Uniform continuity

In ordinary continuity at a point \(c\), \(\delta\) depends on \(\varepsilon\) and also on \(c\). In uniform continuity, a single \(\delta\) serves all points of \(A\). So uniform continuity is a property of the whole set, not of each point, and it always implies ordinary continuity.

The classic example of the difference is \(f(x)=x^2\). On \([0,3]\) we have \(|x^2-y^2|=|x+y||x-y|\le6|x-y|\), so \(\delta=\varepsilon/6\) works. On \(\mathbb R\) the slope grows without bound: with \(x_n=n\) and \(y_n=n+1/n\) the distance between the points tends to \(0\) but \(|f(y_n)-f(x_n)|\ge2\). Something similar happens to \(1/x\) on \((0,1)\), where the trouble is the boundary point missing from the domain.

The Heine-Cantor theorem says every continuous function on a closed bounded interval \([a,b]\) is uniformly continuous. It is proved by contradiction: otherwise there are sequences \(x_n,y_n\) with \(|x_n-y_n|\to0\) but \(|f(x_n)-f(y_n)|\ge\varepsilon_0\); by Bolzano-Weierstrass a subsequence of \((x_n)\) converges to some \(x^*\in[a,b]\), and then \(y_{n_k}\to x^*\) as well, contradicting continuity at \(x^*\).

Other sufficient conditions are being Lipschitz (\(\delta=\varepsilon/L\)) or being differentiable with bounded derivative. A useful property: a uniformly continuous function maps Cauchy sequences to Cauchy sequences, which is why a uniformly continuous function on \((a,b)\) extends continuously to \([a,b]\). A common mistake is to believe \(\sqrt x\) is not uniformly continuous on \([0,\infty)\) because it is not Lipschitz; it is, with \(\delta=\varepsilon^2\).