TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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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 $k \in \mathbb{N}$ let $I_k = [a_k, b_k]$ be a closed bounded interval with $a_k \le b_k$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length), and suppose the family is nested:

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

Write $\ell_k = b_k - a_k \ge 0$ for the length of $I_k$ . Then:

$\displaystyle\bigcap_{k \in \mathbb{N}} I_k$ is nonempty. More precisely, with $a = \sup\{a_k : k \in \mathbb{N}\}$ and $b = \inf\{b_k : k \in \mathbb{N}\}$ , both of which exist, one has $a \le b$ and $\bigcap_{k \in \mathbb{N}} I_k = [a, b].$
$\displaystyle\bigcap_{k \in \mathbb{N}} I_k$ is a single point if and only if $\ell_k \to 0$ (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 $a_k \le b_k$ for every $k \in \mathbb{N}$ 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 $A = \{a_k : k \in \mathbb{N}\}$ and $B = \{b_k : k \in \mathbb{N}\}$ , both nonempty; and $\ell_k = b_k - a_k$ .

[L1]

Closed bounded intervals: $[a,b] = \{x \in \mathbb{R} : a \le x \le b\}$ ; it is nonempty exactly when $a \le b$ , 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 $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L2]

Least-upper-bound property and uniqueness: a nonempty subset of $\mathbb{R}$ bounded above has a unique supremum; the supremum is an upper bound and is $\le$ 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 $\mathbb{R}$ bounded below has a unique infimum; the infimum is a lower bound and is $\ge$ 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: $x_k \le x_{k+1}$ for all $k$ makes $(x_k)$ nondecreasing, and $x_{k+1} \le x_k$ 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 $\mathbb{R}$ (Lower bound, bounded below, bounded set).

[L9]

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

Proof

technique · direct

1.1

Nestedness read on the endpoints: since $a_{k+1} \le b_{k+1}$ , both $a_{k+1}$ and $b_{k+1}$ lie in $I_{k+1} \subseteq I_k$ , so $a_k \le a_{k+1}$ and $b_{k+1} \le b_k$ 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 $k \le m$ and $l \le m$ gives $a_k \le a_m \le b_m \le b_l$ , so $a_k \le b_l$ .

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

$a \le b$ : each $b_l$ is an upper bound of $A$ , so $a \le b_l$ for every $l$ by leastness of the supremum; thus $a$ is a lower bound of $B$ , and $a \le b$ 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 $a_k \le x \le b_k$ 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 $a \le x \le b$ .

step 4.1L1L2L3

6.1

$\ell_k = b_k - a_k \to b - a$ by the algebra of limits.

step 5.2L6

6.2

Since $a \le b$ , 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 $\ell_k \to 0$ 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 $a \le b$ , and $a < b$ would give the two distinct elements $a$ and $b$ . Hence $b - a = 0$ and $\ell_k \to 0$ 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: $\ell_k$ 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 $\ell_k$ is smaller" does need the Archimedean property (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ), 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 $\mathbb{R}$ 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 71 results over 23 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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 000