C[a,b] as a complete metric space
Collect all continuous functions on [a,b] into one set, the space of continuous functions C[a,b]. Give it the distance d(f,g) = ||f - g|| from the sup norm. Then convergence in this metric is precisely [[uniform-convergence|uniform convergence]] — the abstraction we have been building toward. Two earlier theorems together say this space is well behaved.
- Uniform Cauchy criterion (Guide 3): a uniformly Cauchy sequence in C[a,b] has a uniform limit.
- Uniform limit theorem (Guide 4): that limit is itself continuous, so it lies back inside C[a,b].
- Conclusion: C[a,b] is a complete metric space — Cauchy sequences converge, and within the space.
Which families are compact?
In a finite-dimensional space, closed and bounded means compact — every sequence has a convergent subsequence (Bolzano–Weierstrass). In the infinite-dimensional C[a,b], boundedness is not enough. We need two conditions. First, uniform boundedness: a single constant M with |f(x)| <= M for every f in the family and every x. Second, equicontinuity: one delta serves the whole family at once.
Equicontinuity (the key new idea).
A family F of functions on [a,b] is EQUICONTINUOUS if:
for every epsilon > 0 there is ONE delta > 0 such that
for ALL f in F and all x, y with |x - y| < delta,
|f(x) - f(y)| < epsilon.
(Same delta works for every member f at once.)
Why mere boundedness fails: f_n(x) = sin(n x) on [0, 2 pi].
Uniformly bounded: |sin(n x)| <= 1.
But the slopes n -> infinity destroy equicontinuity, and NO
subsequence converges uniformly (the sines never settle).
=> a bounded sequence in C[0,2pi] with no convergent subsequence.
A clean sufficient condition for equicontinuity: a common
Lipschitz bound |f(x) - f(y)| <= L|x - y| for all f in F
(take delta = epsilon / L).The Arzelà–Ascoli theorem
Putting it together gives one of the most useful compactness results in analysis. The Arzelà–Ascoli theorem: a sequence (f_n) in C[a,b] that is uniformly bounded and equicontinuous has a uniformly convergent subsequence. Equivalently, a subset of C[a,b] is compact iff it is closed, bounded, and equicontinuous. This is the correct infinite-dimensional replacement for Bolzano–Weierstrass.
Sketch of the proof (diagonal argument).
Let (f_n) be uniformly bounded and equicontinuous on [a,b].
(1) Pick a countable DENSE set {q_1, q_2, ...} in [a,b] (rationals).
(2) At q_1 the reals f_n(q_1) are bounded; by Bolzano-Weierstrass
take a subsequence converging at q_1. From IT take a further
subsequence converging at q_2. Repeat.
(3) DIAGONAL subsequence g_k = (k-th term of k-th subsequence)
converges at EVERY q_j.
(4) EQUICONTINUITY upgrades pointwise-on-a-dense-set to
uniform-on-all-of-[a,b]: given epsilon, pick delta from
equicontinuity, cover [a,b] by finitely many delta-balls
centered at q's; closeness at those q's + the common delta
force ||g_k - g_l|| < epsilon for large k, l.
(5) g_k is uniformly Cauchy => converges uniformly (C[a,b] complete).