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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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:XC. 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:XC.

[L2]

For complex numbers, z+wz+w and z=0 if and only if z=0 (Conjugation is an involutive real-field automorphism, zz=z2, and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct
1.1

If fnf uniformly, then for ε>0 choose N with fn(x)f(x)<ε/2 for every nN and xX; for m,nN, [L2] gives fm(x)fn(x)<ε, so (fn) is uniformly Cauchy.

L2
1.2

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

L1choose
2.1

Given ε>0, choose N such that fm(x)fn(x)<ε/2 for all m,nN and xX. Fixing nN and passing m in the continuous modulus gives f(x)fn(x)ε/2<ε for every x, so fnf uniformly.

step 1.2L2

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 54 results over 16 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