Uniform Continuity On Compact Metric Spaces

From Pointwise Control to One Global Bound

Continuity controls a function near each individual point. Uniform continuity asks for more: for a chosen output tolerance, one distance threshold must work at every point of the domain. Compactness supplies the key ingredient. If such a global threshold did not exist, we could find pairs of increasingly close domain points whose images stay separated. Compactness then gives a convergent subsequence, and continuity at its limit contradicts that separation.

Throughout, $(X,d_X)$ and $(Y,d_Y)$ are metric spaces, each with its metric topology, and $f:X\to Y$. Recall that $f$ is continuous at $x\in X$ if, for every $\varepsilon>0$, there is a $\delta>0$ such that $d_X(x,z)<\delta$ implies $d_Y(f(x),f(z))<\varepsilon$. This $\delta$ may depend on $x$. A space is compact if every open cover has a finite subcover.

Definition: The map $f:X\to Y$ is uniformly continuous if, for every $\varepsilon>0$, there exists $\delta>0$ such that for all $x,y\in X$, $d_X(x,y)<\delta$ implies $d_Y(f(x),f(y))<\varepsilon$. The same $\delta$ must work for every pair $x,y$.

If $X=\varnothing$, the condition is vacuously true: there are no pairs $x,y\in X$ to test. In the main result, therefore, we may assume $X\ne\varnothing$. We use the standard metric-space compactness result that every sequence in a compact metric space has a convergent subsequence; this equivalence between compactness and sequential compactness was established in the earlier material on compactness in metric spaces.

The Compact-Domain Theorem

Theorem: If $X$ is a compact metric space, $Y$ is a metric space, and $f:X\to Y$ is continuous, then $f$ is uniformly continuous.

Proof. Suppose, for contradiction, that $f$ is not uniformly continuous. Negating the definition gives an $\varepsilon_0>0$ such that for every $\delta>0$, there are $x,y\in X$ satisfying $d_X(x,y)<\delta$ but $d_Y(f(x),f(y))\ge\varepsilon_0$. For each $n\in\mathbb{N}$, use $\delta=1/n$ to choose $x_n,y_n\in X$ with $$ d_X(x_n,y_n)<\frac{1}{n} \qquad\text{and}\qquad d_Y(f(x_n),f(y_n))\ge\varepsilon_0. $$ By compactness, $(x_n)$ has a convergent subsequence $(x_{n_k})$, with $x_{n_k}\to x\in X$. The triangle inequality gives $$ d_X(y_{n_k},x) \le d_X(y_{n_k},x_{n_k})+d_X(x_{n_k},x) <\frac{1}{n_k}+d_X(x_{n_k},x). $$ Both terms on the right tend to $0$, so $y_{n_k}\to x$ as well. By continuity of $f$ at $x$, both $f(x_{n_k})\to f(x)$ and $f(y_{n_k})\to f(x)$. More explicitly, continuity with tolerance $\varepsilon_0/3$ gives a radius $r>0$ such that $d_X(z,x)<r$ implies $d_Y(f(z),f(x))<\varepsilon_0/3$. For all sufficiently large $k$, both $x_{n_k}$ and $y_{n_k}$ are within $r$ of $x$. Therefore $$ d_Y(f(x_{n_k}),f(y_{n_k})) \le d_Y(f(x_{n_k}),f(x))+d_Y(f(x),f(y_{n_k})) <\frac{\varepsilon_0}{3}+\frac{\varepsilon_0}{3} =\frac{2\varepsilon_0}{3} <\varepsilon_0. $$ This contradicts the lower bound $d_Y(f(x_{n_k}),f(y_{n_k}))\ge\varepsilon_0$ for every $k$. Hence $f$ is uniformly continuous. $\square$

The proof uses compactness only to obtain a convergent subsequence of one sequence of points. The second sequence is forced to have the same limit because its distance from the first is less than $1/n$. Continuity at that common limit then makes the image distances small, contradicting the fixed positive separation.

Worked Applications

Worked Example: The Identity on a Closed Interval

Let $X=[-1,2]$ with the usual metric $d_X(x,y)=|x-y|$, and let $f:X\to\mathbb{R}$ be $f(x)=x$, where $\mathbb{R}$ has its usual metric. The interval is compact by the Heine–Borel theorem, and $f$ is continuous. To see the uniform bound directly, let $\varepsilon>0$ and choose $\delta=\varepsilon$. If $x,y\in[-1,2]$ and $|x-y|<\delta$, then $$ |f(x)-f(y)|=|x-y|<\delta=\varepsilon. $$ Thus this single choice works for every pair in the interval.

Check: substituting $f(x)=x$ into the output distance gives $|f(x)-f(y)|=|x-y|$, exactly the input distance. Therefore $\delta=\varepsilon$ gives the required strict inequality, including when one of the points is an endpoint.

Worked Example: A Polynomial on a Compact Interval

Let $X=[-2,3]$ with the usual metric, and define $f(x)=x^2$ as a map to $\mathbb{R}$. The interval is compact by Heine–Borel, and the polynomial is continuous. Given $\varepsilon>0$, set $\delta=\varepsilon/6$. For any $x,y\in[-2,3]$, the sum $x+y$ lies between $-4$ and $6$, so $|x+y|\le6$. Factoring the difference of squares gives $$ |f(x)-f(y)| =|x^2-y^2| =|x-y||x+y| \le 6|x-y|. $$ Consequently, if $|x-y|<\delta$, then $$ |f(x)-f(y)|\le 6|x-y|<6\delta =6\left(\frac{\varepsilon}{6}\right) =\varepsilon. $$ So $f$ is uniformly continuous on this interval.

Check: the bound $|x+y|\le6$ includes the pair $x=y=3$, for which $|x+y|=|3+3|=6$. The estimate allows equality at this intermediate step, but the assumption $|x-y|<\delta$ makes the final output inequality strict.

Worked Example: The Square-Root Function at the Endpoint

Let $X=[0,1]$ with the usual metric, and define $f(x)=\sqrt{x}$ into $\mathbb{R}$. The function is continuous on this compact interval. For $x,y\in[0,1]$, suppose first that $x\ge y$. Both square roots are nonnegative, and $$ (\sqrt{x}-\sqrt{y})^2 \le(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) =x-y. $$ The inequality holds because $\sqrt{x}-\sqrt{y}\ge0$ and $\sqrt{x}+\sqrt{y}\ge\sqrt{x}-\sqrt{y}$. Taking nonnegative square roots gives $|\sqrt{x}-\sqrt{y}|\le\sqrt{x-y}$. If $y\ge x$, interchanging $x$ and $y$ gives the same estimate. Thus, for all $x,y\in[0,1]$, $$ |f(x)-f(y)|\le\sqrt{|x-y|}. $$ Given $\varepsilon>0$, choose $\delta=\varepsilon^2$. If $|x-y|<\delta$, then $$ |f(x)-f(y)|\le\sqrt{|x-y|}<\sqrt{\delta} =\sqrt{\varepsilon^2}=\varepsilon. $$ This proves uniform continuity, including for pairs near $0$.

Check: the chosen $\delta$ is positive because $\varepsilon>0$. The strict input inequality $|x-y|<\varepsilon^2$ yields $\sqrt{|x-y|}<\varepsilon$, not merely a non-strict bound.

Why Compactness Matters

Continuity alone does not supply one distance threshold for all points. A common error is to apply continuity at each $x\in X$, obtain a radius $\delta_x$, and then assume that the radii have a positive lower bound. Pointwise continuity does not provide such a bound on a noncompact domain. The sequential proof avoids this unsupported step: compactness acts on a sequence of counterexample points and produces one limit at which continuity can be used.

Worked Example: A Continuous Function That Is Not Uniformly Continuous

Consider $f:\mathbb{R}\to\mathbb{R}$, $f(x)=x^2$, with the usual metric on both spaces. This function is continuous: at a fixed $a\in\mathbb{R}$, $$ |x^2-a^2|=|x-a||x+a|. $$ If $|x-a|<1$, then $|x+a|\le |x-a|+2|a|<1+2|a|$. Given $\varepsilon>0$, taking $|x-a|<\min\{1,\varepsilon/(2|a|+1)\}$ therefore gives $|x^2-a^2|<\varepsilon$.

To test uniform continuity, let $\varepsilon_0=1$. For any proposed $\delta>0$, choose $n\in\mathbb{N}$ with $n>1/\delta$, and set $x_n=n$ and $y_n=n+1/n$. Then $$ |x_n-y_n|=\frac{1}{n}<\delta, $$ while $$ |f(x_n)-f(y_n)| =\left|(n+1/n)^2-n^2\right| =2+\frac{1}{n^2} >1=\varepsilon_0. $$ Thus every proposed $\delta$ fails for this same output tolerance, and $f$ is not uniformly continuous on $\mathbb{R}$.

Check: the input distances $1/n$ can be made smaller than any prescribed positive $\delta$, whereas the output distances $2+1/n^2$ are always greater than $1$. The contradiction is exactly the negation of uniform continuity. The domain $\mathbb{R}$ is not compact, so this example does not conflict with the theorem.

Common pitfall. In the example above, pointwise continuity gives a radius depending on $a$; it does not justify choosing one positive radius for every $a\in\mathbb{R}$. The concrete pairs $n,n+1/n$ show why such a choice would fail: their input distance tends to $0$, but their output distance stays above $1$. On a compact metric domain, the theorem supplies the missing global conclusion, through the convergent-subsequence argument rather than an unjustified minimum of pointwise radii.

Check Your Understanding

For each question, keep the quantifiers in the definition visible: a uniform $\delta$ must work for every pair of domain points.

Write the negation of uniform continuity using an $\varepsilon_0>0$ and pairs $x,y\in X$.
In the theorem’s proof, why do $x_{n_k}\to x$ and $d_X(x_{n_k},y_{n_k})<1/n_k$ imply $y_{n_k}\to x$?
For $f(x)=3x+1$ on $[0,1]$, find a choice of $\delta$ in terms of $\varepsilon$ and verify the required inequality.
Why does the estimate for $\sqrt{x}$ still work when one of $x,y$ is $0$?
For $f(x)=x^2$ on $\mathbb{R}$, explain why the pairs $n,n+1/n$ disprove uniform continuity for $\varepsilon_0=1$.