TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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In a complete metric space nested nonempty closed sets whose diameters tend to $0$ meet in exactly one point, and this property characterises completeness

Statement

Assume the Axiom of Countable Choice for assertion 1. Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric). Call a sequence $(F_{k})_{k \in N}$ of subsets of $X$ a Cantor chain if every $F_{k}$ is nonempty, closed (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement) and bounded, $F_{k + 1} \subseteq F_{k}$ for every $k$ , and $diam (F_{k}) \to 0$ in $R$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Limits and Cauchy sequences of reals). Then:

If $(X, d)$ is complete (Complete metric space: every Cauchy sequence converges in the space), every Cantor chain in $X$ has an intersection $⋂_{k \in N} F_{k}$ with exactly one element.
Conversely, if every Cantor chain in $X$ has nonempty intersection, then $(X, d)$ is complete.

Boundedness of each $F_{k}$ is part of the definition of a Cantor chain because $diam$ is defined for nonempty bounded sets only in this library (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space); it is not an extra hypothesis but the precondition for writing the diameter condition down.

Facts & Assumptions

Given: For assertion 1, the Axiom of Countable Choice; a metric space $(X, d)$ ; a Cantor chain $(F_{k})$ in $X$ ; a real $ε > 0$ .

[A1]

Completeness of $(X, d)$ : every Cauchy sequence in $X$ converges to a point of $X$ (Complete metric space: every Cauchy sequence converges in the space, Cauchy sequence in a metric space).

[A2]

The converse hypothesis: every Cantor chain in $X$ has nonempty intersection.

[L1]

For nonempty bounded $A \subseteq X$ , $diam (A) = \sup {d (a, b) : a, b \in A}$ , so $d (a, b) \leq diam (A)$ for all $a, b \in A$ , and $diam (A) \geq 0$ ; a set of reals bounded above has a least upper bound, and any upper bound of that set dominates it (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Complete ordered field (least-upper-bound property)).

[L2]

Closure by adherent points: $x \in \overline{A}$ means $B (x, r) \cap A \neq \emptyset$ for every real $r > 0$ ; $A \subseteq \overline{A}$ ; $\overline{A}$ is closed and is the smallest closed superset of $A$ , and $A$ is closed exactly when $A = \overline{A}$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset, Open ball, closed ball and sphere in a metric space).

[L3]

A closed set is sequentially closed: a sequence in it that converges in $X$ has its limit in it (A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed).

[L4]

Countable choice: a family $(A_{k})_{k \in N}$ of nonempty sets admits $k \mapsto a_{k}$ with $a_{k} \in A_{k}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L5]

Triangle inequality (M3), symmetry (M2) and separation (M1) of a metric, and nonnegativity of a metric (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, Nonnegativity of a metric is a consequence of the other axioms, not an axiom).

[L6]

Limits of reals preserve non-strict inequalities, and a constant sequence converges to that constant (Limits preserve non-strict inequalities, Limits and Cauchy sequences of reals).

[L7]

Convergence and Cauchyness may be tested with real $ε > 0$ rather than rational (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , Cauchy sequence in a metric space, The rationals embed densely in the reals).

[L8]

The range of a Cauchy sequence is bounded (Every Cauchy sequence in a metric space is bounded), and a subset of a bounded set is bounded (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L9]

Induction on $N$ (The principle of mathematical induction).

Proof

technique · direct

1.1

Nestedness propagates: for $k \leq l$ one has $F_{l} \subseteq F_{k}$ , by induction on $l$ from $F_{l + 1} \subseteq F_{l}$ and transitivity of inclusion.

1.2

Assume [A1] and let $(F_{k})$ be a Cantor chain. Every $F_{k}$ is nonempty, so [L4] supplies a sequence $(x_{k})$ with $x_{k} \in F_{k}$ for every $k$ .

A1L4choose

1.3

A preliminary about closures, used in claim 2: let $A \subseteq X$ be nonempty and bounded, let $u, v \in \overline{A}$ and let $η > 0$ be real; then $B (u, η)$ and $B (v, η)$ meet $A$ , so there are $a, b \in A$ with $d (u, a) < η$ and $d (v, b) < η$ , whence $d (u, v) \leq d (u, a) + d (a, b) + d (b, v) < diam (A) + 2 η$ .

L1L2L5

1.4

If $x, y \in ⋂_{k} F_{k}$ then $d (x, y) \leq diam (F_{k})$ for every $k$ by [L1]; the constant sequence with value $d (x, y)$ converges to $d (x, y)$ and $diam (F_{k}) \to 0$ , so $d (x, y) \leq 0$ , and $d (x, y) \geq 0$ forces $d (x, y) = 0$ and $x = y$ .

L1L5L6

1.5

For claim 2 assume [A2] and let $(x_{k})$ be a Cauchy sequence in $X$ ; put $A_{k} : = {x_{j} : j \geq k}$ and $F_{k} : = \overline{A_{k}}$ .

A2construct

2.1

Since $d (u, v) < diam (A) + 2 η$ for every real $η > 0$ , we get $d (u, v) \leq diam (A)$ : were $d (u, v) > diam (A)$ , the value $η : = (d (u, v) - diam (A)) / 3$ would be positive and would give $d (u, v) < diam (A) / 3 + 2 d (u, v) / 3 < d (u, v)$ .

step 1.3algebra

2.2

Back to claim 1: for any $K \in N$ and all $m, n \geq K$ we have $x_{m} \in F_{m} \subseteq F_{K}$ and $x_{n} \in F_{n} \subseteq F_{K}$ , so $d (x_{m}, x_{n}) \leq diam (F_{K})$ .

step 1.1step 1.2L1

2.3

$A_{k + 1} \subseteq A_{k}$ , and $\overline{A_{k}}$ is a closed superset of $A_{k + 1}$ , so $F_{k + 1} \subseteq F_{k}$ by minimality of the closure.

step 1.5L2

3.1

Hence $diam (A)$ is an upper bound of ${d (u, v) : u, v \in \overline{A}}$ ; fixing $u \in \overline{A}$ , which exists since $A \neq \emptyset$ and $A \subseteq \overline{A}$ , gives $\overline{A} \subseteq B (u, diam (A) + 1)$ , so $\overline{A}$ is nonempty and bounded and $diam (\overline{A}) \leq diam (A)$ . And ${d (a, b) : a, b \in A} \subseteq {d (u, v) : u, v \in \overline{A}}$ gives $diam (A) \leq diam (\overline{A})$ , so the two diameters are equal.

step 2.1L1L2

3.2

Given a real $ε > 0$ , the convergence $diam (F_{k}) \to 0$ supplies $K$ with $diam (F_{K}) < ε$ , so $d (x_{m}, x_{n}) < ε$ for all $m, n \geq K$ ; hence $(x_{k})$ is Cauchy, and by [A1] it converges to some $x \in X$ .

step 2.2A1L6L7

4.1

Fix $K \in N$ . For every $k \geq K$ we have $x_{k} \in F_{k} \subseteq F_{K}$ , and the tail $(x_{K + j})_{j \in N}$ converges to $x$ because $(x_{k})$ does; since $F_{K}$ is closed it is sequentially closed, so $x \in F_{K}$ . As $K$ was arbitrary, $x \in ⋂_{k} F_{k}$ .

step 1.1step 3.2L3L7

4.2

Each $A_{k}$ is nonempty and is contained in the bounded range of $(x_{k})$ , hence bounded; so each $F_{k}$ is nonempty, closed and, by step 3.1, bounded with $diam (F_{k}) = diam (A_{k})$ .

step 3.1step 1.5L2L8

5.1

Claim 1 is established: the intersection contains $x$ by step 4.1 and no second point by step 1.4.

step 4.1step 1.4

5.2

Given a real $ε > 0$ , Cauchyness supplies $K$ with $d (x_{m}, x_{n}) < ε / 2$ for all $m, n \geq K$ ; then $ε / 2$ is an upper bound of ${d (a, b) : a, b \in A_{k}}$ for every $k \geq K$ , so $0 \leq diam (A_{k}) \leq ε / 2 < ε$ for $k \geq K$ . Hence $diam (F_{k}) \to 0$ and $(F_{k})$ is a Cantor chain.

step 4.2step 2.3L1L5L7

6.1

By [A2] there is $x \in ⋂_{k} F_{k}$ . Given a real $ε > 0$ , take $K$ as in step 5.2 for $ε$ ; since $x \in F_{K} = \overline{A_{K}}$ , the ball $B (x, ε / 2)$ meets $A_{K}$ , so there is $j \geq K$ with $d (x, x_{j}) < ε / 2$ , and then for every $k \geq K$ we get $d (x, x_{k}) \leq d (x, x_{j}) + d (x_{j}, x_{k}) < ε / 2 + ε / 2 = ε$ .

step 5.2A2L2L5

7.1

So $x_{k} \to x$ with $x \in X$ , every Cauchy sequence in $X$ converges, and $(X, d)$ is complete; this is claim 2, and claim 1 is step 5.1.

step 5.1step 6.1L7∎

Remarks

The diameter hypothesis cannot be dropped, and neither can it be weakened to "the diameters are bounded". On $N$ with a metric taking values just above $1$ the tails ${n, n + 1, \dots}$ are nested, closed, bounded and complete with empty intersection (On $N$ with $d (m, n) = 1 + 1 / (m + n)$ for $m \neq n$ the sets ${n, n + 1, \dots}$ are nested, closed, bounded and complete with empty intersection ↗); what fails there is exactly $diam (F_{k}) \to 0$ .
Why the equality $diam (\overline{A}) = diam (A)$ is proved and not assumed. Claim 2 builds its Cantor chain out of the tails of a Cauchy sequence, which are almost never closed, so it must close them; and closing a set could in principle enlarge its diameter. Steps 1.3, 2.1 and 3.1 are the proof that it cannot, and they are the only place in this item where the definition of the closure by adherent points is used at full strength.
Where choice enters. Only at step 1.2, which picks one point from each $F_{k}$ ; that is ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Claim 2 is choice free apart from what A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed is not asked to supply here: step 6.1 uses the definition of the closure directly rather than a sequence extracted from it.
Relation to the nested interval property. For $X = R$ and $F_{k} = [a_{k}, b_{k}]$ this is the nested interval property with the extra hypothesis that the lengths tend to $0$ , which is what buys uniqueness of the common point. The general statement replaces "interval" by "closed set" and "length" by "diameter", and completeness is what replaces the least-upper-bound property.

Depends on

Used by

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59 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

Cantor's intersection theorem (Wikipedia) (standard reference, not scraped)
Complete metric space (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)

In a complete metric space nested nonempty closed sets whose diameters tend to 00 meet in exactly one point, and this property characterises completeness