TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$

Statement

For each $k \in N$ let $I_{k} = [a_{k}, b_{k}]$ be a closed bounded interval with $a_{k} \leq b_{k}$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length), and suppose the family is nested:

$I_{k + 1} \subseteq I_{k} (k \in N) .$

Write $ℓ_{k} = b_{k} - a_{k} \geq 0$ for the length of $I_{k}$ . Then:

$⋂_{k \in N} I_{k}$ is nonempty. More precisely, with $a = \sup {a_{k} : k \in N}$ and $b = \inf {b_{k} : k \in N}$ , both of which exist, one has $a \leq b$ and $⋂_{k \in N} I_{k} = [a, b] .$
$⋂_{k \in N} I_{k}$ is a single point if and only if $ℓ_{k} \to 0$ (Limits and Cauchy sequences of reals).

Every hypothesis is load bearing. Dropping closedness makes the intersection empty; dropping boundedness does the same; and dropping nonemptiness of the individual intervals is vacuously fatal.

Facts & Assumptions

Given: Closed bounded intervals $I_{k} = [a_{k}, b_{k}]$ with $a_{k} \leq b_{k}$ for every $k \in N$ and $I_{k + 1} \subseteq I_{k}$ for every $k$ ; the sequences $(a_{k})$ and $(b_{k})$ of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences); their ranges $A = {a_{k} : k \in N}$ and $B = {b_{k} : k \in N}$ , both nonempty; and $ℓ_{k} = b_{k} - a_{k}$ .

[L1]

Closed bounded intervals: $[a, b] = {x \in R : a \leq x \leq b}$ ; it is nonempty exactly when $a \leq b$ , it is the singleton ${a}$ when $a = b$ , it has two distinct elements $a$ and $b$ when $a < b$ , and its length is $b - a$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L2]

Least-upper-bound property and uniqueness: a nonempty subset of $R$ bounded above has a unique supremum; the supremum is an upper bound and is $\leq$ every upper bound (Complete ordered field (least-upper-bound property), Suprema and infima are unique).

[L3]

Greatest-lower-bound property and uniqueness: a nonempty subset of $R$ bounded below has a unique infimum; the infimum is a lower bound and is $\geq$ every lower bound (Every nonempty set bounded below has an infimum, Suprema and infima are unique).

[L4]

Monotone sequences, and the fact that consecutive comparisons suffice: $x_{k} \leq x_{k + 1}$ for all $k$ makes $(x_{k})$ nondecreasing, and $x_{k + 1} \leq x_{k}$ for all $k$ makes it nonincreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L5]

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

[L6]

Algebra of limits: if $u_{k} \to u$ and $v_{k} \to v$ then $u_{k} - v_{k} \to u - v$ (Algebra of limits: sums, scalar multiples, products and quotients).

[L7]

A sequence of reals has at most one limit (A sequence has at most one limit).

[L8]

Bounded above and bounded below, for a subset of $R$ (Lower bound, bounded below, bounded set).

[L9]

The order on $N$ is total and transitive, so any two indices $k, l$ admit an index $m$ with $k \leq m$ and $l \leq m$ , namely the larger of the two (Order on the natural numbers, $\leq$ is a linear order on $N$ ).

Proof

technique · direct

1.1

Nestedness read on the endpoints: since $a_{k + 1} \leq b_{k + 1}$ , both $a_{k + 1}$ and $b_{k + 1}$ lie in $I_{k + 1} \subseteq I_{k}$ , so $a_{k} \leq a_{k + 1}$ and $b_{k + 1} \leq b_{k}$ for every $k$ .

givenL1

2.1

Hence $(a_{k})$ is nondecreasing and $(b_{k})$ is nonincreasing.

step 1.1L4

3.1

For all indices $k$ and $l$ : choosing $m$ with $k \leq m$ and $l \leq m$ gives $a_{k} \leq a_{m} \leq b_{m} \leq b_{l}$ , so $a_{k} \leq b_{l}$ .

step 2.1givenL9

4.1

Every $b_{l}$ is therefore an upper bound of $A$ and every $a_{k}$ a lower bound of $B$ ; both sets are nonempty, so $a : = \sup A$ and $b : = \inf B$ exist and are unique.

step 3.1L2L3L8

5.1

$a \leq b$ : each $b_{l}$ is an upper bound of $A$ , so $a \leq b_{l}$ for every $l$ by leastness of the supremum; thus $a$ is a lower bound of $B$ , and $a \leq b$ by greatestness of the infimum.

step 4.1L2L3

5.2

By monotone convergence, $a_{k} \to a$ and $b_{k} \to b$ .

step 2.1step 4.1L5

5.3

The intersection is exactly $[a, b]$ : a real $x$ lies in every $I_{k}$ exactly when $a_{k} \leq x \leq b_{k}$ for every $k$ , that is exactly when $x$ is an upper bound of $A$ and a lower bound of $B$ , and by leastness of $a$ and greatestness of $b$ that holds exactly when $a \leq x \leq b$ .

step 4.1L1L2L3

6.1

$ℓ_{k} = b_{k} - a_{k} \to b - a$ by the algebra of limits.

step 5.2L6

6.2

Since $a \leq b$ , the interval $[a, b]$ is nonempty, so the intersection is nonempty; together with step 5.3 this is claim 1.

step 5.1step 5.3L1

7.1

If $ℓ_{k} \to 0$ then $b - a = 0$ by uniqueness of limits, so $a = b$ and the intersection is $[a, a] = {a}$ , a single point.

step 5.3step 6.1L1L7

7.2

Conversely, if the intersection is a single point then $a = b$ : it equals $[a, b]$ with $a \leq b$ , and $a < b$ would give the two distinct elements $a$ and $b$ . Hence $b - a = 0$ and $ℓ_{k} \to 0$ by step 6.1.

step 5.1step 5.3step 6.1L1

8.1

Claim 1 is step 6.2 and claim 2 is the pair of implications in steps 7.1 and 7.2, so a nested sequence of nonempty closed bounded intervals has nonempty intersection, equal to $[a, b]$ , and that intersection is a single point exactly when the lengths tend to $0$ .

step 6.2step 7.1step 7.2∎

Remarks

No Archimedean input is needed. The lengths are handled entirely by the algebra of limits and the uniqueness of limits: $ℓ_{k}$ always converges, to $b - a$ , and the two directions of claim 2 are then the two directions of " $b - a = 0$ ". A proof that instead argues "if $b - a > 0$ then some $ℓ_{k}$ is smaller" does need the Archimedean property (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ), and it is avoidable, so it is avoided.
Nestedness gives more than it is usually stated to give. The intersection is not merely nonempty; it is the closed interval $[a, b]$ , and $a$ and $b$ are the limits of the endpoint sequences. The single-point case is exactly the case in which those two limits agree, and that is what makes the nested interval property usable as a construction of a real number, as in The nested intervals $[0, 1 / k]$ intersect in exactly ${0}$ ↗.
This is one of the standard equivalents of completeness. Nested intervals together with the Archimedean property imply the least-upper-bound property, so the implication proved here is not reversible for free: it is half of an equivalence whose other half needs the Archimedean hypothesis separately. Two independent proofs that $R$ is Cauchy complete, and why the library records both records where this library stands on those routes.
The witnesses for the two deleted hypotheses are The nested open intervals $(0, 1 / k)$ have empty intersection ↗, which keeps boundedness and drops closedness, and The nested closed unbounded sets $[k, \infty)$ have empty intersection, so boundedness cannot be dropped ↗, which keeps closedness and drops boundedness. Neither is used above; each shows that the corresponding hypothesis cannot be removed.

Depends on

Used by

Dependency tree · two levels

45 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)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (Thm 2.38) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §6.4 (standard reference, not scraped)
J. Lebl, Basic Analysis I, §1.4 (standard reference, not scraped)

A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to 00