Try
Suppose a sequence of continuous response curves is drawn on [0,2]. Every curve obeys the same slope bound: its outputs differ by at most the distance between inputs. Its value at 0 stays between −2 and 2. Must the entire sequence settle to one curve? No: it could alternate between two different constant functions. A more useful question is whether some infinite subsequence settles uniformly, with one error bound over the whole input interval.
For a worked family, consider fₙ(x)=cₙ+dₙx, where |cₙ|≤2 and |dₙ|≤1. Then |fₙ(x)|≤4 and |fₙ(x)−fₙ(y)|≤|x−y| on [0,2]. A convergent subsequence of coefficient pairs would give a uniform limit because
sup over [0,2] of |fₙ(x)−(c+dx)| ≤ |cₙ−c|+2|dₙ−d|.
But general continuous functions do not come with just two coefficients. We will replace coefficient coordinates by values on a countable dense set, then prove why a finite selection of those values controls the entire domain at each requested accuracy.
Predict
Predict which statement is enough: each function is continuous, or all functions share a continuity neighborhood for each point and tolerance? The second is equicontinuity. For a family F on a metric space K, it means that for every x∈K and ε>0 there is δ>0 such that every f∈F and every y∈K with d(x,y)<δ satisfy |f(x)−f(y)|<ε. The radius cannot depend on the particular function.
The other hypothesis is pointwise boundedness: for each x, the set of values f(x), as f ranges over F, is bounded in R. Initially the bound may depend on x. The preceding lesson proves that on compact K, these hypotheses provide common boundedness and uniform equicontinuity. The latter means one δ works for every x,y and every family member.
Neither hypothesis says the sequence itself converges. Our conclusion will be: every sequence in this family has a uniformly convergent subsequence, whose limit is continuous. The limiting function need not be a member of the original family; that distinction will matter in the next lesson on closures.
Experiment
Use the worked 1-Lipschitz family on [0,2]. Place sample centers at 0, 0.02, 0.04, and so on through 2. Every point is within distance at most 0.01 of a center. Suppose two curves differ by at most 0.01 at every center. Predict their largest possible discrepancy between centers.
A finite numerical certificate when a common Lipschitz bound is known| Step | Difference controlled | Upper bound |
|---|
| Move first curve to a nearby center d | |f(x)−f(d)| | 0.01 |
|---|
| Compare curves at the center | |f(d)−g(d)| | 0.01 |
|---|
| Move second curve back to x | |g(d)−g(x)| | 0.01 |
|---|
| Triangle inequality | |f(x)−g(x)| | 0.03 |
|---|
Repeat the calculation with common Lipschitz constant 2 and center discrepancy 0.005. The bound becomes 0.02+0.005+0.02=0.045. A denser grid would improve the spatial terms. Without a common spatial bound, matching at these centers does not control a narrow peak placed between them.
This finite calculation is an error certificate under its stated hypotheses. It is not by itself a proof that infinitely many curves admit compatible center comparisons. The extraction argument supplies that missing infinite step.
Understand
Let K be a nonempty compact metric space, and let (fₙ) be a sequence from a pointwise bounded, equicontinuous family of real continuous functions. We prove the compact-domain Arzelà–Ascoli extraction theorem.
Make a countable list of sample coordinates
For every positive integer m, compactness supplies a finite 1/m-net Fₘ with centers in K. Their union D is countable and dense: any ball of positive radius contains a center from a sufficiently fine net. List all the finite nets in succession as d₁,d₂,…, allowing repetitions. If K is finite, or consists of one point, this repeated list still works. We do not assume K has infinitely many distinct points.
Extract at one coordinate at a time
At d₁, the scalar sequence fₙ(d₁) is bounded, so Bolzano–Weierstrass supplies an infinite convergent subsequence. Write I₁ for its set of original indices. Within that subsequence, extract again so that values at d₂ converge, obtaining an infinite I₂ contained in I₁. Continue: Iⱼ is an infinite subset of Iⱼ₋₁ whose values at dⱼ converge. Previously established convergence survives further subsequences.
Choose n₁ in I₁. Recursively choose nⱼ in Iⱼ with nⱼ>nⱼ₋₁; this is possible because an infinite subset of the positive integers is unbounded. For any fixed i, every j≥i has nⱼ∈Iⱼ⊆Iᵢ. Consequently f at indices nⱼ converges at dᵢ. This proves convergence at every point of the list, one point at a time. Choosing the minimum of each Iⱼ would not work: nested sets could all contain 1, producing a constant index rather than a subsequence.
Upgrade dense-coordinate convergence to uniform Cauchy control
First recall why compactness gives one spatial radius for the family. For a chosen small tolerance, equicontinuity at each center gives a neighborhood controlling all functions relative to that center. Finitely many half-radius neighborhoods cover K. Taking the minimum of their half-radii ensures two sufficiently close points both lie in one original neighborhood. The triangle inequality through its center controls their outputs uniformly over the family. This is the common-radius result proved in the preceding lesson.
Given ε>0, choose such a δ so that d(x,y)<δ makes every family difference |f(x)−f(y)| less than ε/3. The δ-balls centered in D cover K because D is dense. Compactness selects finitely many centers d at which the diagonal values converge. At each selected center they are Cauchy; take the maximum of these finitely many cutoffs. Beyond it, every pair of selected functions differs by less than ε/3 at every chosen center.
For an arbitrary x∈K, choose one of those centers within δ. The three-error ledger now bounds the difference at x by two spatial errors and one center error, each less than ε/3. Thus one cutoff works for all x, proving that the diagonal subsequence is uniformly Cauchy. No maximum over infinitely many unrelated convergence cutoffs was taken.
Use the existing complete function space
The earlier uniform Cauchy criterion gives a uniform real-valued limit. The uniform-limit continuity theorem makes it continuous. Equivalently, invoke the proved completeness of C(K) under the supremum norm. This completes the theorem: a single increasing subsequence converges uniformly on all of K.
Real scalar values matter. Scalar boundedness supplied convergent coordinate subsequences; the same claim would not hold automatically for bounded values in an arbitrary infinite-dimensional normed space.
Apply
Use the written tasks to test each hypothesis independently. Constants fₙ=n have perfect common spatial control but no pointwise bound. Powers xⁿ on [0,1] are bounded, but their steepening behavior near 1 defeats equicontinuity. Every subsequence retains the same discontinuous pointwise limit, so none can converge uniformly.
On R, translated tents hₙ(x)=max(0,1−|x−n|) are all bounded by 1 and share Lipschitz constant 1. Distinct tents have supremum distance 1: evaluate at one center. No subsequence is uniformly Cauchy on R. Yet on each fixed compact interval they eventually vanish. This contrasts global uniform convergence with local uniform convergence and identifies the missing compact-domain hypothesis.
Finally, prove why continuous pointwise limits alone do not rescue the finite-net argument. The practice places shrinking peaks at 1/n on [0,1]. Every fixed point eventually misses them, while each function still has height 1 somewhere. The correct obstruction is failure of a common spatial radius, not failure of continuity of any individual curve.
Continue
Retain the proof in four pieces: a dense sample list, nested coordinate subsequences, an increasing diagonal choice, and a finite-net uniform Cauchy estimate. A plot may illustrate the common-radius condition, but it cannot verify every function in an infinite family or choose the infinite subsequence for you.
The next lesson proves the converse compactness statements and checks whether the uniform limit remains inside a specified family. A family can have convergent subsequences in C(K) without containing those limits itself. Closedness is a separate condition, not an extra conclusion hidden inside extraction.
