Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16
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A sequence of complex-valued functions converges uniformly if and only if it is uniformly Cauchy

Statement

Let X be a set and fn:X→C. Then (fn) converges uniformly on X if and only if it is uniformly Cauchy (Uniform convergence and the uniformly Cauchy condition for complex-valued functions, with the componentwise dictionary). This includes X=∅.

Facts & Assumptions

Given: A set X and functions fn:X→C.

[L2]

For complex numbers, ∣z+w∣≤∣z∣+∣w∣ and ∣z∣=0 if and only if z=0 (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct
1.1L2

If fn→f uniformly, then for ε>0 choose N with ∣fn(x)−f(x)∣<ε/2 for every n≥N and x∈X; for m,n≥N, [L2] gives ∣fm(x)−fn(x)∣<ε, so (fn) is uniformly Cauchy.

1.2L1choose

Conversely, suppose (fn) is uniformly Cauchy. For each x∈X, the sequence (fn(x)) is Cauchy and hence has a limit f(x)∈C by [L1]; this defines f:X→C, including the unique empty function when X=∅.

2.1step 1.2L2∎

Given ε>0, choose N such that ∣fm(x)−fn(x)∣<ε/2 for all m,n≥N and x∈X. Fixing n≥N and passing m→∞ in the continuous modulus gives ∣f(x)−fn(x)∣≤ε/2<ε for every x, so fn→f uniformly.

Depends on

Used by

Dependency tree · two levels

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Sources