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TheoremStatement: 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 continuous real-valued functions on a metric space is continuous

Statement

Let (X,d) be a metric space and let fk:XR be continuous for every kN, where R has its usual metric dR(s,t)=st. If fkf uniformly on X, then f:XR is continuous.

Facts & Assumptions

Given: A metric space (X,d), continuous functions fk:XR, and uniform convergence fkf.

[A1]

Uniform convergence gives, for every real η>0, one index N such that fk(x)f(x)<η for every kN and every xX (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

[A2]

Continuity of fN at a means that for every real η>0 there is δ>0 such that d(x,a)<δ implies fN(x)fN(a)<η (Continuity of a map between metric spaces, at a point and globally, in the ε-δ form, The absolute value makes R a metric space: d(x,y)=xy is a metric, its open balls are the intervals (xr,x+r), and it is unbounded).

[L1]

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

Proof

technique · direct
1.1

Fix aX and a real ε>0. By uniform convergence choose N such that fN(x)f(x)<ε/3 for every xX.

A1choose
1.2

By continuity of fN at a, choose δ>0 such that d(x,a)<δ implies fN(x)fN(a)<ε/3.

A2choose
2.1

If d(x,a)<δ, then f(x)f(a)f(x)fN(x)+fN(x)fN(a)+fN(a)f(a)<ε.

step 1.1step 1.2L1algebra
3.1

Thus f is continuous at the arbitrary point a, and hence continuous on X.

step 2.1A2

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 38 results over 10 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