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

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The unrestricted nested interval property fails in $R ((t^{- 1}))$

Statement refuted

Refuted claim: the unrestricted nested interval property holds in $K = R ((t^{- 1}))$ , that is, every nested sequence $I_{0} \supseteq I_{1} \supseteq \dots$ of closed intervals $I_{n} = [a_{n}, b_{n}]_{K}$ of $K$ (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field) has $⋂_{n} I_{n} \neq \emptyset$ .

The witness is

$a_{n} : = ι (n) t^{- 1}, b_{n} : = ι ⁣ (\frac{1}{n + 1}) (n \in N),$

where $ι (c)$ is the constant series with value $c$ at index $0$ and $ι (n)$ abbreviates $ι (n \cdot 1_{R})$ (The formal Laurent series $R ((t^{- 1}))$ : support bounded below, valuation, leading coefficient). The intervals $[a_{n}, b_{n}]_{K}$ are nested and their intersection is empty: a common point would have to be an infinitesimal of valuation $\geq 1$ , because it lies below every positive real constant, and simultaneously not such an element, because it lies above every multiple $ι (n) t^{- 1}$ of $t^{- 1}$ .

This refutes only the unrestricted form. The shrinking form, with the additional hypothesis that the lengths tend to $0$ in $K$ , is true ( $R ((t^{- 1}))$ has the nested interval property for lengths tending to $0$ ), and the lengths here do not tend to $0$ .

Facts & Assumptions

Given: $K = R ((t^{- 1}))$ with its valuation $v$ , leading coefficient $lc$ , monomials $t^{- a}$ and constants $ι (c)$ ; and the elements $a_{n} = ι (n) t^{- 1}$ , $b_{n} = ι (1 / (n + 1))$ for $n \in N$ .

[L1]

For nonzero $h \in K$ : $h (k) = 0$ for $k < v (h)$ and $h (v (h)) = lc (h) \neq 0$ ; $t^{- a}$ is nonzero with $v (t^{- a}) = a$ and $lc (t^{- a}) = 1$ ; and $ι (c)$ for $c \neq 0$ is nonzero with $v (ι (c)) = 0$ and $lc (ι (c)) = c$ (The formal Laurent series $R ((t^{- 1}))$ : support bounded below, valuation, leading coefficient, $R ((t^{- 1}))$ is an ordered field, ordered by the sign of the leading coefficient).

[L2]

$K$ is an ordered field in which $f < g$ holds exactly when $g - f \neq 0_{K}$ and $lc (g - f) > 0$ ; exactly one of $f < g$ , $f = g$ , $g < f$ holds; $ι$ is a ring homomorphism, so $ι (c) + ι (d) = ι (c + d)$ and $ι (c) ι (d) = ι (c d)$ ( $R ((t^{- 1}))$ is an ordered field, ordered by the sign of the leading coefficient, Ordered field).

[L3]

For nonzero $f, g \in K$ : $f g \neq 0_{K}$ with $v (f g) = v (f) + v (g)$ and $lc (f g) = lc (f) lc (g)$ ; $- f \neq 0_{K}$ with $v (- f) = v (f)$ and $lc (- f) = - lc (f)$ ; if $v (f) < v (g)$ then $f + g \neq 0_{K}$ with $lc (f + g) = lc (f)$ ; and if $v (f) = v (g)$ with $lc (f) + lc (g) \neq 0$ then $f + g \neq 0_{K}$ with $v (f + g) = v (f)$ and $lc (f + g) = lc (f) + lc (g)$ (Valuation and leading coefficient in $R ((t^{- 1}))$ : $v (f g) = v (f) + v (g)$ , and the behaviour of $v$ under sums, $R ((t^{- 1}))$ is a commutative ring: the product is a finite sum and both operations preserve support bounded below).

[L4]

$0_{K} < t^{- (k + 1)} < t^{- k}$ for every $k \in Z$ ; and if $∣ h ∣ < t^{- k}$ then $h (j) = 0$ for every $j < k$ ( $R ((t^{- 1}))$ is non-Archimedean, and the monomials $t^{- k}$ are cofinal below its positive elements).

[L5]

$[a, b]_{K} = {x \in K : a \leq x \leq b}$ for $a \leq b$ ; a sequence of closed intervals is nested when $I_{n + 1} \subseteq I_{n}$ for every $n$ ; and its lengths tend to $0$ in $K$ when for every $ε > 0$ in $K$ they are eventually $< ε$ (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field).

[L6]

$R$ is a complete ordered field, hence Archimedean: for every real $c$ there is a natural $n$ with $c < n \cdot 1_{R}$ , and for every real $c > 0$ there is a natural $n \geq 1$ with $1 / n < c$ (The reals form a totally ordered field, The Cauchy-sequence reals have the least-upper-bound property, Every complete ordered field is Archimedean, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

Counterexample

technique · direct

1.1

For $n \geq 1$ the element $a_{n} = ι (n) t^{- 1}$ is nonzero with $v (a_{n}) = 0 + 1 = 1$ and $lc (a_{n}) = n \cdot 1 = n > 0$ , while $a_{0} = ι (0) t^{- 1} = 0_{K}$ ; and for every $n$ the element $b_{n} = ι (1 / (n + 1))$ is nonzero with $v (b_{n}) = 0$ and $lc (b_{n}) = 1 / (n + 1) > 0$ . Also $a_{1} = ι (1) t^{- 1} = t^{- 1}$ .

givenL1L2L3

2.1

$a_{n} \leq b_{n}$ for every $n$ : for $n = 0$ this is $0_{K} < ι (1) = 1_{K}$ , which holds since $lc (1_{K}) = 1 > 0$ ; and for $n \geq 1$ we have $v (b_{n}) = 0 < 1 = v (- a_{n})$ by [step 1.1] and [L3], so $b_{n} - a_{n}$ is nonzero with leading coefficient $1 / (n + 1) > 0$ , that is $a_{n} < b_{n}$ . So each $[a_{n}, b_{n}]_{K}$ is a closed interval.

step 1.1L1L2L3L5

2.2

The sequence is nested: $a_{n + 1} - a_{n} = (ι (n + 1) - ι (n)) t^{- 1} = ι (1) t^{- 1} = t^{- 1} > 0_{K}$ by [L2] and [L4], so $a_{n} < a_{n + 1}$ ; and $b_{n} - b_{n + 1} = ι (\frac{1}{n + 1} - \frac{1}{n + 2}) = ι (\frac{1}{(n + 1) (n + 2)})$ , which is nonzero with positive leading coefficient, so $b_{n + 1} < b_{n}$ . Hence $a_{n} \leq a_{n + 1}$ and $b_{n + 1} \leq b_{n}$ , and every $x$ with $a_{n + 1} \leq x \leq b_{n + 1}$ satisfies $a_{n} \leq x \leq b_{n}$ , that is $I_{n + 1} \subseteq I_{n}$ .

step 1.1L1L2L3L4L5

3.1

Suppose $x \in ⋂_{n} I_{n}$ . Then $a_{1} \leq x$ , and $a_{1} = t^{- 1} > 0_{K}$ by [step 1.1] and [L4], so $x > 0_{K}$ ; hence $x \neq 0_{K}$ and $lc (x) > 0$ by [L2]. Write $q : = v (x)$ and $c : = lc (x) > 0$ .

step 1.1step 2.1L1L2L4L5

4.1

$q \geq 1$ . If $q < 0$ then $v (x) < 0 = v (- b_{0})$ by [step 1.1] and [L3], so $x - b_{0}$ is nonzero with leading coefficient $c > 0$ and $x > b_{0}$ , contradicting $x \leq b_{0}$ . If $q = 0$ , use [L6] to fix a natural $n \geq 1$ with $1 / n < c$ and set $n^{'} : = n - 1$ , so that $1 / (n^{'} + 1) < c$ ; then $x$ and $- b_{n^{'}}$ both have valuation $0$ with leading coefficients summing to $c - 1 / (n^{'} + 1) \neq 0$ , so by [L3] $x - b_{n^{'}}$ is nonzero with leading coefficient $c - 1 / (n^{'} + 1) > 0$ , giving $x > b_{n^{'}}$ and contradicting $x \leq b_{n^{'}}$ . By trichotomy on $Z$ the remaining case is $q \geq 1$ .

step 3.1L1L2L3L5L6

4.2

$q < 1$ . If $q > 1$ then $v (- a_{1}) = 1 < q$ by [step 1.1] and [L3], so $x - a_{1}$ is nonzero with leading coefficient $lc (- a_{1}) = - 1 < 0$ , giving $x < a_{1}$ and contradicting $a_{1} \leq x$ . If $q = 1$ , use [L6] to fix a natural $n$ with $c < n \cdot 1_{R}$ , so $n \geq 1$ ; then $x$ and $- a_{n}$ both have valuation $1$ with leading coefficients summing to $c - n \neq 0$ , so by [L3] $x - a_{n}$ is nonzero with leading coefficient $c - n < 0$ , giving $x < a_{n}$ and contradicting $a_{n} \leq x$ . Hence $q \neq 1$ and $q ≯ 1$ .

step 3.1L1L2L3L5L6

5.1

Steps 4.1 and 4.2 are incompatible, so no $x$ lies in every $I_{n}$ : the nested sequence $(I_{n})$ of [step 2.1] and [step 2.2] has $⋂_{n} I_{n} = \emptyset$ , which refutes the unrestricted nested interval property for $K$ .

step 2.1step 2.2step 4.1step 4.2L5

6.1

Consistency with $R ((t^{- 1}))$ has the nested interval property for lengths tending to $0$ : the lengths here do not tend to $0$ in $K$ . Indeed $(b_{n} - a_{n}) (0) = 1 / (n + 1) \neq 0$ by [step 1.1], so by [L4] the inequality $∣ b_{n} - a_{n} ∣ < t^{- 1}$ fails for every $n$ ; taking $ε = t^{- 1}$ shows the shrinking hypothesis of [L5] is not satisfied.

step 1.1step 2.1L4L5∎

Remarks

What the counterexample really exhibits. It is a gap in $K$ . Each of the two requirements is satisfiable on its own — $ι (1)$ lies above every $ι (n) t^{- 1}$ , and $0_{K}$ lies below every $ι (1 / (n + 1))$ — yet steps 4.1 and 4.2 show that nothing in $K$ satisfies both at once. Both sides of the gap are approached along countable sequences, which is why intervals indexed by $N$ can straddle it, and the lengths cannot shrink across it: they stay of valuation $0$ while the left endpoints stay of valuation $1$ .
Why this does not contradict Cauchy completeness. $(a_{n})$ is not Cauchy in $K$ : consecutive terms differ by exactly $t^{- 1}$ , so the Cauchy condition fails at $ε = t^{- 1}$ . Cauchy completeness (Every Cauchy sequence in $R ((t^{- 1}))$ converges: $K$ is sequentially Cauchy complete) constrains sequences whose terms crowd together in the order of $K$ , and neither endpoint sequence here does.
Consequence for the equivalence of completeness properties. Since $K$ is Cauchy complete but has neither the least-upper-bound property ( $R ((t^{- 1}))$ does not have the least-upper-bound property; its canonical naturals have no supremum) nor the unrestricted nested interval property, any statement of the form "nested intervals imply least upper bounds" has to say which nested interval property it means. The form that $K$ does satisfy is the shrinking one, and that is the form for which $K$ is a counterexample to the implication.

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Sources

Nested intervals (Wikipedia) (standard reference, not scraped)
Ordered field (Wikipedia) (standard reference, not scraped)

The unrestricted nested interval property fails in R((t−1))R((t−1))