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CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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)(X,d) be a metric space. If each fk:XRf_k:X\to\mathbb{R} is uniformly continuous and fkff_k\to f uniformly on XX, then ff is uniformly continuous.

Facts & Assumptions

Given: A metric space (X,d)(X,d), uniformly continuous functions fk:XRf_k:X\to\mathbb{R}, and uniform convergence fkff_k\to 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 fNf_N means that for every real η>0\eta>0 there is δ>0\delta>0 such that d(x,y)<δd(x,y)<\delta implies fN(x)fN(y)<η|f_N(x)-f_N(y)|<\eta for all x,yXx,y\in X (Uniform continuity of a map of metric spaces: one δ\delta serving every point, The absolute value makes R\mathbb{R} a metric space: d(x,y)=xyd(x,y) = |x-y| is a metric, its open balls are the intervals (xr,x+r)(x-r, x+r), and it is unbounded).

[L1]

For reals u,vu,v, u+vu+v|u+v|\le|u|+|v| (The triangle inequality).

Proof

technique · direct
1.1

Let ε>0\varepsilon>0 be real. Choose NN such that fN(x)f(x)<ε/3|f_N(x)-f(x)|<\varepsilon/3 for every xXx\in X.

A1choose
1.2

By uniform continuity of fNf_N, choose δ>0\delta>0 such that d(x,y)<δd(x,y)<\delta implies fN(x)fN(y)<ε/3|f_N(x)-f_N(y)|<\varepsilon/3 for every x,yXx,y\in X.

A2choose
2.1

If d(x,y)<δd(x,y)<\delta, then f(x)f(y)f(x)fN(x)+fN(x)fN(y)+fN(y)f(y)<ε|f(x)-f(y)|\le|f(x)-f_N(x)|+|f_N(x)-f_N(y)|+|f_N(y)-f(y)|<\varepsilon.

step 1.1step 1.2L1algebra
3.1

The same δ\delta serves every pair x,yx,y, so ff is uniformly continuous.

step 2.1A2

Depends on

Used by

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