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The uniform limit of continuous real-valued functions on a metric space is continuous
Statement
Let be a metric space and let be continuous for every , where has its usual metric . If uniformly on , then is continuous.
Facts & Assumptions
Given: A metric space , continuous functions , and uniform convergence .
Uniform convergence gives, for every real , one index such that for every and every (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).
Continuity of at means that for every real there is such that implies (Continuity of a map between metric spaces, at a point and globally, in the - form, The absolute value makes a metric space: is a metric, its open balls are the intervals , and it is unbounded).
For reals , (The triangle inequality).
Proof
Fix and a real . By uniform convergence choose such that for every .
By continuity of at , choose such that implies .
If , then .
Thus is continuous at the arbitrary point , and hence continuous on .
Depends on
- Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions
- Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$-$\delta$ form
- The absolute value makes $\mathbb{R}$ 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
- The triangle inequality
Used by
- The sum of a real power series is continuous at every point strictly inside its interval of convergence Corollary
- Rudin's bounded rational spikes are not equicontinuous and have no uniformly convergent subsequence Counterexample
- The Koch curve is a uniform limit of polygonal paths of lengths (4/3)ⁿ but is not rectifiable Counterexample
- Fredholm alternative for an integral equation Example
- The sine harmonics are pointwise bounded but have no uniformly convergent subsequence Example
- A uniformly convergent sequence of continuous functions, together with its limit, is equicontinuous Lemma
- Functions satisfying a fixed local Lipschitz bound somewhere form a closed subset of C([0,1]) Lemma
- A uniform limit of continuous complex-valued functions is continuous Theorem
- C(K,ℝ) is complete in the supremum metric for every nonempty compact metric space K Theorem
- If continuously differentiable functions converge at one point and their derivatives converge uniformly on a closed interval, then the functions converge uniformly to a differentiable function whose derivative is the derivative limit Theorem
- Picard iteration from 1 produces the exponential partial sums Theorem
- Schwartz space is Fréchet Theorem
- The classical Weierstrass series converges uniformly to a continuous function Theorem
- The Takagi series converges uniformly to a continuous nowhere differentiable function Theorem
Dependency tree · two levels
20 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
- Stanford Math 63CM, Additional Lecture Notes, Theorem 1.16 (standard reference, not scraped)
- W. Trench, Introduction to Real Analysis (standard reference, not scraped)