Continuity is local preservation of limits. Uniform continuity is a global version that makes one choice of scale work everywhere on the domain.
Pointwise continuity and sequences
In metric spaces, continuity can be read from convergent sequences. This is the local form; the uniform form below controls pairs of points that may move through the domain.
Definition. A map \(f:X\to Y\) between metric spaces is continuous at \(x\) if for every \(\varepsilon>0\) there is \(\delta>0\) such that \(d_X(x,z)\lt \delta\Rightarrow d_Y(f(x),f(z))\lt \varepsilon\).
Theorem. \(f\) is continuous at \(x\) if and only if \(x_n\to x\) implies \(f(x_n)\to f(x)\).
Proof. The epsilon-delta definition directly gives the sequential implication. Conversely, if continuity fails, there is \(\varepsilon>0\) such that for every \(n\) one can choose \(x_n\) with \(d(x_n,x)\lt 1/n\) but \(d(f(x_n),f(x))\ge\varepsilon\), contradicting sequential continuity.
Uniform continuity
Uniform continuity removes dependence of \(\delta\) on the base point.
Definition. \(f:X\to Y\) is uniformly continuous if for every \(\varepsilon>0\) there is \(\delta>0\) such that \(d_X(x,z)\lt \delta\Rightarrow d_Y(f(x),f(z))\lt \varepsilon\) for all \(x,z\in X\).
Theorem. Every continuous map from a compact metric space to a metric space is uniformly continuous.
Proof. For each \(x\), choose \(r_x>0\) such that \(d(x,z)\lt 2r_x\Rightarrow d(f(x),f(z))\lt \varepsilon/2\). Finitely many balls \(B(x_i,r_{x_i})\) cover compact \(X\). Let \(\delta=\min_i r_{x_i}\). If \(d(y,z)\lt \delta\) and \(y\in B(x_i,r_{x_i})\), then both \(y\) and \(z\) lie within \(2r_{x_i}\) of \(x_i\), so \(d(f(y),f(z))\lt \varepsilon\).
Proposition. Uniformly continuous maps send Cauchy sequences to Cauchy sequences. An ordinary continuous map need not do so.
Proof. Given \(\varepsilon>0\), choose one \(\delta>0\) from uniform continuity. If \((x_n)\) is Cauchy, then \(d_X(x_m,x_n)<\delta\) for all sufficiently large \(m,n\), whence \(d_Y(f(x_m),f(x_n))<\varepsilon\). For failure under mere continuity, \(f(x)=1/x\) on \((0,1)\) is continuous, while the Cauchy sequence \(x_n=1/n\) has the non-Cauchy image \(n\).
Continuous extension from dense sets
Uniform continuity is exactly what allows extension from approximating domains.
Theorem. If \(A\) is dense in a metric space \(X\), \(Y\) is complete, and \(f:A\to Y\) is uniformly continuous, then \(f\) extends uniquely to a continuous map \(\overline f:X\to Y\).
Proof. For \(x\in X\), choose \(a_n\in A\) with \(a_n\to x\). The sequence \((a_n)\) is Cauchy, so uniform continuity makes \((f(a_n))\) Cauchy in complete \(Y\). Define \(\overline f(x)=\lim f(a_n)\). If \(b_n\to x\) as well, then \(d_X(a_n,b_n)\to0\), and uniform continuity gives \(d_Y(f(a_n),f(b_n))\to0\); the definition is independent of the approximation.
To preserve the same modulus, fix \(\varepsilon>0\) and choose \(\delta\) so that \(d_X(a,b)<\delta\) implies \(d_Y(f(a),f(b))<\varepsilon\). If \(d_X(x,y)<\delta/3\), choose \(a_n\to x\), \(b_n\to y\); eventually \(d_X(a_n,b_n)<\delta\), and passage to the limit gives \(d_Y(\overline f(x),\overline f(y))\le\varepsilon\). Thus \(\overline f\) is uniformly continuous. Any continuous extension agrees with \(f\) on dense \(A\), hence agrees with \(\overline f\) by taking limits.
Example (completion from rational data). Every Lipschitz map \(f:\mathbb Q\to\mathbb R\) has a unique Lipschitz extension to \(\mathbb R\) with the same constant. In particular, defining a function first on rational approximations is legitimate only after an estimate independent of the chosen approximation has been proved.
Moduli of continuity
Quantitative continuity is often more useful than qualitative continuity.
Definition. A modulus of continuity for \(f\) is a nondecreasing function \(\omega:[0,\infty)\to[0,\infty]\) with \(\omega(t)\downarrow0\) as \(t\downarrow0\) and \[ d_Y(f(x),f(y))\le\omega(d_X(x,y)). \] The least possible modulus is \[ \omega_f(t)=\sup\{d_Y(f(x),f(y)):d_X(x,y)\le t\}. \]
Example (Lipschitz and Hölder scales). Lipschitz maps satisfy \(\omega_f(t)\le Lt\), while \(\alpha\)-Hölder maps satisfy \(\omega_f(t)\le Ct^\alpha\), \(0<\alpha\le1\). On \([0,1]\), \(f(x)=x^\alpha\) is \(\alpha\)-Hölder because concavity gives \((x+h)^\alpha-x^\alpha\le h^\alpha\). For \(\alpha<1\), it is not Lipschitz near \(0\).
Proposition. A map is uniformly continuous exactly when \(\omega_f(t)\to0\) as \(t\downarrow0\). If \(X\) is compact and \(f\) is continuous, then \(\omega_f(t)<\infty\) and this limit holds.
Proof. The first equivalence is just the quantifier definition written through the supremum. On compact \(X\), the function \((x,y)\mapsto d_Y(f(x),f(y))\) is continuous on compact \(X\times X\), hence bounded; Heine-Cantor gives the vanishing limit.
Uniform limits
Uniform convergence preserves continuity because the error can be controlled before taking the point limit.
Theorem. If \(f_n:X\to Y\) are continuous, \(Y\) is metric, and \(f_n\to f\) uniformly, then \(f\) is continuous.
Proof. Fix \(x\) and \(\varepsilon>0\). Choose \(n\) such that \(\sup_z d(f_n(z),f(z))\lt \varepsilon/3\). By continuity of \(f_n\) at \(x\), choose \(\delta\) such that \(d(x,y)\lt \delta\Rightarrow d(f_n(x),f_n(y))\lt \varepsilon/3\). The triangle inequality gives \(d(f(x),f(y))\lt \varepsilon\).