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$R ((t^{- 1}))$ has the nested interval property for lengths tending to $0$

Statement

Let $K = R ((t^{- 1}))$ and let $(I_{n})_{n \in N}$ with $I_{n} = [a_{n}, b_{n}]_{K}$ be a nested sequence of closed intervals in $K$ whose lengths tend to $0$ in $K$ , that is, for every $ε > 0$ in $K$ there is $N$ with $b_{n} - a_{n} < ε$ for all $n \geq N$ (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field). Then

$⋂_{n \in N} I_{n}$

contains exactly one element of $K$ .

The hypothesis that the lengths tend to $0$ may not be dropped: this is the nested interval property in its shrinking form only, and nothing on this page establishes the unrestricted form for $K$ . The remarks below record what happens without the hypothesis.

Facts & Assumptions

Given: A nested sequence $(I_{n})_{n \in N}$ of closed intervals $I_{n} = [a_{n}, b_{n}]_{K}$ in $K$ , so $a_{n} \leq b_{n}$ and $I_{n + 1} \subseteq I_{n}$ for every $n$ , whose lengths tend to $0$ in $K$ .

[L1]

$[a, b]_{K} = {x \in K : a \leq x \leq b}$ for $a \leq b$ ; a sequence $(x_{n})$ in $K$ is Cauchy in $K$ when for every $ε > 0$ in $K$ there is $N$ with $∣ x_{n} - x_{m} ∣ < ε$ for all $n, m \geq N$ , and converges to $L$ when for every $ε > 0$ in $K$ there is $N$ with $∣ x_{n} - L ∣ < ε$ for all $n \geq N$ ; the lengths $b_{n} - a_{n}$ tend to $0$ when for every $ε > 0$ in $K$ they are eventually $< ε$ (Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field).

[L2]

Every Cauchy sequence in $K$ converges in $K$ (Every Cauchy sequence in $R ((t^{- 1}))$ converges: $K$ is sequentially Cauchy complete).

[L3]

$K$ is an ordered field ( $R ((t^{- 1}))$ is an ordered field, ordered by the sign of the leading coefficient, Ordered field), so its order is total and transitive and $x < y$ means $0 < y - x$ . Compatibility with addition is used below in its NONSTRICT form, $x \leq y \Rightarrow x + z \leq y + z$ , whereas Order is preserved by adding a constant and by adding inequalities states the STRICT forms and only those ( $x < y \Rightarrow x + z < y + z$ , and $x < y$ with $z < w$ giving $x + z < y + w$ ); the nonstrict form is the first strict form together with the case $x = y$ , where the two sides are equal, the order being total (Ordered field).

[L4]

$∣ z ∣ \geq 0$ , $∣ z ∣ = 0$ only for $z = 0$ , and $∣ z ∣$ equals $z$ or $- z$ ; so $∣ z ∣ = z$ when $z \geq 0$ (Basic properties of the absolute value, Absolute value in an ordered field).

[L5]

The order on $N$ is total ( $\leq$ is a linear order on $N$ , Order on the natural numbers) and induction is available (The principle of mathematical induction, The natural numbers $N$ (von Neumann)).

Proof

technique · direct

1.1

For each $n$ , the endpoints $a_{n + 1}$ and $b_{n + 1}$ belong to $I_{n + 1}$ because $a_{n + 1} \leq b_{n + 1}$ , and $I_{n + 1} \subseteq I_{n}$ , so both belong to $I_{n}$ ; by [L1] this says $a_{n} \leq a_{n + 1}$ and $b_{n + 1} \leq b_{n}$ . Hence $a_{n} \leq a_{n + 1} \leq b_{n + 1} \leq b_{n}$ .

givenL1L3

1.2

The intersection contains at most one element. Suppose $x, y \in ⋂_{n} I_{n}$ with $x \neq y$ , so $∣ x - y ∣ > 0$ by [L4]. For each $n$ both $x$ and $y$ lie in $[a_{n}, b_{n}]_{K}$ , so $x - y \leq b_{n} - a_{n}$ and $y - x \leq b_{n} - a_{n}$ by [L1] and [L3], and since $∣ x - y ∣$ is one of $x - y$ , $y - x$ by [L4] we get $∣ x - y ∣ \leq b_{n} - a_{n}$ for every $n$ . Applying the shrinking hypothesis with $ε : = ∣ x - y ∣$ produces some $n$ with $b_{n} - a_{n} < ∣ x - y ∣$ , a contradiction.

givenL1L3L4

2.1

Whenever $n \leq m$ one has $a_{n} \leq a_{m} \leq b_{m} \leq b_{n}$ : this is [step 1.1] for $m = n + 1$ , it is trivial for $m = n$ , and the general case follows by induction on $m$ using transitivity of the order.

step 1.1L3L5

3.1

$(a_{n})_{n \in N}$ is Cauchy in $K$ . Let $ε > 0$ in $K$ and take $N$ with $b_{n} - a_{n} < ε$ for all $n \geq N$ . Let $n, m \geq N$ ; by [L5] we may assume $n \leq m$ , the other case being the same with the roles exchanged. By [step 2.1], $a_{n} \leq a_{m} \leq b_{m} \leq b_{n}$ , so $0 \leq a_{m} - a_{n} \leq b_{n} - a_{n} < ε$ , and $∣ a_{m} - a_{n} ∣ = a_{m} - a_{n} < ε$ by [L4].

step 2.1givenL1L3L4L5

4.1

By [L2] there is $L \in K$ with $a_{n} \to L$ in $K$ .

step 3.1L2

5.1

$a_{n} \leq L$ for every $n$ . Otherwise $L < a_{n}$ for some $n$ ; put $ε : = a_{n} - L > 0$ and use [step 4.1] to fix $N$ with $∣ a_{m} - L ∣ < ε$ for all $m \geq N$ . Pick $m$ with $m \geq N$ and $m \geq n$ ([L5]). By [step 2.1], $a_{n} \leq a_{m}$ , so $a_{m} - L \geq a_{n} - L = ε > 0$ and hence $∣ a_{m} - L ∣ = a_{m} - L \geq ε$ by [L4], contradicting $∣ a_{m} - L ∣ < ε$ .

step 2.1step 4.1L1L3L4L5

5.2

$L \leq b_{n}$ for every $n$ . Otherwise $b_{n} < L$ for some $n$ ; put $ε : = L - b_{n} > 0$ and fix $N$ with $∣ a_{m} - L ∣ < ε$ for all $m \geq N$ . Pick $m$ with $m \geq N$ and $m \geq n$ . By [step 2.1], $a_{m} \leq b_{m} \leq b_{n}$ , so $L - a_{m} \geq L - b_{n} = ε > 0$ and hence $∣ a_{m} - L ∣ = L - a_{m} \geq ε$ by [L4], again a contradiction.

step 2.1step 4.1L1L3L4L5

6.1

By [step 5.1] and [step 5.2], $a_{n} \leq L \leq b_{n}$ for every $n$ , so $L \in ⋂_{n} I_{n}$ by [L1] and the intersection is nonempty; by [step 1.2] it has no second element. Hence $⋂_{n} I_{n} = {L}$ .

step 5.1step 5.2step 1.2L1∎

Remarks

This is the shrinking form, and the restriction is real. The unrestricted nested interval property — every nested sequence of nonempty closed intervals meets — is false in $K$ , and The unrestricted nested interval property fails in $R ((t^{- 1}))$ exhibits a nested sequence with empty intersection. So the hypothesis here is not a convenience of the proof, and no item on this page may be cited for the unrestricted form.
A trap in the hypothesis: "lengths $2 / n$ " does not mean shrinking. The condition is that the lengths tend to $0$ in the order of $K$ , tested against every positive $ε \in K$ , not merely against positive real constants. A nested sequence whose $n$ -th length is the constant series $ι (2 / (n + 1))$ does not satisfy it: since $ι (c)$ takes the nonzero value $c$ at index $0$ , clause 4 of $R ((t^{- 1}))$ is non-Archimedean, and the monomials $t^{- k}$ are cofinal below its positive elements forbids $∣ ι (c) ∣ < t^{- 1}$ , so no such length ever gets below $ε = t^{- 1}$ . Real-indexed shrinking is strictly weaker than shrinking in $K$ , and a proof that assumed the former would be proving a different theorem.
Where completeness enters. Exactly once, at [step 4.1]. Everything before it is monotonicity bookkeeping valid in any ordered field, and everything after it uses only the order and the absolute value. That is why the corollary is a corollary of Every Cauchy sequence in $R ((t^{- 1}))$ converges: $K$ is sequentially Cauchy complete and not an independent argument about series.

Depends on

Used by

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47 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Nested intervals (Wikipedia) (standard reference, not scraped)
Cauchy sequence (Wikipedia) (standard reference, not scraped)
Cantor theorem (Encyclopedia of Mathematics) (standard reference, not scraped)
Cauchy sequences in ordered fields (University of Tennessee notes) (standard reference, not scraped)

R((t−1))R((t−1)) has the nested interval property for lengths tending to 00