron, iteration for the square root, and it is the standard illustration of A nondecreasing sequence bounded above converges to the supremum of its range, and a nonincreasing sequence bounded below to the infimum: monotone plus bounded delivers a limit, and the recursion then identifies the limit, because the limit must be a fixed point of the map that produced the sequence.

Indexing. Sequences in this library are functions on $N$ , which starts at $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences). The family above, indexed from $1$ , is realised as $x_{k} : = g_{k - 1}$ for the sequence $(g_{j})_{j \in N}$ with $g_{0} = 2$ and $g_{j + 1} = (g_{j} + 2 / g_{j}) / 2$ , and the verification below works with $(g_{j})$ . The shift changes nothing: convergence and monotonicity read the same under it (Convergence depends only on the tail).

Facts & Assumptions

Given: The set $A : = {u \in R : u > 0}$ , the element $2 \in A$ , and the function $f : A \to A$ with $f (u) = (u + 2 / u) / 2$ , which does land in $A$ because $u > 0$ gives $2 / u > 0$ and hence $f (u) > 0$ ; by the recursion theorem (The recursion theorem) the unique $g : N \to A$ with $g_{0} = 2$ and $g_{j + 1} = f (g_{j})$ . We write $x_{k} : = g_{k - 1}$ for $k \geq 1$ , so $x_{1} = 2$ and $x_{k + 1} = f (x_{k})$ .

[L1]

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

[L2]

Square roots: every $a \geq 0$ has a unique $s \geq 0$ with $s^{2} = a$ ; in particular $(\sqrt{2})^{2} = 2$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Integer powers $a^{m}$ ).

[L3]

Powers and order: for $a, b \geq 0$ , $a < b$ exactly when $a^{2} < b^{2}$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L4]

A nonzero square is positive: $t \neq 0$ gives $t^{2} > 0$ (Squares of nonzero elements are positive).

[L5]

Order and arithmetic: $1 > 0$ , hence $2 = 1 + 1 > 0$ and $2 > 1$ ; sums of positives are positive; adding a constant preserves the order; a positive has a positive inverse, and a quotient of positives is positive (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Inverses of positives are positive, and reciprocation reverses order, Ordered field, Complete ordered field (least-upper-bound property)).

[L6]

Monotone sequences, with consecutive comparisons sufficing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences), and boundedness below of a subset of $R$ (Lower bound, bounded below, bounded set).

[L7]

Monotone convergence: a nonincreasing sequence whose range is bounded below converges, to the infimum of its range (A nondecreasing sequence bounded above converges to the supremum of its range, and a nonincreasing sequence bounded below to the infimum, Limits and Cauchy sequences of reals).

[L8]

Algebra of limits, including the quotient case when the denominators and the limit are nonzero (Algebra of limits: sums, scalar multiples, products and quotients); limits preserve non-strict inequalities (Limits preserve non-strict inequalities); a sequence and its tails have the same limits (Convergence depends only on the tail); limits are unique (A sequence has at most one limit).

Verification

technique · direct

1.1

Every term is positive, since $g$ takes values in $A$ by construction.

givenL5

1.2

By induction, $(g_{j})^{2} > 2$ for every $j$ . Base: $(g_{0})^{2} = 4 > 2$ . Step: assuming $(g_{j})^{2} > 2$ , the identity $(g_{j + 1})^{2} - 2 = (\frac{(g_{j})^{2} - 2}{2 g_{j}})^{2}$ holds by field arithmetic, and its right-hand side is the square of a nonzero element, since $(g_{j})^{2} - 2 > 0$ and $2 g_{j} > 0$ , hence is $> 0$ .

givenL1L4L5

2.1

Every term satisfies $g_{j} > \sqrt{2}$ : both $g_{j}$ and $\sqrt{2}$ are $\geq 0$ , and $(\sqrt{2})^{2} = 2 < (g_{j})^{2}$ .

step 1.1step 1.2L2L3

2.2

The sequence is strictly decreasing: $g_{j} - g_{j + 1} = g_{j} - \frac{1}{2} (g_{j} + 2 / g_{j}) = \frac{(g_{j})^{2} - 2}{2 g_{j}} > 0$ , since the numerator is positive by step 1.2 and the denominator by step 1.1; consecutive comparisons then give strict decrease, hence also that $(g_{j})$ is nonincreasing.

step 1.1step 1.2L5L6

3.1

The range of $(g_{j})$ is bounded below by $\sqrt{2}$ , so by monotone convergence $(g_{j})$ converges; write $L$ for its limit.

step 2.1step 2.2L6L7

4.1

$L \geq \sqrt{2}$ : the inequality $g_{j} > \sqrt{2}$ holds at every index, so it passes to the limit in its non-strict form. In particular $L > 0$ , since $\sqrt{2} > 0$ by $(\sqrt{2})^{2} = 2 > 0$ and $\sqrt{2} \geq 0$ .

step 2.1step 3.1L2L4L8

5.1

By the algebra of limits, using $g_{j} \neq 0$ for every $j$ and $L \neq 0$ , the sequence $j \mapsto \frac{1}{2} (g_{j} + 2 / g_{j})$ converges to $\frac{1}{2} (L + 2 / L)$ .

step 1.1step 3.1step 4.1L8

6.1

The sequence $j \mapsto g_{j + 1}$ is the first tail of $(g_{j})$ , so it also converges to $L$ ; and it is the same sequence as in step 5.1, by the recursion clause.

step 3.1L8

7.1

By uniqueness of limits, $L = \frac{1}{2} (L + 2 / L)$ , hence $2 L = L + 2 / L$ , hence $L = 2 / L$ and $L^{2} = 2$ .

step 5.1step 6.1L8L5

8.1

Since $L \geq 0$ and $L^{2} = 2$ , uniqueness of the nonnegative square root gives $L = \sqrt{2}$ . So $(g_{j})$ , and with it $(x_{k})$ , is strictly decreasing, stays above $\sqrt{2}$ , and converges to $\sqrt{2}$ .

step 4.1step 7.1step 2.1step 2.2L2∎

Remarks

The identification of the limit is the interesting half. Monotone convergence produces $L$ but says only that it is the infimum of the range, which is not a usable description. Passing to the limit in the recursion turns the description into an equation, $L = (L + 2 / L) / 2$ , and that equation has exactly one nonnegative solution. The step that makes this legitimate is Convergence depends only on the tail: the shifted sequence $(g_{j + 1})$ has the same limit as $(g_{j})$ , so the two sides of the recursion may be compared in the limit.
The hypothesis $L \neq 0$ is not free. The quotient case of Algebra of limits: sums, scalar multiples, products and quotients requires it, and it is supplied by step 4.1, not assumed. Had the sequence been allowed to approach $0$ the argument would break exactly there, and this is the usual place where a proof of this example is incomplete.
Convergence is very fast, though nothing above uses that. The identity in step 1.2 also gives $g_{j + 1} - \sqrt{2} = (g_{j} - \sqrt{2})^{2} / (2 g_{j})$ , so the error is squared at each step: the iteration is Newton's method applied to $u^{2} - 2$ . The contractive estimate of Every contractive sequence is Cauchy, hence converges, with error bound $∣ x - x_{k} ∣ \leq c^{k - 1} ∣ x_{2} - x_{1} ∣ / (1 - c)$ for $k \geq 1$ would give only geometric decay, so it is a weaker tool here, and the monotone route is both shorter and sharper.
Nothing in the argument is special to $2$ . The same proof with $2$ replaced by any $a > 0$ , starting from any $x_{1} > 0$ with $x_{1}^{2} > a$ , converges to $\sqrt{a}$ . The starting value $2$ is chosen because $2^{2} = 4 > 2$ makes the base case of step 1.2 immediate.

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

The sequence $x_{1} = 1$ , $x_{k + 1} = \sqrt{2 + x_{k}}$ increases to $2$

Example

Let $(x_{k})_{k \geq 1}$ be given by

$x_{1} = 1, x_{k + 1} = \sqrt{2 + x_{k}} (k \geq 1) .$

Then $(x_{k})$ is strictly increasing, every term satisfies $0 \leq x_{k} < 2$ , and

$\lim_{k} x_{k} = 2.$

Informally this is the value of the nested radical $\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}$ , and the point of the example is that the expression means nothing until the sequence is shown to converge; only then does passing to the limit in the recursion identify the value.

Indexing. As in The Babylonian sequence $x_{1} = 2$ , $x_{k + 1} = (x_{k} + 2 / x_{k}) / 2$ decreases to $\sqrt{2}$ , the family indexed from $1$ is realised as $x_{k} : = g_{k - 1}$ for the sequence $(g_{j})_{j \in N}$ with $g_{0} = 1$ and $g_{j + 1} = \sqrt{2 + g_{j}}$ , and the verification works with $(g_{j})$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Convergence depends only on the tail).

Facts & Assumptions

Given: The set $A : = {u \in R : 0 \leq u < 2}$ , the element $1 \in A$ , and the function $f : A \to A$ with $f (u) = \sqrt{2 + u}$ ; by the recursion theorem (The recursion theorem) the unique $g : N \to A$ with $g_{0} = 1$ and $g_{j + 1} = f (g_{j})$ . We write $x_{k} : = g_{k - 1}$ for $k \geq 1$ , so $x_{1} = 1$ and $x_{k + 1} = \sqrt{2 + x_{k}}$ .

[L1]

Recursion theorem (The recursion theorem).

[L2]

Square roots: every $a \geq 0$ has a unique $s \geq 0$ with $s^{2} = a$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Integer powers $a^{m}$ ).

[L3]

Powers and order: for $a, b \geq 0$ , $a < b$ exactly when $a^{2} < b^{2}$ , and $a \leq b$ exactly when $a^{2} \leq b^{2}$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L4]

Order and arithmetic: $1 > 0$ , so $2 = 1 + 1 > 0$ and $4 = 2 \cdot 2$ ; adding a constant preserves the order, and inequalities may be added (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Ordered field, Complete ordered field (least-upper-bound property)).

[L5]

A field has no zero divisors: $u v = 0$ forces $u = 0$ or $v = 0$ (A field has no zero divisors: $a b = 0 \Rightarrow a = 0$ or $b = 0$ ).

[L6]

Monotone sequences, with consecutive comparisons sufficing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences); boundedness above of a subset of $R$ (Lower bound, bounded below, bounded set).

[L7]

Monotone convergence: a nondecreasing sequence whose range is bounded above converges, to the supremum of its range (A nondecreasing sequence bounded above converges to the supremum of its range, and a nonincreasing sequence bounded below to the infimum, Limits and Cauchy sequences of reals).

[L8]

Algebra of limits (Algebra of limits: sums, scalar multiples, products and quotients); limits preserve non-strict inequalities (Limits preserve non-strict inequalities); a sequence and its tails have the same limits (Convergence depends only on the tail); limits are unique (A sequence has at most one limit).

Verification

technique · direct

1.1

The function $f$ does map $A$ into $A$ , so the construction is legitimate: for $0 \leq u < 2$ we have $2 \leq 2 + u < 4$ , hence $\sqrt{2 + u} \geq 0$ and $(\sqrt{2 + u})^{2} = 2 + u < 4 = 2^{2}$ , which gives $\sqrt{2 + u} < 2$ .

givenL2L3L4

1.2

Consequently every term satisfies $0 \leq g_{j} < 2$ , since $g$ takes its values in $A$ .

givenL1

2.1

The sequence is strictly increasing. Fix $j$ . From $0 \leq g_{j} < 2$ we get $g_{j} - 2 < 0$ and $g_{j} + 1 > 0$ , so $(g_{j} - 2) (g_{j} + 1) < 0$ , that is $(g_{j})^{2} - g_{j} - 2 < 0$ , that is $(g_{j})^{2} < 2 + g_{j} = (g_{j + 1})^{2}$ . Since $g_{j} \geq 0$ and $g_{j + 1} \geq 0$ , this gives $g_{j} < g_{j + 1}$ ; consecutive comparisons then give strict increase, hence also that $(g_{j})$ is nondecreasing.

step 1.2L2L3L4L6

3.1

The range of $(g_{j})$ is bounded above by $2$ , so by monotone convergence $(g_{j})$ converges; write $L$ for its limit.

step 1.2step 2.1L6L7

4.1

$1 \leq L \leq 2$ : the inequalities $1 = g_{0} \leq g_{j} \leq 2$ hold at every index, by step 2.1 and step 1.2, and pass to the limit in their non-strict form.

step 1.2step 2.1step 3.1L8

4.2

The sequence $j \mapsto g_{j + 1}$ is the first tail of $(g_{j})$ , so it converges to $L$ , and therefore $j \mapsto (g_{j + 1})^{2}$ converges to $L^{2}$ by the algebra of limits.

step 3.1L8

4.3

The same sequence satisfies $(g_{j + 1})^{2} = 2 + g_{j}$ for every $j$ , and $j \mapsto 2 + g_{j}$ converges to $2 + L$ .

step 3.1L8L2

5.1

By uniqueness of limits, $L^{2} = 2 + L$ , that is $(L - 2) (L + 1) = 0$ .

step 4.2step 4.3L8

6.1

Since $L \geq 1$ we have $L + 1 \geq 2 > 0$ , so $L + 1 \neq 0$ , and a field has no zero divisors, so $L - 2 = 0$ and $L = 2$ . Thus $(g_{j})$ , and with it $(x_{k})$ , is strictly increasing, lies in $[0, 2)$ , and converges to $2$ .

step 4.1step 5.1step 1.2step 2.1L4L5∎

Remarks

The limit is not attained. Every term is strictly below $2$ and the limit is $2$ , which is the supremum of the range and does not belong to it. That is the ordinary situation for a strictly increasing convergent sequence, and it is why A nondecreasing sequence bounded above converges to the supremum of its range, and a nonincreasing sequence bounded below to the infimum is stated with a supremum rather than a maximum.
The quadratic has two roots and only one is admissible. The limit equation $L^{2} = 2 + L$ is solved by $L = 2$ and by $L = - 1$ . The second is excluded by step 4.1, which is why the bound $L \geq 1$ is proved rather than waved through: without it the argument would identify the limit only up to a sign, and a reader who writes down the limit equation without checking the range of $L$ has proved strictly less than the example claims.
Squaring the recursion avoids a continuity argument. Passing to the limit in $g_{j + 1} = \sqrt{2 + g_{j}}$ directly would need continuity of the square root, which this library has not proved at this point. Squaring first turns the recursion into $(g_{j + 1})^{2} = 2 + g_{j}$ , in which only the algebra of limits is required. The device is worth remembering: an identity between polynomials in the terms passes to the limit for free, whereas an identity involving a function does not.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passverified 2026-08-05 (claude-sonnet-5)Open item page →

The nested intervals $[0, 1 / k]$ intersect in exactly ${0}$

Example

For $k \geq 1$ let $I_{k} : = [0, 1 / k]$ , a nonempty closed bounded interval (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). The family is nested, its lengths tend to $0$ , and

$⋂_{k \geq 1} [0, \frac{1}{k}] = {0} .$

This is the standard instance of the single-point case of A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ , and the intersection is computed twice over: once by the theorem, which says the intersection is a single point, and once by inspection, which says that point is $0$ .

Indexing. Written on $N$ , the family is $J_{j} : = [0, 1 / (j + 1)]$ for $j \in N$ , which is the same family under the substitution $k = j + 1$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences). The verification uses $(J_{j})$ .

Facts & Assumptions

Given: For $j \in N$ the closed bounded interval $J_{j} : = [0, 1 / (j + 1)]$ , where $j + 1$ denotes the canonical natural $(j + 1) \cdot 1_{R}$ , which is positive and hence invertible; and the lengths $ℓ_{j} : = 1 / (j + 1) - 0 = 1 / (j + 1)$ .

[L1]

Intervals: $[a, b] = {x : a \leq x \leq b}$ is a closed bounded interval, nonempty exactly when $a \leq b$ , of length $b - a$ ; and $[a, a] = {a}$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L2]

Nested interval property: a nested sequence of nonempty closed bounded intervals has nonempty intersection, and that intersection is a single point exactly when the lengths tend to $0$ (A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ ).

[L3]

Canonical naturals: $n \cdot 1_{R} > 0$ for $n \geq 1$ , and $m \mapsto m \cdot 1_{R}$ is strictly increasing (Canonical naturals are positive and strictly increasing).

[L4]

Reciprocals: $a > 0$ gives $1 / a > 0$ , and $0 < a < b$ gives $0 < 1 / b < 1 / a$ (Inverses of positives are positive, and reciprocation reverses order).

[L5]

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

[L6]

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

[L7]

Convergence of a sequence of reals to $0$ ; it suffices to test a real $ε > 0$ (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L8]

Trichotomy of the order on $R$ (Complete ordered field (least-upper-bound property), Ordered field).

Verification

technique · direct

1.1

Each $J_{j}$ is a nonempty closed bounded interval: $j + 1 \geq 1$ gives $1 / (j + 1) > 0$ , so $0 \leq 1 / (j + 1)$ and [L1] applies; its length is $ℓ_{j} = 1 / (j + 1)$ .

givenL1L3L4

2.1

The family is nested: $0 < (j + 1) < (j + 2)$ gives $0 < 1 / (j + 2) < 1 / (j + 1)$ , so $0 \leq x \leq 1 / (j + 2)$ implies $0 \leq x \leq 1 / (j + 1)$ , that is $J_{j + 1} \subseteq J_{j}$ .

step 1.1L3L4

2.2

The lengths tend to $0$ . Let $ε > 0$ be real and use [L5] to fix a natural $n \geq 1$ with $1 / n < ε$ . For every $j \geq n$ we have $0 < n \leq j + 1$ , hence $0 < 1 / (j + 1) \leq 1 / n < ε$ , and $∣ ℓ_{j} - 0 ∣ = ℓ_{j} < ε$ since $ℓ_{j} > 0$ .

step 1.1L3L4L5L6L7L8

3.1

By [L2] applied to steps 1.1, 2.1 and 2.2, the intersection $⋂_{j} J_{j}$ is nonempty and is a single point.

step 1.1step 2.1step 2.2L2

4.1

That point is $0$ : indeed $0 \in J_{j}$ for every $j$ , since $0 \leq 0 \leq 1 / (j + 1)$ , so $0$ lies in the intersection, and a set that is a single point and contains $0$ is ${0}$ .

step 3.1step 1.1L1

5.1

Hence $⋂_{j} J_{j} = {0}$ , which in the notation of the statement is $⋂_{k \geq 1} [0, 1 / k] = {0}$ .

step 3.1step 4.1L1∎

Remarks

The two computations are independent and both are needed. That the intersection is a single point comes from A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ and uses that the lengths are null; that the point is $0$ comes from inspection. Without the theorem one would still have to rule out the intersection being larger than ${0}$ , which is exactly the content of the length condition.
The Archimedean property is the whole of step 2.2. In a non-Archimedean ordered field the same family has lengths that do not tend to $0$ , and the intersection contains every positive infinitesimal, so it is not a single point. What makes the example come out as stated is For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ .
Compare the open version. Removing the left endpoint gives $(0, 1 / k)$ , whose intersection is empty (The nested open intervals $(0, 1 / k)$ have empty intersection). The two computations differ in exactly one respect, whether the common point $0$ belongs to the sets, and that is the hypothesis of closedness in the theorem.

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

The sequence $(- 1)^{k} (1 + 1 / k)$ is bounded with subsequential limit set exactly ${- 1, 1}$

Example

For $k \geq 1$ let

$x_{k} = (- 1)^{k} (1 + \frac{1}{k}) .$

Then $(x_{k})$ is bounded, with $1 < ∣ x_{k} ∣ \leq 2$ at every index, it does not converge, and its subsequential limit set (Subsequential limit of a real sequence, and the subsequential limit set) is exactly

$SL (x) = {- 1, 1} .$

The example separates two things that a first reading of Bolzano-Weierstrass can run together. A bounded sequence must have a subsequential limit; it may have several; and having several is exactly what stops it converging. Here there are two, and neither is a value of the sequence, since $∣ x_{k} ∣ > 1$ always.

Indexing and the sign. Written on $N$ the sequence is $u_{j} : = t_{j} (1 + 1 / (j + 1))$ for $j \in N$ , where $t_{j} : = - s_{j}$ and $(s_{k})$ 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 $s \circ o \equiv - 1$ . Since $s_{σ (j)} = - s_{j}$ , the sequence $(t_{j})$ is $j \mapsto s_{j + 1}$ , so $u_{j} = x_{j + 1}$ and $(u_{j})$ is the family above under the substitution $k = j + 1$ . The verification uses $(u_{j})$ .

Facts & Assumptions

Given: The alternating sequence $(s_{k})$ and the index maps $e, o$ 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 sequence $t_{j} : = - s_{j}$ ; the sequence $p_{j} : = 1 + 1 / (j + 1)$ , where $j + 1$ denotes the canonical natural $(j + 1) \cdot 1_{R}$ ; and $u_{j} : = t_{j} p_{j}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

The alternating sequence: $∣ s_{k} ∣ = 1$ for every $k$ , $s_{e_{i}} = 1$ and $s_{o_{i}} = - 1$ for every $i$ , and $e$ and $o$ are strictly increasing (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]

Canonical naturals: $n \cdot 1_{R} > 0$ for $n \geq 1$ , and $m \mapsto m \cdot 1_{R}$ is strictly increasing (Canonical naturals are positive and strictly increasing).

[L3]

Reciprocals: $a > 0$ gives $1 / a > 0$ , and $0 < a \leq b$ gives $0 < 1 / b \leq 1 / a$ (Inverses of positives are positive, and reciprocation reverses order).

[L4]

[L5]

Absolute value: $∣ a b ∣ = ∣ a ∣ ∣ b ∣$ , $∣ t ∣ \geq 0$ , $∣ t ∣ = t$ for $t \geq 0$ , and $∣ t ∣ = 1$ forces $t = 1$ or $t = - 1$ (Basic properties of the absolute value, Absolute value in an ordered field).

[L6]

Algebra of limits (Algebra of limits: sums, scalar multiples, products and quotients); subsequences inherit the limit (Subsequences inherit the limit); the absolute value is compatible with limits (The absolute value is compatible with limits); limits are unique (A sequence has at most one limit).

[L7]

Convergence and boundedness of a sequence of reals; it suffices to test a real $ε > 0$ (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L8]

Subsequential limits: $L \in SL (u)$ exactly when some subsequence of $(u_{j})$ converges to $L$ (Subsequential limit of a real sequence, and the subsequential limit set).

[L9]

Trichotomy of the order on $R$ (Complete ordered field (least-upper-bound property), Ordered field).

Verification

technique · direct

1.1

For every $j$ : $j + 1 \geq 1$ gives $0 < 1 / (j + 1) \leq 1$ , so $1 < p_{j} \leq 2$ ; and $∣ t_{j} ∣ = ∣ - s_{j} ∣ = ∣ s_{j} ∣ = 1$ .

givenL1L2L3L5

1.2

The sequence $(p_{j})$ converges to $1$ : given a real $ε > 0$ , [L4] supplies a natural $n \geq 1$ with $1 / n < ε$ , and for $j \geq n$ we have $0 < 1 / (j + 1) \leq 1 / n < ε$ , so $∣ p_{j} - 1 ∣ = 1 / (j + 1) < ε$ .

givenL2L3L4L5L7

2.1

$(u_{j})$ is bounded, with $1 < ∣ u_{j} ∣ = ∣ t_{j} ∣ p_{j} = p_{j} \leq 2$ at every index.

step 1.1L5L7

2.2

Along the even index map: $t_{e_{i}} = - s_{e_{i}} = - 1$ , so $u_{e_{i}} = - p_{e_{i}}$ ; since $e$ is strictly increasing, $(p_{e_{i}})_{i}$ is a subsequence of $(p_{j})$ and so converges to $1$ , whence $u_{e_{i}} \to - 1$ .

step 1.2L1L6L8

2.3

Along the odd index map: $t_{o_{i}} = - s_{o_{i}} = 1$ , so $u_{o_{i}} = p_{o_{i}} \to 1$ by the same argument.

step 1.2L1L6L8

3.1

Hence $- 1 \in SL (u)$ and $1 \in SL (u)$ .

step 2.2step 2.3L8

3.2

Conversely, let $L \in SL (u)$ and fix a strictly increasing $n$ with $u_{n_{i}} \to L$ . Then $∣ u_{n_{i}} ∣ \to ∣ L ∣$ ; but $∣ u_{n_{i}} ∣ = p_{n_{i}}$ by step 2.1, and $(p_{n_{i}})_{i}$ is a subsequence of $(p_{j})$ , so it converges to $1$ .

step 1.2step 2.1L6L8

4.1

By uniqueness of limits $∣ L ∣ = 1$ , so $L = 1$ or $L = - 1$ .

step 3.2L5L6L9

5.1

Combining, $SL (u) = {- 1, 1}$ ; and $(u_{j})$ does not converge, since a convergent sequence has exactly one subsequential limit. Bounded by step 2.1, the sequence $(u_{j})$ , that is $(x_{k})$ , therefore has the asserted properties.

step 2.1step 3.1step 4.1L6L8∎

Remarks

Where Bolzano-Weierstrass sits. Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence guarantees that a bounded sequence has at least one subsequential limit; this example computes the whole set and finds two. Nothing above uses the theorem, since the two subsequential limits are exhibited directly, and that is the honest order of business: an existence theorem is not needed once a witness is in hand.
The perturbation is what makes the example more than the alternating sequence. With $p_{j}$ replaced by the constant $1$ the sequence is $\pm 1$ alternating and each subsequential limit is attained infinitely often. Here $∣ x_{k} ∣ > 1$ at every index, so neither subsequential limit is ever a value of the sequence. Subsequential limits are limits, not values.
The converse direction is where the partition earns its keep. Step 4.1 rules out every other candidate by taking absolute values, which collapses the sign and leaves the single null perturbation to handle. The alternative, splitting an arbitrary subsequence according to how many of its indices are even, needs the disjointness half 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 $s \circ o \equiv - 1$ and is longer.
This is the standard example of a bounded divergent sequence with a computable $lim sup$ and $lim inf$ , namely $1$ and $- 1$ . Those notions are developed on the next page of the track, and this item is written so that it can be cited there without change.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-08-06 (claude-sonnet-5)Open item page →

The sequence $x_{k + 1} = (x_{k} + 1) / 3$ is contractive with $c = 1 / 3$ and converges to $1 / 2$

Example

Fix any $a \in R$ and let $(x_{k})$ be the sequence with $x_{0} = a$ and

$x_{k + 1} = \frac{x_{k} + 1}{3} (k \in N) .$

Then $(x_{k})$ is contractive with contraction constant $c = 1 / 3$ (Contractive sequence: $∣ x_{k + 2} - x_{k + 1} ∣ \leq c ∣ x_{k + 1} - x_{k} ∣$ for a fixed $0 < c < 1$ ), and

$\lim_{k} x_{k} = \frac{1}{2},$

whatever the starting value $a$ is. Moreover Every contractive sequence is Cauchy, hence converges, with error bound $∣ x - x_{k} ∣ \leq c^{k - 1} ∣ x_{2} - x_{1} ∣ / (1 - c)$ for $k \geq 1$ supplies the error bound $∣ 1 / 2 - x_{k} ∣ \leq 3^{- (k - 1)} ∣ x_{2} - x_{1} ∣ / (1 - 1 / 3)$ for $k \geq 1$ , which is computable from the first three terms alone.

The limit $1 / 2$ is the unique solution of $L = (L + 1) / 3$ , that is the unique fixed point of the map $u \mapsto (u + 1) / 3$ . This is the smallest honest instance of the Banach fixed point theorem: a contraction on $R$ has one fixed point, and every orbit converges to it.

Facts & Assumptions

Given: A real $a$ , and by the recursion theorem (The recursion theorem) applied to $R$ , the element $a$ and the function $f (u) = (u + 1) / 3$ , the unique sequence $(x_{k})$ of reals with $x_{0} = a$ and $x_{k + 1} = f (x_{k})$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences); the constant $c : = 1 / 3$ .

[L1]

Recursion theorem (The recursion theorem).

[L2]

Order and arithmetic: $1 > 0$ , so $3 = 1 + 1 + 1 > 1 > 0$ ; a positive has a positive inverse; and $0 < r < 1$ exactly when $1 / r > 1$ , so $0 < 1 / 3 < 1$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Inverses of positives are positive, and reciprocation reverses order, Reciprocals and order: $1 / r$ against $1$ , Ordered field, Complete ordered field (least-upper-bound property)).

[L3]

Field arithmetic: $(u + 1) / 3 - (v + 1) / 3 = (u - v) / 3$ , and $3 L = L + 1$ is equivalent to $2 L = 1$ (Field).

[L4]

Absolute value: $∣ a b ∣ = ∣ a ∣ ∣ b ∣$ , and $∣ t ∣ = t$ for $t \geq 0$ , so $∣ (u - v) / 3 ∣ = ∣ u - v ∣ / 3$ (Basic properties of the absolute value).

[L5]

Contractive sequences: a constant $c$ with $0 < c < 1$ and $∣ x_{k + 2} - x_{k + 1} ∣ \leq c ∣ x_{k + 1} - x_{k} ∣$ at every index (Contractive sequence: $∣ x_{k + 2} - x_{k + 1} ∣ \leq c ∣ x_{k + 1} - x_{k} ∣$ for a fixed $0 < c < 1$ ).

[L6]

Every contractive sequence is Cauchy and converges, with the stated error bound for $k \geq 1$ (Every contractive sequence is Cauchy, hence converges, with error bound $∣ x - x_{k} ∣ \leq c^{k - 1} ∣ x_{2} - x_{1} ∣ / (1 - c)$ for $k \geq 1$ , Limits and Cauchy sequences of reals).

[L7]

Algebra of limits (Algebra of limits: sums, scalar multiples, products and quotients); a sequence and its tails have the same limits (Convergence depends only on the tail); limits are unique (A sequence has at most one limit).

Verification

technique · direct

1.1

The constant $c = 1 / 3$ satisfies $0 < c < 1$ .

givenL2

1.2

For every $k$ : $x_{k + 2} - x_{k + 1} = \frac{x_{k + 1} + 1}{3} - \frac{x_{k} + 1}{3} = \frac{x_{k + 1} - x_{k}}{3}$ , so $∣ x_{k + 2} - x_{k + 1} ∣ = \frac{1}{3} ∣ x_{k + 1} - x_{k} ∣$ .

givenL1L3L4

2.1

Hence $(x_{k})$ is contractive with contraction constant $c = 1 / 3$ , the inequality of [L5] holding with equality at every index.

step 1.1step 1.2L5

3.1

By [L6] the sequence converges; write $L$ for its limit, and the error bound of [L6] holds for it with $c = 1 / 3$ .

step 2.1L6

4.1

The sequence $k \mapsto x_{k + 1}$ is the first tail of $(x_{k})$ , so it converges to $L$ ; and by the algebra of limits $k \mapsto (x_{k} + 1) / 3$ converges to $(L + 1) / 3$ .

step 3.1L7

5.1

The two sequences of step 4.1 are the same sequence, by the recursion clause, so uniqueness of limits gives $L = (L + 1) / 3$ , hence $3 L = L + 1$ , hence $2 L = 1$ and $L = 1 / 2$ .

step 4.1L1L3L7

6.1

So for every starting value $a$ the sequence is contractive with $c = 1 / 3$ and converges to $1 / 2$ , with the error bound of step 3.1.

step 2.1step 3.1step 5.1∎

Remarks

The starting value is genuinely arbitrary. Nothing in the verification uses anything about $a$ , and the limit does not depend on it. What does depend on it is the error bound, through $∣ x_{2} - x_{1} ∣$ , which for this recursion equals $∣ a - 1 / 2 ∣ \cdot 2 / 9$ .
Contractivity is exact here, not an estimate. Step 1.2 gives equality, $∣ x_{k + 2} - x_{k + 1} ∣ = \frac{1}{3} ∣ x_{k + 1} - x_{k} ∣$ , so for every starting value $a \neq 1 / 2$ the constant $c = 1 / 3$ is the smallest admissible one; for $a = 1 / 2$ the sequence is constant, every gap is $0$ , and every $c \in (0, 1)$ is admissible (Contractive sequence: $∣ x_{k + 2} - x_{k + 1} ∣ \leq c ∣ x_{k + 1} - x_{k} ∣$ for a fixed $0 < c < 1$ ). The error bound of Every contractive sequence is Cauchy, hence converges, with error bound $∣ x - x_{k} ∣ \leq c^{k - 1} ∣ x_{2} - x_{1} ∣ / (1 - c)$ for $k \geq 1$ is as sharp as that theorem can make it. Contrast $x_{k + 1} = x_{k} + 1 / x_{k}$ from $x_{1} = 1$ has strictly decreasing consecutive gaps and diverges, so no uniform $c < 1$ exists, where the ratio of consecutive gaps tends to $1$ and no admissible constant exists at all.
Why a fixed point is forced. The limit satisfies the recursion equation because a sequence and its shift have the same limit (Convergence depends only on the tail) and the algebra of limits transports the right-hand side. This is the same move as in The Babylonian sequence $x_{1} = 2$ , $x_{k + 1} = (x_{k} + 2 / x_{k}) / 2$ decreases to $\sqrt{2}$ and The sequence $x_{1} = 1$ , $x_{k + 1} = \sqrt{2 + x_{k}}$ increases to $2$ ; the difference is that here convergence comes from contractivity rather than from monotonicity, and no monotonicity is available, since for $a > 1 / 2$ the sequence decreases and for $a < 1 / 2$ it increases while for $a = 1 / 2$ it is constant.

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-07-26Open item page →

The truncated decimal approximations of $\sqrt{2}$ form a Cauchy sequence of rationals with no rational limit

Statement refuted

Refuted claim: every Cauchy sequence of rationals converges to a rational; equivalently, $Q$ is complete (FALSE: the rationals are complete, The rationals as equivalence classes of pairs of integers).

The witness is the sequence of truncated decimal approximations of $\sqrt{2}$ , $s_{0} = 1$ , $s_{1} = 1.4$ , $s_{2} = 1.41$ , $s_{3} = 1.414$ , and so on: $s_{n} = k_{n} / 10^{n}$ where $k_{n}$ is the largest natural number with $k_{n}^{2} \leq 2 \cdot 10^{2 n}$ . That this sequence is Cauchy in $Q$ and has no rational limit is proved in full in FALSE: the rationals are complete and is not repeated here.

What this item adds is the view from $R$ , which is what makes the witness informative rather than merely negative: the same sequence converges in $R$ , and its limit is $\sqrt{2}$ . So the defect is not in the sequence but in $Q$ , and the contrast is exactly The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges, which says that a Cauchy sequence of reals never behaves this way.

Facts & Assumptions

Given: For $n \in N$ the rational $s_{n} = k_{n} / 10^{n}$ , where $k_{n}$ is the largest natural with $k_{n}^{2} \leq 2 \cdot 10^{2 n}$ , together with the properties established for it in FALSE: the rationals are complete; and the real $\sqrt{2}$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ). Rationals are identified with their images in $R$ under the embedding $q \mapsto \hat{q}$ (The rationals embed densely in the reals), so $(s_{n})$ is also a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

The construction and its properties: $s_{n} \geq 0$ , $s_{n}^{2} \leq 2 < (s_{n} + 10^{- n})^{2}$ , and $(s_{n})$ is a Cauchy sequence of rationals with no rational limit (FALSE: the rationals are complete).

[L2]

The embedding $q \mapsto \hat{q}$ is an injective, order-preserving field homomorphism of $Q$ into $R$ , so every identity and inequality between rationals holds between their images and conversely (The rationals embed densely in the reals, The rationals as equivalence classes of pairs of integers).

[L3]

Square roots: $\sqrt{2} \geq 0$ is the unique nonnegative real with $(\sqrt{2})^{2} = 2$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Integer powers $a^{m}$ ).

[L4]

Powers and order: for $a, b \geq 0$ , $a \leq b$ exactly when $a^{2} \leq b^{2}$ , and $a < b$ exactly when $a^{2} < b^{2}$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ); and $(1 / t)^{n} = 1 / t^{n} = t^{- n}$ for $t \neq 0$ (Laws of integer exponents).

[L5]

Order: $10 > 1 > 0$ , so $0 < 1 / 10 < 1$ (Reciprocals and order: $1 / r$ against $1$ , Inverses of positives are positive, and reciprocation reverses order, Ordered field, Complete ordered field (least-upper-bound property)); and $∣ t ∣ = t$ for $t \geq 0$ (Basic properties of the absolute value).

[L6]

For $∣ r ∣ < 1$ the sequence $(r^{n})$ converges to $0$ (For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ ).

[L7]

Squeeze theorem (The squeeze theorem) and the algebra of limits (Algebra of limits: sums, scalar multiples, products and quotients); a constant sequence converges to its value (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

[L8]

Every convergent sequence of reals is Cauchy (Every convergent sequence is Cauchy); every Cauchy sequence of reals converges (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges); limits are unique (A sequence has at most one limit).

[L9]

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

Counterexample

technique · direct

1.1

The inequalities of [L1] hold verbatim in $R$ , since the embedding preserves the order and the field operations.

givenL1L2

1.2

$0 < 1 / 10 < 1$ and $(1 / 10)^{n} = 10^{- n}$ for every $n$ .

givenL4L5

2.1

From $s_{n} \geq 0$ , $\sqrt{2} \geq 0$ and $s_{n}^{2} \leq 2 = (\sqrt{2})^{2}$ we get $s_{n} \leq \sqrt{2}$ ; from $(\sqrt{2})^{2} = 2 < (s_{n} + 10^{- n})^{2}$ with both bases $\geq 0$ we get $\sqrt{2} < s_{n} + 10^{- n}$ . Hence $0 \leq \sqrt{2} - s_{n} < 10^{- n}$ for every $n$ .

step 1.1L3L4

2.2

The sequence $(10^{- n}) = ((1 / 10)^{n})$ converges to $0$ .

step 1.2L6

3.1

The constant sequence $0$ and the sequence $(10^{- n})$ both converge to $0$ , and $0 \leq \sqrt{2} - s_{n} \leq 10^{- n}$ at every index, so the squeeze theorem gives $\sqrt{2} - s_{n} \to 0$ ; by the algebra of limits $s_{n} = \sqrt{2} - (\sqrt{2} - s_{n}) \to \sqrt{2}$ .

step 2.1step 2.2L7

4.1

In particular $(s_{n})$ converges in $R$ and is therefore Cauchy as a sequence of reals; this is the behaviour The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges guarantees for every Cauchy sequence of reals, and it is what fails in $Q$ .

step 3.1L8

4.2

Suppose $(s_{n})$ converged to a rational $q$ . Then in $R$ it converges to $\hat{q}$ , so $\hat{q} = \sqrt{2}$ by uniqueness of limits, hence ${\hat{q}}^{2} = 2$ ; the embedding is injective and preserves squaring, so $q^{2} = 2$ in $Q$ .

step 3.1L2L3L8

5.1

No rational squares to $2$ , so no such $q$ exists: $(s_{n})$ is a Cauchy sequence of rationals with no rational limit, and the claim that $Q$ is complete is refuted.

step 4.2L1L9∎

Remarks

The limit exists; it is merely not rational. That is the entire content of the counterexample and the reason the construction of $R$ is worth doing. The sequence is Cauchy in $Q$ , so $Q$ "should" have a limit for it, and the point at which it converges lies outside $Q$ .
Decimal truncation is a convenience, not the mechanism. Any sequence of rationals converging to any irrational does the same job, for instance the Babylonian iterates of The Babylonian sequence $x_{1} = 2$ , $x_{k + 1} = (x_{k} + 2 / x_{k}) / 2$ decreases to $\sqrt{2}$ started at $2$ , which are all rational and converge to $\sqrt{2}$ . Truncated decimals are chosen because the two-sided estimate $s_{n} \leq \sqrt{2} < s_{n} + 10^{- n}$ is immediate from the definition of $k_{n}$ and turns into convergence with one application of the squeeze theorem.
Note which completeness is which. FALSE: the rationals are complete refutes Cauchy completeness of $Q$ . $Q$ also fails the least-upper-bound property, on the same underlying fact that $\sqrt{2} \notin Q$ , and the two failures are not the same statement: Cauchy completeness and Dedekind completeness differ in general, and coincide only in the presence of the Archimedean property. Two independent proofs that $R$ is Cauchy complete, and why the library records both records where this library stands on that.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passverified 2026-08-05 (claude-sonnet-5)Open item page →

The nested open intervals $(0, 1 / k)$ have empty intersection

Statement refuted

Refuted claim: a nested sequence of nonempty bounded open intervals has nonempty intersection (FALSE: a nested sequence of nonempty bounded open intervals has nonempty intersection, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

The witness is $J_{k} = (0, 1 / k)$ for $k \geq 1$ : each is a nonempty bounded open interval, the family is nested, and

$⋂_{k \geq 1} (0, \frac{1}{k}) = \emptyset .$

The refutation is carried out in full in FALSE: a nested sequence of nonempty bounded open intervals has nonempty intersection and is recorded here as the named counterexample. The comparison worth keeping in view is the closed family $[0, 1 / k]$ , which differs only by the inclusion of the left endpoint and intersects in ${0}$ ; that computation is The nested intervals $[0, 1 / k]$ intersect in exactly ${0}$ .

Facts & Assumptions

Given: For $j \in N$ the open interval $J_{j} : = {x \in R : 0 < x < 1 / (j + 1)}$ , which is the family $(0, 1 / k)$ for $k \geq 1$ under the substitution $k = j + 1$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

The family $(J_{j})$ consists of nonempty bounded open intervals, is nested, and has empty intersection (FALSE: a nested sequence of nonempty bounded open intervals has nonempty intersection, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L2]

Canonical naturals are positive and strictly increasing in the index (Canonical naturals are positive and strictly increasing); reciprocals of positives are positive and reciprocation reverses the order (Inverses of positives are positive, and reciprocation reverses order).

[L3]

[L4]

Trichotomy of the order on $R$ (Complete ordered field (least-upper-bound property), Ordered field).

[L5]

The refuted claim: a nested sequence of nonempty bounded open intervals has nonempty intersection.

Counterexample

technique · direct

1.1

Each $J_{j}$ is a nonempty bounded open interval and $J_{j + 1} \subseteq J_{j}$ , so the family is an instance of the claim, which asserts that its intersection is nonempty.

givenL1L5

2.1

Suppose $x$ belonged to every $J_{j}$ . Then $x > 0$ , and $x < 1 / (j + 1)$ for every $j \in N$ .

step 1.1L1

3.1

Since $x > 0$ , fix a natural $n \geq 1$ with $1 / n < x$ , and write $n = j + 1$ with $j \in N$ ; step 2.1 then gives $x < 1 / n$ as well, which trichotomy forbids.

step 2.1L2L3L4

4.1

So no such $x$ exists: $⋂_{j} J_{j} = \emptyset$ , and the claim is refuted by a family of nonempty bounded open intervals.

step 1.1step 3.1L1L5∎

Remarks

Exactly one hypothesis of A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ is missing. The intervals here are nonempty, bounded and nested; they are not closed. The true theorem is therefore not contradicted, and the counterexample shows that its closedness hypothesis cannot be dropped.
The candidate point exists and is excluded by a hair. With $J_{j} = (a_{j}, b_{j})$ one still has $a_{j} \to 0$ and $b_{j} \to 0$ , and the only possible common point is $0$ ; but $0 \notin J_{j}$ for any $j$ , because the left endpoint is excluded. Closedness is precisely the hypothesis that puts the limiting endpoint into each set. Compare The nested intervals $[0, 1 / k]$ intersect in exactly ${0}$ , where it is present and the intersection is ${0}$ .
The Archimedean property is doing the work in step 3.1. In a non-Archimedean ordered field a positive infinitesimal lies in every $(0, 1 / n)$ , and the intersection is nonempty. So this is a counterexample about $R$ , supplied by For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , and not a formal consequence of openness alone.
Boundedness is a separate hypothesis and fails separately. The nested closed unbounded sets $[k, \infty)$ have empty intersection, so boundedness cannot be dropped keeps closedness and drops boundedness, with the same empty intersection, so neither hypothesis implies the other.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

The nested closed unbounded sets $[k, \infty)$ have empty intersection, so boundedness cannot be dropped

Statement refuted

Refuted claim: a nested sequence of nonempty closed intervals has nonempty intersection, boundedness being unnecessary (A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

The witness is $I_{k} = [k, \infty)$ for $k \in N$ , where $k$ denotes the canonical natural of $R$ . Each is a nonempty closed interval, the family is nested, and

$⋂_{k \in N} [k, \infty) = \emptyset .$

A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ therefore cannot be improved by deleting "bounded" from its hypotheses. Together with the open-interval counterexample on this page, which deletes "closed" instead, this shows that the two hypotheses are independent and that neither is an artefact of the proof.

Facts & Assumptions

Given: For $k \in N$ the set $I_{k} : = {x \in R : k \leq x}$ , where $k$ denotes the canonical natural $k \cdot 1_{R}$ ; this is the closed interval $[k, \infty)$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

Intervals: $[a, \infty) = {x : a \leq x}$ is a closed interval, it is not bounded above, and it is nonempty since it contains $a$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L2]

Canonical naturals: $m \mapsto m \cdot 1_{R}$ is strictly increasing, so $k \leq k + 1$ in $R$ for every $k \in N$ (Canonical naturals are positive and strictly increasing).

[L3]

Archimedean property: for every real $x$ there is a natural $n \geq 1$ with $x < n \cdot 1_{R}$ (Every complete ordered field is Archimedean).

[L4]

Trichotomy and transitivity of the order on $R$ (Complete ordered field (least-upper-bound property), Ordered field).

[L5]

Nested interval property, for nonempty closed bounded intervals (A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ ).

[L6]

The refuted claim: a nested sequence of nonempty closed intervals has nonempty intersection.

Counterexample

technique · direct

1.1

Each $I_{k}$ is a nonempty closed interval, containing the canonical natural $k$ ; and none of them is bounded, since $[a, \infty)$ has no upper bound.

givenL1

2.1

The family is nested: $k \leq k + 1$ in $R$ , so $k + 1 \leq x$ implies $k \leq x$ , that is $I_{k + 1} \subseteq I_{k}$ .

step 1.1L2L4

3.1

Suppose $x$ belonged to every $I_{k}$ . Then $k \cdot 1_{R} \leq x$ for every $k \in N$ .

step 2.1L1

4.1

By the Archimedean property fix a natural $n \geq 1$ with $x < n \cdot 1_{R}$ ; step 3.1 applied to $k = n$ gives $n \cdot 1_{R} \leq x$ , which trichotomy forbids.

step 3.1L3L4

5.1

So no such $x$ exists: the family $(I_{k})$ consists of nonempty closed intervals, is nested, and has empty intersection. The claim is refuted, and boundedness cannot be dropped from A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ .

step 1.1step 2.1step 4.1L5L6∎

Remarks

What fails is the existence of the supremum, not the argument's bookkeeping. In the proof of A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ the intersection is computed as $[\sup a_{k}, \inf b_{k}]$ . Here $a_{k} = k$ and the set ${k : k \in N}$ has no supremum in $R$ , precisely because $R$ is Archimedean, so there is no candidate point at all. This is a different failure mode from The nested open intervals $(0, 1 / k)$ have empty intersection, where the candidate point exists and is merely not a member.
The sets are intervals but not bounded intervals. Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length admits $[a, \infty)$ as one of its nine forms and assigns it no length, which is exactly why the length hypothesis of the nested interval property has nothing to say about this family.
The Archimedean property is again what makes the intersection empty. In a non-Archimedean ordered field an element exceeding every canonical natural lies in every $I_{k}$ , so the intersection is nonempty there. The counterexample is a statement about $R$ .

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

$x_{k} = \sqrt{k}$ has $x_{k + 1} - x_{k} \to 0$ and is not Cauchy

Statement refuted

Refuted claim: a sequence of reals whose consecutive differences tend to $0$ is Cauchy (FALSE: if $∣ x_{k + 1} - x_{k} ∣ \to 0$ then $(x_{k})$ is Cauchy, Limits and Cauchy sequences of reals).

The witness is $x_{k} = \sqrt{k}$ for $k \in N$ . Its consecutive differences satisfy

$x_{k + 1} - x_{k} = \sqrt{k + 1} - \sqrt{k} = \frac{1}{\sqrt{k + 1} + \sqrt{k}} ⟶ 0,$

while the sequence itself is unbounded, hence not Cauchy (Every Cauchy sequence of reals is bounded). The refutation is carried out in full in FALSE: if $∣ x_{k + 1} - x_{k} ∣ \to 0$ then $(x_{k})$ is Cauchy; this item records the witness and adds the sharper statement that $\sqrt{k}$ diverges to $+ \infty$ (Divergence to $+ \infty$ and to $- \infty$ ).

Facts & Assumptions

Given: The sequence $(x_{k})$ of reals with $x_{k} : = \sqrt{k}$ , where $k$ denotes the canonical natural $k \cdot 1_{R}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[L1]

The witness and its two properties: $x_{k + 1} - x_{k} = 1 / (\sqrt{k + 1} + \sqrt{k})$ tends to $0$ , and $(x_{k})$ is unbounded and not Cauchy (FALSE: if $∣ x_{k + 1} - x_{k} ∣ \to 0$ then $(x_{k})$ is Cauchy).

[L2]

Square roots, and the factorisation $b^{2} - a^{2} = (b - a) (b + a)$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Factorisation of $b^{n} - a^{n}$ , and the resulting Lipschitz estimate, Integer powers $a^{m}$ ).

[L3]

Powers and order: for $a, b \geq 0$ and $n \geq 1$ , $a < b$ exactly when $a^{n} < b^{n}$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L4]

Canonical naturals: positive for $n \geq 1$ , and strictly increasing in the index (Canonical naturals are positive and strictly increasing); reciprocals of positives are positive and reciprocation reverses the order (Inverses of positives are positive, and reciprocation reverses order).

[L5]

Archimedean property, both forms (Every complete ordered field is Archimedean, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L6]

Absolute value: $∣ t ∣ = t$ for $t \geq 0$ , and $∣ t ∣ \geq t$ (Basic properties of the absolute value).

[L7]

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

[L8]

Divergence to $+ \infty$ : for every real $M$ there is $K$ with $x_{k} > M$ for all $k \geq K$ (Divergence to $+ \infty$ and to $- \infty$ ).

[L9]

Trichotomy and transitivity of the order on $R$ (Complete ordered field (least-upper-bound property), Ordered field).

Counterexample

technique · direct

1.1

The sequence $(x_{k})$ satisfies the hypothesis of the refuted claim, its consecutive differences tending to $0$ , and it is not Cauchy.

givenL1L7

1.2

The failure is as strong as possible: $(x_{k})$ diverges to $+ \infty$ . Let $M \in R$ and put $M^{'} : = ∣ M ∣ \geq M$ , so $M^{'} \geq 0$ . By [L5] fix a natural $n \geq 1$ with $(M^{'})^{2} < n$ .

givenL5L6

2.1

It therefore refutes the claim: having null consecutive differences does not make a sequence Cauchy.

step 1.1L1

3.1

For every $k \geq n$ : $(x_{k})^{2} = k \geq n > (M^{'})^{2}$ with $x_{k} \geq 0$ and $M^{'} \geq 0$ , so $x_{k} > M^{'} \geq M$ . Since $M$ was arbitrary, $x_{k} \to + \infty$ .

step 1.2L2L3L4L8L9∎

Remarks

The gaps are null and their sums are not. The differences are about $1 / (2 \sqrt{k})$ , so they tend to $0$ ; but they telescope, and $x_{m} - x_{k} = \sqrt{m} - \sqrt{k}$ is as large as one likes for $m$ large. The Cauchy condition constrains $∣ x_{m} - x_{k} ∣$ for all large pairs, and no hypothesis about consecutive pairs alone can deliver that.
What the correct hypothesis looks like. Geometric decay of the gaps, with a single ratio $c < 1$ at every index, does suffice (Contractive sequence: $∣ x_{k + 2} - x_{k + 1} ∣ \leq c ∣ x_{k + 1} - x_{k} ∣$ for a fixed $0 < c < 1$ , Every contractive sequence is Cauchy, hence converges, with error bound $∣ x - x_{k} ∣ \leq c^{k - 1} ∣ x_{2} - x_{1} ∣ / (1 - c)$ for $k \geq 1$ ), because then the telescoped sums are dominated by a convergent geometric bound. Merely shrinking gaps are not enough either, which is $x_{k + 1} = x_{k} + 1 / x_{k}$ from $x_{1} = 1$ has strictly decreasing consecutive gaps and diverges, so no uniform $c < 1$ exists.
The same sequence separates two notions that are easy to confuse. Its gaps are null, so it is "eventually almost constant" in a naive reading; it is nevertheless unbounded and divergent to $+ \infty$ . Nothing about the local behaviour of a sequence controls its global behaviour.

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-08-06 (claude-sonnet-5)Open item page →

The sequence $1, 1, 2, 1, 3, 1, 4, \dots$ is unbounded and has a convergent subsequence

Statement refuted

Refuted claim: a sequence of reals with a convergent subsequence is bounded, which is the converse of Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence (FALSE: a sequence with a convergent subsequence is bounded (the converse of Bolzano-Weierstrass), Sequences of reals: bounded, eventually, frequently, tails, subsequences).

The witness is the interleaving

$1, 1, 2, 1, 3, 1, 4, \dots$

whose terms at even indices are $1, 2, 3, \dots$ and whose terms at odd indices are all $1$ . It is unbounded, and its odd-indexed subsequence is constant, hence convergent. The refutation is carried out in full in FALSE: a sequence with a convergent subsequence is bounded (the converse of Bolzano-Weierstrass); this item records the witness and adds the computation of its subsequential limit set.

Facts & Assumptions

Given: The strictly increasing index maps $e, o$ 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 $s \circ o \equiv - 1$ , whose ranges partition $N$ , and the sequence $(y_{n})$ of reals with $y_{n} : = (j + 1) \cdot 1_{R}$ when $n = e_{j}$ and $y_{n} : = 1$ when $n = o_{j}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

The witness: $(y_{n})$ is well defined, the subsequence $(y_{o_{j}})$ is constantly $1$ and converges to $1$ , and $(y_{n})$ is unbounded (FALSE: a sequence with a convergent subsequence is bounded (the converse of Bolzano-Weierstrass)).

[L2]

The index maps are strictly increasing and their ranges partition $N$ (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 $s \circ o \equiv - 1$ ).

[L3]

Canonical naturals: positive for $m \geq 1$ and strictly increasing in the index (Canonical naturals are positive and strictly increasing); the Archimedean property (Every complete ordered field is Archimedean).

[L4]

Boundedness of a sequence and of a subset of $R$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Lower bound, bounded below, bounded set); a constant sequence converges to its value (Limits and Cauchy sequences of reals); and every convergent sequence of reals is bounded (Every convergent sequence is bounded).

[L5]

Subsequential limits (Subsequential limit of a real sequence, and the subsequential limit set).

[L6]

Absolute value: $∣ t ∣ = t$ for $t \geq 0$ (Basic properties of the absolute value); trichotomy of the order (Complete ordered field (least-upper-bound property), Ordered field).

[L7]

Bolzano-Weierstrass: every bounded sequence of reals has a convergent subsequence (Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence).

[L8]

The refuted claim: a sequence of reals with a convergent subsequence is bounded.

Counterexample

technique · direct

1.1

$(y_{n})$ has a convergent subsequence, namely the constant subsequence along $o$ , which converges to $1$ ; so it satisfies the hypothesis of the claim.

givenL1L2L4L5L8

1.2

$(y_{n})$ is unbounded: no real $M$ satisfies $∣ y_{n} ∣ \leq M$ at every index.

givenL1L3L4L6

2.1

The claim is therefore refuted: a convergent subsequence does not force boundedness, and the converse of Bolzano-Weierstrass fails.

step 1.1step 1.2L7L8

3.1

The subsequential limit set of $(y_{n})$ is exactly ${1}$ . It contains $1$ by step 1.1. Conversely, let $(y_{n_{i}})$ converge; a convergent sequence is bounded, and along the even indices the values $y_{e_{j}} = j + 1$ exceed every bound, so only finitely many $n_{i}$ can lie in the range of $e$ ; all later $n_{i}$ lie in the range of $o$ , where the value is $1$ , so the subsequence is eventually constantly $1$ and its limit is $1$ .

step 1.1step 1.2L2L3L4L5∎

Remarks

A one-point subsequential limit set does not imply convergence. By step 3.1 the set is ${1}$ , yet the sequence is unbounded and hence divergent. What is true is the converse implication: a convergent sequence has exactly one subsequential limit (Subsequential limit of a real sequence, and the subsequential limit set). So the subsequential limit set being a single point is necessary and not sufficient for convergence, and the missing hypothesis is boundedness.
The interleaving is what makes the example work, and it needs the partition. Defining $(y_{n})$ by cases on whether $n$ is even or odd is legitimate only because every natural number is exactly one of the two (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 $s \circ o \equiv - 1$ ). That is the same fact that makes the two subsequences between them exhaust the sequence.
Compare the bounded case. With the unbounded branch replaced by a second constant $c \neq 1$ , the same interleaving is bounded and divergent, its two subsequential limits being $1$ and $c$ . Bolzano-Weierstrass applies there and produces one of the two; here it does not apply at all, and the conclusion nevertheless happens to hold, which is exactly why the converse is not a theorem. A different bounded sequence with two subsequential limits is worked out in full at The sequence $(- 1)^{k} (1 + 1 / k)$ is bounded with subsequential limit set exactly ${- 1, 1}$ , which reaches its two limits by an alternating sign carrying a null perturbation rather than by interleaving constants, so that neither limit is a value of the sequence.

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

$x_{k + 1} = x_{k} + 1 / x_{k}$ from $x_{1} = 1$ has strictly decreasing consecutive gaps and diverges, so no uniform $c < 1$ exists

Statement refuted

Refuted claim: a sequence whose consecutive gaps are strictly decreasing,

$∣ x_{k + 2} - x_{k + 1} ∣ < ∣ x_{k + 1} - x_{k} ∣ for every k,$

is contractive, or at least converges (Contractive sequence: $∣ x_{k + 2} - x_{k + 1} ∣ \leq c ∣ x_{k + 1} - x_{k} ∣$ for a fixed $0 < c < 1$ , Every contractive sequence is Cauchy, hence converges, with error bound $∣ x - x_{k} ∣ \leq c^{k - 1} ∣ x_{2} - x_{1} ∣ / (1 - c)$ for $k \geq 1$ ).

The witness is $x_{1} = 1$ , $x_{k + 1} = x_{k} + 1 / x_{k}$ . Its gaps are $x_{k + 1} - x_{k} = 1 / x_{k}$ , strictly decreasing because $(x_{k})$ is strictly increasing; and the sequence diverges to $+ \infty$ (Divergence to $+ \infty$ and to $- \infty$ ). Since a contractive sequence converges (Every contractive sequence is Cauchy, hence converges, with error bound $∣ x - x_{k} ∣ \leq c^{k - 1} ∣ x_{2} - x_{1} ∣ / (1 - c)$ for $k \geq 1$ ), no contraction constant $c < 1$ can exist for it: the ratios of consecutive gaps are all below $1$ but have no bound below $1$ that works at every index.

Indexing. Written on $N$ the sequence is $(g_{j})$ with $g_{0} = 1$ and $g_{j + 1} = g_{j} + 1 / g_{j}$ , and $x_{k} : = g_{k - 1}$ for $k \geq 1$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

Facts & Assumptions

Given: The set $A : = {u \in R : u \geq 1}$ , the element $1 \in A$ , and the function $f : A \to A$ with $f (u) = u + 1 / u$ , which lands in $A$ because $u \geq 1 > 0$ gives $1 / u > 0$ and hence $f (u) > u \geq 1$ ; by the recursion theorem (The recursion theorem) the unique $g : N \to A$ with $g_{0} = 1$ and $g_{j + 1} = f (g_{j})$ ; and the gaps $d_{j} : = g_{j + 1} - g_{j}$ .

[L1]

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

[L2]

Order and arithmetic: $1 > 0$ , so $2 = 1 + 1 > 0$ ; sums of positives are positive; adding a constant preserves the order; a positive has a positive inverse, and $0 < a < b$ gives $0 < 1 / b < 1 / a$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Inverses of positives are positive, and reciprocation reverses order, Ordered field, Complete ordered field (least-upper-bound property)).

[L3]

Powers: $(u + v)^{2} = u^{2} + 2 u v + v^{2}$ and $u \cdot (1 / u) = 1$ , so $(u + 1 / u)^{2} = u^{2} + 2 + 1 / u^{2}$ ; and for $a, b \geq 0$ , $a < b$ exactly when $a^{2} < b^{2}$ (Integer powers $a^{m}$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L4]

Canonical naturals: positive for $n \geq 1$ , and strictly increasing in the index (Canonical naturals are positive and strictly increasing); the Archimedean property (Every complete ordered field is Archimedean).

[L5]

Absolute value: $∣ t ∣ = t$ for $t \geq 0$ and $∣ t ∣ \geq t$ (Basic properties of the absolute value).

[L6]

Monotone sequences, with consecutive comparisons sufficing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L7]

Divergence to $+ \infty$ (Divergence to $+ \infty$ and to $- \infty$ ); a convergent sequence is bounded (Every convergent sequence is bounded); convergence (Limits and Cauchy sequences of reals).

[L8]

Contractive sequences and their convergence: a contractive sequence, with a constant $c$ satisfying $0 < c < 1$ at every index, converges (Contractive sequence: $∣ x_{k + 2} - x_{k + 1} ∣ \leq c ∣ x_{k + 1} - x_{k} ∣$ for a fixed $0 < c < 1$ , Every contractive sequence is Cauchy, hence converges, with error bound $∣ x - x_{k} ∣ \leq c^{k - 1} ∣ x_{2} - x_{1} ∣ / (1 - c)$ for $k \geq 1$ ).

Counterexample

technique · direct

1.1

Every term satisfies $g_{j} \geq 1 > 0$ , since $g$ takes values in $A$ by construction.

givenL2

2.1

The sequence is strictly increasing: $d_{j} = g_{j + 1} - g_{j} = 1 / g_{j} > 0$ , and consecutive comparisons give strict increase.

step 1.1L2L6

2.2

By induction, $(g_{j})^{2} \geq 1 + 2 j$ for every $j$ , where $j$ denotes the canonical natural. Base: $(g_{0})^{2} = 1 = 1 + 2 \cdot 0$ . Step: $(g_{j + 1})^{2} = (g_{j} + 1 / g_{j})^{2} = (g_{j})^{2} + 2 + 1 / (g_{j})^{2} > (g_{j})^{2} + 2 \geq (1 + 2 j) + 2 = 1 + 2 (j + 1)$ .

step 1.1L1L2L3L4

3.1

The gaps are strictly decreasing: $0 < g_{j} < g_{j + 1}$ gives $0 < 1 / g_{j + 1} < 1 / g_{j}$ , that is $0 < d_{j + 1} < d_{j}$ ; and $∣ d_{j + 1} ∣ < ∣ d_{j} ∣$ since both are positive.

step 1.1step 2.1L2L5

3.2

$(g_{j})$ diverges to $+ \infty$ . Let $M \in R$ and put $M^{'} : = ∣ M ∣ \geq M$ , so $M^{'} \geq 0$ . By [L4] fix a natural $n \geq 1$ with $(M^{'})^{2} < n$ . Then $(g_{n})^{2} \geq 1 + 2 n > n > (M^{'})^{2}$ , and since $g_{n} \geq 0$ and $M^{'} \geq 0$ this gives $g_{n} > M^{'} \geq M$ ; for every $k \geq n$ strict increase gives $g_{k} \geq g_{n} > M$ .

step 2.1step 2.2L3L4L5L7

4.1

$(g_{j})$ does not converge: a convergent sequence is bounded, whereas step 3.2 exhibits terms above every real.

step 3.2L7

5.1

No $c$ with $0 < c < 1$ is a contraction constant for $(g_{j})$ : if one were, the sequence would be contractive and would converge, contradicting step 4.1.

step 4.1L8

6.1

So $(g_{j})$ , that is $(x_{k})$ , has strictly decreasing consecutive gaps by step 3.1, diverges to $+ \infty$ by step 3.2, and admits no uniform $c < 1$ by step 5.1: strictly decreasing gaps neither make a sequence contractive nor make it converge.

step 3.1step 3.2step 5.1L8∎

Remarks

The gaps are not merely decreasing, they are null. Since $d_{j} = 1 / g_{j}$ and $g_{j} \to + \infty$ , for any real $ε > 0$ the terms eventually exceed $1 / ε$ , so $0 < d_{j} < ε$ eventually (Inverses of positives are positive, and reciprocation reverses order). So this sequence is also a witness for FALSE: if $∣ x_{k + 1} - x_{k} ∣ \to 0$ then $(x_{k})$ is Cauchy, alongside $\sqrt{k}$ ; the two are close relatives, since $(g_{j})^{2} \geq 1 + 2 j$ says $g_{j}$ grows at least like $\sqrt{2 j}$ .
What the uniform constant is really asking for. The ratio of consecutive gaps here is $d_{j + 1} / d_{j} = g_{j} / g_{j + 1}$ , which is below $1$ at every index and approaches $1$ as $j$ grows. A contraction constant would have to sit strictly between all of those ratios and $1$ , and there is no room: the supremum of the ratios is $1$ itself. This is the precise sense in which Contractive sequence: $∣ x_{k + 2} - x_{k + 1} ∣ \leq c ∣ x_{k + 1} - x_{k} ∣$ for a fixed $0 < c < 1$ asks for more than "each gap smaller than the last".
The comparison with a genuine contraction. In The sequence $x_{k + 1} = (x_{k} + 1) / 3$ is contractive with $c = 1 / 3$ and converges to $1 / 2$ the ratio is exactly $1 / 3$ at every index, so the constant exists and is optimal, and the error bound of Every contractive sequence is Cauchy, hence converges, with error bound $∣ x - x_{k} ∣ \leq c^{k - 1} ∣ x_{2} - x_{1} ∣ / (1 - c)$ for $k \geq 1$ applies. The difference between the two examples is not the speed at which the gaps shrink at any given index but whether the shrinking is uniform.

Monotone Sequences, Bolzano-Weierstrass, and Cauchy Completeness: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The Babylonian sequence x1=2x1​=2, xk+1=(xk+2/xk)/2xk+1​=(xk​+2/xk​)/2 decreases to 22​

Example

Facts & Assumptions

Verification

Remarks

The sequence x1=1x1​=1, xk+1=2+xkxk+1​=2+xk​​ increases to 22

Example

Facts & Assumptions

Verification

Remarks

The nested intervals [0,1/k][0,1/k] intersect in exactly {0}{0}

Example

Facts & Assumptions

Verification

Remarks

The sequence (−1)k(1+1/k)(−1)k(1+1/k) is bounded with subsequential limit set exactly {−1,1}{−1,1}

Example

Facts & Assumptions

Verification

Remarks

The sequence xk+1=(xk+1)/3xk+1​=(xk​+1)/3 is contractive with c=1/3c=1/3 and converges to 1/21/2

Example

Facts & Assumptions

Verification

Remarks

The truncated decimal approximations of 22​ form a Cauchy sequence of rationals with no rational limit

Statement refuted

Facts & Assumptions

Counterexample

Remarks

The nested open intervals (0,1/k)(0,1/k) have empty intersection

Statement refuted

Facts & Assumptions

Counterexample

Remarks

The nested closed unbounded sets [k,∞)[k,∞) have empty intersection, so boundedness cannot be dropped

Statement refuted

Facts & Assumptions

Counterexample

Remarks

xk=kxk​=k​ has xk+1−xk→0xk+1​−xk​→0 and is not Cauchy

Statement refuted

Facts & Assumptions

Counterexample

Remarks

The sequence 1,1,2,1,3,1,4,…1,1,2,1,3,1,4,… is unbounded and has a convergent subsequence

Statement refuted

Facts & Assumptions

Counterexample

Remarks

xk+1=xk+1/xkxk+1​=xk​+1/xk​ from x1=1x1​=1 has strictly decreasing consecutive gaps and diverges, so no uniform c<1c<1 exists

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Sources

The Babylonian sequence $x_{1} = 2$ , $x_{k + 1} = (x_{k} + 2 / x_{k}) / 2$ decreases to $\sqrt{2}$

The sequence $x_{1} = 1$ , $x_{k + 1} = \sqrt{2 + x_{k}}$ increases to $2$

The nested intervals $[0, 1 / k]$ intersect in exactly ${0}$

The sequence $(- 1)^{k} (1 + 1 / k)$ is bounded with subsequential limit set exactly ${- 1, 1}$

The sequence $x_{k + 1} = (x_{k} + 1) / 3$ is contractive with $c = 1 / 3$ and converges to $1 / 2$

The truncated decimal approximations of $\sqrt{2}$ form a Cauchy sequence of rationals with no rational limit

The nested open intervals $(0, 1 / k)$ have empty intersection

The nested closed unbounded sets $[k, \infty)$ have empty intersection, so boundedness cannot be dropped

$x_{k} = \sqrt{k}$ has $x_{k + 1} - x_{k} \to 0$ and is not Cauchy

The sequence $1, 1, 2, 1, 3, 1, 4, \dots$ is unbounded and has a convergent subsequence

$x_{k + 1} = x_{k} + 1 / x_{k}$ from $x_{1} = 1$ has strictly decreasing consecutive gaps and diverges, so no uniform $c < 1$ exists