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CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)
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The uniform limit of uniformly continuous real-valued functions is uniformly continuous

Statement

Let (X,d) be a metric space. If each fk:X→R is uniformly continuous and fk→f uniformly on X, then f is uniformly continuous.

Facts & Assumptions

Given: A metric space (X,d), uniformly continuous functions fk:X→R, and uniform convergence fk→f.

[A1]

Uniform convergence gives one index serving every point for any prescribed positive real error (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

[A2]

Uniform continuity of fN means that for every real η>0 there is δ>0 such that d(x,y)<δ implies ∣fN(x)−fN(y)∣<η for all x,y∈X (Uniform continuity of a map of metric spaces: one δ serving every point, The absolute value makes R a metric space: d(x,y)=∣x−y∣ is a metric, its open balls are the intervals (x−r,x+r), and it is unbounded).

[L1]

For reals u,v, ∣u+v∣≤∣u∣+∣v∣ (The triangle inequality).

Proof

technique · direct
1.1

Let ε>0 be real. Choose N such that ∣fN(x)−f(x)∣<ε/3 for every x∈X.

A1choose
1.2

By uniform continuity of fN, choose δ>0 such that d(x,y)<δ implies ∣fN(x)−fN(y)∣<ε/3 for every x,y∈X.

A2choose
2.1

If d(x,y)<δ, then ∣f(x)−f(y)∣≤∣f(x)−fN(x)∣+∣fN(x)−fN(y)∣+∣fN(y)−f(y)∣<ε.

step 1.1step 1.2L1algebra
3.1

The same δ serves every pair x,y, so f is uniformly continuous.

step 2.1A2∎

Depends on

Used by

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Dependency tree · two levels

21 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