Proof of Continuity on a Closed Interval Implies Uniform Continuity

Let $a,b\in\mathbb{R}$ with $a<b$, and let $f:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$. We prove that $f$ is uniformly continuous on $[a,b]$.

Suppose, to the contrary, that $f$ is not uniformly continuous on $[a,b]$. Then by Uniform Continuity on a Subset of the Real Numbers, there exists $\varepsilon_0>0$ such that for every $\delta>0$ there exist $x,y\in[a,b]$ with $|x-y|<\delta$ and

$$|f(x)-f(y)|\ge \varepsilon_0.$$

We will construct a nested sequence of closed intervals

$$I_n=[\alpha_n,\beta_n]\subseteq [a,b]$$

with the following properties:

$$I_{n+1}\subseteq I_n, \qquad \beta_n-\alpha_n=\frac{b-a}{2^n},$$

and for each $n$ the restriction of $f$ to $I_n$ is not uniformly continuous.

Set $I_0=[a,b]$. Since $f$ is not uniformly continuous on $I_0$, at least one of the two halves

$$\left[a,\frac{a+b}{2}\right], \qquad \left[\frac{a+b}{2},b\right]$$

also has the property that the restriction of $f$ to that half is not uniformly continuous; otherwise both restrictions would be uniformly continuous, and then $f$ would be uniformly continuous on all of $[a,b]$. Choose that half to be $I_1$. Continuing inductively, once $I_n$ is chosen, bisect it into two halves and let $I_{n+1}$ be a half on which the restriction of $f$ is not uniformly continuous.

The left endpoints $(\alpha_n)$ form an increasing sequence, and every right endpoint $\beta_n$ is an upper bound for the set $\{\alpha_n:n\in\mathbb{N}\}$. By the least upper bound property Least Upper Bound Property of the Real Numbers, there exists

$$c = \sup\{\alpha_n:n\in\mathbb{N}\}\in\mathbb{R}.$$

Since $\alpha_n\le c$ for all $n$, and since $\beta_n$ is an upper bound for the set of left endpoints, one also has $c\le \beta_n$ for all $n$. Thus

$$c\in I_n$$

for every $n$.

Now $f$ is continuous at $c$ because it is continuous on $[a,b]$. Hence there exists $\delta>0$ such that whenever $|t-c|<\delta$, one has

$$|f(t)-f(c)|<\frac{\varepsilon_0}{2}.$$

Choose $n$ so large that the length of $I_n$ is less than $\delta$. Since $c\in I_n$, every point $t\in I_n$ satisfies $|t-c|<\delta$. Therefore for all $x,y\in I_n$,

$$|f(x)-f(y)|\le |f(x)-f(c)|+|f(c)-f(y)|<\varepsilon_0.$$

This shows that the restriction of $f$ to $I_n$ is uniformly continuous, contradicting the construction of $I_n$.

The contradiction shows that $f$ is uniformly continuous on $[a,b]$.