Uniform continuity strengthens ordinary continuity by requiring one $\delta$ to work throughout the whole domain. The focus keyword uniform continuity is important because it explains why a function may behave well on a compact interval and badly on an unbounded or punctured domain. Distance functions provide a central source of uniformly continuous functions in every metric space. :::definition[Uniform Continuity] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. A function $f:X\to Y$ is called uniformly continuous if for every $\epsilon>0$, there exists $\delta>0$ such that $d_Y(f(x_1),f(x_2))<\epsilon$ whenever $x_1,x_2\in X$ and $d_X(x_1,x_2)<\delta$. ::: :::remark In ordinary continuity, $\delta$ may depend on the point. In uniform continuity, $\delta$ depends only on $\epsilon$ and works throughout the entire domain. ::: :::example[Square Function on the Real Line] The function $f(x)=x^2$ on $\mathbb R$ is continuous but not uniformly continuous. Let $x_n=n$ and $y_n=n+1/n$. Then $|x_n-y_n|\to0$, but $|f(y_n)-f(x_n)|=2+1/n^2$, which does not tend to $0$. ::: :::definition[Distance from a Set] Let $(X,d)$ be a metric space and let $A\subseteq X$ be nonempty. For $x\in X$, define $d(x,A)=\inf\{d(x,a):a\in A\}$. This is called the distance from $x$ to the set $A$. ::: :::theorem[Distance from a Set is Uniformly Continuous] Let $A$ be a nonempty subset of a metric space $(X,d)$. The function $x\mapsto d(x,A)$ is uniformly continuous on $X$. ::: :::proof Given that $f(x)=d(x,A)$. To prove uniform continuity. For $x,z\in X$ and $a\in A$, $d(x,a)\leq d(x,z)+d(z,a)$. Taking infimum over $a\in A$ gives $d(x,A)\leq d(x,z)+d(z,A)$, so $d(x,A)-d(z,A)\leq d(x,z)$. Interchanging $x$ and $z$ gives $|d(x,A)-d(z,A)|\leq d(x,z)$. Given $\epsilon>0$, choose $\delta=\epsilon$. Hence, the function is uniformly continuous. $\square$ ::: :::example[Reciprocal Function on an Open Interval] The function $f(x)=1/x$ on $(0,1)$ is continuous but not uniformly continuous. Take $x_n=1/n$ and $y_n=1/(n+1)$. Then $|x_n-y_n|\to0$, but $|f(x_n)-f(y_n)|=1$. ::: :::example[Square Function on a Closed Interval] The function $f(x)=x^2$ on $[0,1]$ is uniformly continuous because $|x^2-y^2|=|x-y||x+y|\leq2|x-y|$. ::: :::theorem[Closure Characterization by Distance] Let $A$ be a nonempty subset of $(X,d)$. Then $x\in\overline A$ if and only if $d(x,A)=0$. ::: :::proof Given $x\in X$. To prove the equivalence. If $x\in\overline A$, every ball around $x$ meets $A$, so for every $\epsilon>0$ there exists $a\in A$ with $d(x,a)<\epsilon$. Thus $d(x,A)=0$. Conversely, if $d(x,A)=0$, then for every $\epsilon>0$ there exists $a\in A$ with $d(x,a)<\epsilon$. Hence every ball around $x$ meets $A$, and $x\in\overline A$. $\square$ ::: :::theorem[Separation of Disjoint Closed Sets] If $A$ and $B$ are disjoint closed subsets of a metric space $X$, then there is a continuous $f:X\to\mathbb R$ with $f=0$ on $A$, $f=1$ on $B$, and $0\leq f\leq1$. ::: :::proof Define $f(x)=d(x,A)/(d(x,A)+d(x,B))$. Since $A$ and $B$ are closed and disjoint, the denominator is positive. Distance functions are continuous, so $f$ is continuous. If $x\in A$, then $f(x)=0$. If $x\in B$, then $f(x)=1$. $\square$ :::