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Uniform continuity

Move the quantifiers and continuity becomes uniform — a single delta works everywhere. We see the difference, the failure of x^2, the rescue on compact sets, and Lipschitz as a strong sufficient condition.

One delta to rule them all

Ordinary continuity lets delta depend on the point a. Uniform continuity forbids that dependence: f is uniformly continuous on a set S if for every epsilon > 0 there is one delta > 0 (good for the whole set) such that for all x, y in S, |x - y| < delta implies |f(x) - f(y)| < epsilon. The change is purely in the order of quantifiers — delta is chosen before x and y, not after.

Claim: f(x) = x^2 is NOT uniformly continuous on [0, infinity).
  Fix epsilon = 1. We show NO single delta works.
  Take the pair  x_n = n + 1/n,  y_n = n.
    |x_n - y_n| = 1/n -> 0   (so eventually < any delta),
  but
    |f(x_n) - f(y_n)| = (n + 1/n)^2 - n^2 = 2 + 1/n^2 > 2 > 1 = epsilon.
  So points get arbitrarily close while their images stay > 1 apart.
  No delta can force the images within epsilon = 1.  Hence not uniformly continuous.
x^2 is continuous everywhere yet not uniformly continuous on the half-line: the slope grows without bound.

When continuity upgrades for free

On a closed bounded interval the distinction vanishes: a function continuous on [a, b] is automatically uniformly continuous there. This is sometimes called the Heine–Cantor theorem and is really a statement about compactness — the interval is compact by the Heine–Borel theorem. The proof by contradiction uses Bolzano–Weierstrass in the same way as the extreme value theorem.

Theorem (Heine-Cantor): f continuous on compact [a,b] => f uniformly continuous on [a,b].
Proof (contradiction):
  Suppose not. Then for some epsilon > 0, for every n, there are points
     x_n, y_n in [a,b]  with  |x_n - y_n| < 1/n   but   |f(x_n) - f(y_n)| >= epsilon.
  By Bolzano-Weierstrass, pass to a subsequence x_{n_k} -> p in [a,b].
  Since |x_{n_k} - y_{n_k}| < 1/n_k -> 0,  also  y_{n_k} -> p.
  By continuity at p:  f(x_{n_k}) -> f(p)  and  f(y_{n_k}) -> f(p),
     so  |f(x_{n_k}) - f(y_{n_k})| -> 0.
  But that quantity is >= epsilon for all k -- contradiction.  QED
Compactness closes the gap: continuity on [a,b] is the same as uniform continuity there.

A clean sufficient condition for uniform continuity is Lipschitz continuity: if there is a constant K with |f(x) - f(y)| <= K|x - y| for all x, y, then just take delta = epsilon/K and the uniform definition is satisfied at once. Every Lipschitz function is uniformly continuous; the converse is false (sqrt(x) on [0, 1] is uniformly continuous but not Lipschitz, since its slope blows up at 0). These counterexamples map out exactly how the three notions — continuous, uniformly continuous, Lipschitz — sit strictly inside one another.