Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Every Cauchy sequence in a metric space is bounded

Statement

Let (X,d) be a metric space (Metric space: d(x,y)=0 iff x=y, symmetry, and the triangle inequality; pseudometric and ultrametric) and let (xk) be a Cauchy sequence in X (Cauchy sequence in a metric space). Then its range R:={ xk:k∈N } is a bounded subset of X (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space): there are a point x0∈X and a real r>0 with R⊆B(x0,r) (Open ball, closed ball and sphere in a metric space).

Consequently R is nonempty and bounded, so diam⁡(R) exists (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

Facts & Assumptions

Given: A metric space (X,d) and a Cauchy sequence (xk) in X; write R:={xk:k∈N}.

[A1]

Cauchyness at the real value 1: there is N∈N with d(xm,xn)<1 for all m,n≥N (Cauchy sequence in a metric space, The rationals embed densely in the reals).

[L1]

A nonempty finite set of reals has a maximum, and every element of the set is at most that maximum (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L3]

Membership in a ball: y∈B(x,r) means d(x,y)<r, and the radius is a positive real (Open ball, closed ball and sphere in a metric space).

[L4]

A subset A⊆X is bounded when A=∅ or A⊆B(x0,r) for some x0∈X and real r>0 (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

Proof

technique · direct
1.1

Fix N as in [A1], so that d(xm,xn)<1 whenever m,n≥N; in particular d(xk,xN)<1 for every k≥N.

A1
1.2

The set S:={ d(xj,xN):j≤N }∪{1} is a nonempty finite set of reals, so it has a maximum M:=max⁡S, and M≥1>0.

L1L2
2.1

Put r:=M+1, a real with r>1>0. For j≤N we have d(xj,xN)≤M<r, and for k≥N we have d(xk,xN)<1≤M<r; every index is of one of the two kinds, so d(xk,xN)<r for every k∈N.

step 1.1step 1.2choose
3.1

Hence xk∈B(xN,r) for every k, that is R⊆B(xN,r) with xN∈X and r>0, so R is bounded; and R is nonempty because it contains x0.

step 2.1L3L4∎

Remarks

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Sources