Uniform continuity controls all nearby pairs in the domain. The focus keyword uniformly continuous mappings is especially useful because such mappings preserve Cauchy sequences. This result distinguishes uniform continuity from ordinary continuity and is essential when extending functions from dense subsets into complete spaces. :::theorem[Composition of Uniformly Continuous Mappings] Let $f:X\to Y$ and $g:Y\to Z$ be uniformly continuous mappings between metric spaces. Then $g\circ f:X\to Z$ is uniformly continuous. ::: :::proof Given that $f$ and $g$ are uniformly continuous. To prove that $g\circ f$ is uniformly continuous. Let $\epsilon>0$. Choose $\eta>0$ such that $d_Z(g(y_1),g(y_2))<\epsilon$ whenever $d_Y(y_1,y_2)<\eta$. Choose $\delta>0$ such that $d_Y(f(x_1),f(x_2))<\eta$ whenever $d_X(x_1,x_2)<\delta$. Then $d_X(x_1,x_2)<\delta$ implies $d_Z(g(f(x_1)),g(f(x_2)))<\epsilon$. $\square$ ::: :::example[Continuous Mapping Not Preserving Cauchy Sequences] The function $f(x)=1/x$ on $X=(0,1)$ is continuous. The sequence $x_n=1/n$ is Cauchy in $X$, but $f(x_n)=n$ is not Cauchy in $\mathbb R$. ::: :::theorem[Uniformly Continuous Mappings Preserve Cauchy Sequences] Let $f:X\to Y$ be uniformly continuous. If $\{x_n\}$ is a Cauchy sequence in $X$, then $\{f(x_n)\}$ is a Cauchy sequence in $Y$. ::: :::proof Given that $f$ is uniformly continuous and $\{x_n\}$ is Cauchy. To prove that $\{f(x_n)\}$ is Cauchy. Let $\epsilon>0$. Choose $\delta>0$ such that $d_Y(f(x),f(y))<\epsilon$ whenever $d_X(x,y)<\delta$. Since $\{x_n\}$ is Cauchy, choose $n_0$ such that $d_X(x_n,x_m)<\delta$ for all $m,n\geq n_0$. Therefore $d_Y(f(x_n),f(x_m))<\epsilon$ for all $m,n\geq n_0$. Hence, $\{f(x_n)\}$ is Cauchy. $\square$ ::: :::theorem[Uniformly Continuous Extension from a Dense Subset] Let $A$ be dense in $(X,d_X)$ and let $(Y,d_Y)$ be complete. If $f:A\to Y$ is uniformly continuous, then $f$ has a unique continuous extension $g:X\to Y$. Moreover, $g$ is uniformly continuous. ::: :::proof Given that $A$ is dense, $Y$ is complete, and $f$ is uniformly continuous. To construct the extension. For $x\in X$, choose $a_n\in A$ with $a_n\to x$. Then $\{a_n\}$ is Cauchy, so $\{f(a_n)\}$ is Cauchy in $Y$ and converges. Define $g(x)=\lim f(a_n)$. Uniform continuity shows this definition is independent of the approximating sequence. If $x\in A$, using the constant sequence gives $g(x)=f(x)$. Passing the uniform continuity estimate to limits gives uniform continuity of $g$. Uniqueness follows from agreement on dense $A$. $\square$ ::: :::remark Completeness of $Y$ is necessary. The identity map from $\mathbb Q$ to $\mathbb Q$ is uniformly continuous, but it cannot extend continuously from $\mathbb Q$ to $\mathbb R$ with values in $\mathbb Q$. ::: :::exercise 1. Prove every Lipschitz mapping is uniformly continuous. 2. Give a continuous map that does not preserve Cauchy sequences. 3. Explain why completeness of $Y$ is needed in the extension theorem. ::: :::answer 1. Choose $\delta=\epsilon/L$ when the Lipschitz constant is $L>0$. 2. $x\mapsto1/x$ on $(0,1)$. 3. The image Cauchy sequence must converge in $Y$. ::: :::faq[Frequently Asked Questions] Q: Does every continuous function preserve Cauchy sequences? A: No. Uniform continuity is sufficient. Q: Why is dense-domain extension unique? A: Continuous maps agreeing on a dense subset agree everywhere. :::