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

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 uniformly continuous map sends Cauchy sequences to Cauchy sequences

Statement

Let $(X, d_{X})$ and $(Y, d_{Y})$ be metric spaces (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), let $f : X \to Y$ be uniformly continuous (Uniform continuity of a map of metric spaces: one $δ$ serving every point) and let $(x_{k})$ be a Cauchy sequence in $X$ (Cauchy sequence in a metric space). Then $(f (x_{k}))$ is a Cauchy sequence in $Y$ .

Continuity alone does not suffice, and the failure is not marginal: a continuous map can send a Cauchy sequence to an unbounded one. The witness is named in the remarks below.

Facts & Assumptions

Given: Metric spaces $(X, d_{X})$ and $(Y, d_{Y})$ , a uniformly continuous $f : X \to Y$ , a Cauchy sequence $(x_{k})$ in $X$ , and a real $ε > 0$ .

[A1]

Uniform continuity: for every real $ε > 0$ there is a real $δ > 0$ with $d_{Y} (f (x), f (x^{'})) < ε$ whenever $d_{X} (x, x^{'}) < δ$ , for all $x, x^{'} \in X$ (Uniform continuity of a map of metric spaces: one $δ$ serving every point).

[A2]

Cauchyness of $(x_{k})$ : for every real $δ > 0$ there is $K$ with $d_{X} (x_{m}, x_{n}) < δ$ for all $m, n \geq K$ (Cauchy sequence in a metric space, The rationals embed densely in the reals).

[L1]

Cauchyness in $Y$ is established by producing, for every real $ε > 0$ , an index $K$ with $d_{Y} (f (x_{m}), f (x_{n})) < ε$ for all $m, n \geq K$ (Cauchy sequence in a metric space, The rationals embed densely in the reals).

Proof

technique · direct

1.1

Apply [A1] to $ε$ to obtain a real $δ > 0$ such that $d_{Y} (f (x), f (x^{'})) < ε$ for every pair $x, x^{'} \in X$ with $d_{X} (x, x^{'}) < δ$ .

A1choose

1.2

Apply [A2] to that $δ$ to obtain $K \in N$ with $d_{X} (x_{m}, x_{n}) < δ$ for all $m, n \geq K$ .

2.1

For all $m, n \geq K$ the pair $(x_{m}, x_{n})$ satisfies the hypothesis of step 1.1, so $d_{Y} (f (x_{m}), f (x_{n})) < ε$ .

step 1.1step 1.2

3.1

Since $ε > 0$ was an arbitrary real, $(f (x_{k}))$ is Cauchy in $(Y, d_{Y})$ .

step 2.1L1∎

Remarks

This is the exact point where uniform continuity is indispensable. The $δ$ produced in step 1.1 is chosen before any index is known, and it is then fed to the Cauchy condition. With ordinary continuity the $δ$ would depend on a base point, and there is no base point available: the sequence has no limit in $X$ to serve as one. That is not a defect of the proof but the reason the statement is false for continuous maps; the witness is $x \mapsto 1 / x$ on $(0, 1)$ ( $x \mapsto 1 / x$ is continuous on $(0, 1)$ and sends the Cauchy sequence $(1 / (k + 2))_{k \geq 0}$ to an unbounded one ↗).
No completeness and no surjectivity is assumed, and $f$ need not be injective. The theorem is a statement about a single map and a single sequence.
Consequences used later on this page. Uniformly equivalent metrics have the same Cauchy sequences (Topologically, uniformly and Lipschitz equivalent metrics on a set), because uniform equivalence says exactly that both identity maps are uniformly continuous; and a uniformly continuous map defined on a dense subspace transports the Cauchy sequences that approximate a point, which is what makes A uniformly continuous map from a dense subspace into a complete metric space extends uniquely to a uniformly continuous map on the whole space possible.

Depends on

Used by

Dependency tree · two levels

25 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

Uniform continuity (Wikipedia) (standard reference, not scraped)
Cauchy sequence (Wikipedia) (standard reference, not scraped)