Extreme Value Theorem
If \(I\) is a closed, bounded interval. If \(f:I \rightarrow \mathbb{R}\) is continuous on \(I\), then \(f\) is uniformly continuous on \(I\).

Proof

Suppose for the sake of contradiction that it wasn’t uniformly continuous on $I$. Then by negating the definition of uniform continuity, there exists an \(\epsilon_0 > 0\) such that for all \(\delta > 0\), there exists points \(x,y \in I\) with \(|x - y| < \delta\) and \(|f(x) - f(y)| \geq \epsilon_0\).

Now choose \(\delta_n = \frac{1}{n}\). Then, for each \(n \in \mathbb{N}\), there exists points \(x_n, y_n \in I\) such that

$$ \begin{align*} |x_n - y_n| \leq \frac{1}{n} \quad \quad \quad \quad (1) \end{align*} $$

and

$$ \begin{align*} |f(x_n) - f(y_n)| \geq \epsilon_0 \quad \quad \quad (2) \end{align*} $$

Since \(I\) is closed and bounded, then by Bolzano-Weirestrass, both \(\{x_n\}\) and \(\{y_n\}\) have convergent subsequences \(x_{n_k} \rightarrow x\) and \(y_{n_k} \rightarrow y\) where \(x, y \in I\). But recall that \(f\) is continuous, then by the Sequential Characterization of Continuity, we know that

$$ \begin{align*} f(x_{n_k}) \rightarrow f(x) \quad \text{ and } \quad f(y_{n_k}) \rightarrow f(y) \quad \quad (3) \end{align*} $$

Observe that as \(k \rightarrow \infty\), then \(\frac{1}{n_k}\) also approaches \(0\) so equation 1 becomes

$$ \begin{align*} \lim\limits_{k \to \infty} 0 &\leq \lim\limits_{k \to \infty} |x_{n_k} - y_{n_k}| \leq \lim\limits_{k \to \infty} \frac{1}{n_k} \\ 0 &\leq \lim\limits_{k \to \infty} |x_{n_k} - y_{n_k}| \leq 0 \end{align*} $$

Thus, \(\lim\limits_{k \to \infty} |x_{n_k} - y_{n_k}| = 0\). But since \(x_{n_k} - y_{n_k} \rightarrow 0\), then we must have that \(x = y\). However this means by (3), that

$$ \begin{align*} f(x_{n_k}) \rightarrow f(x) \quad \text{ and } \quad f(y_{n_k}) \rightarrow f(y)=f(x) \end{align*} $$

so as \(k \to \infty\)

$$ \begin{align*} |f(x_{n_k}) - f(y_{n_k}) \rightarrow 0 \end{align*} $$

This is a contradiction since by (2) we know that for all \(k\)

$$ \begin{align*} |f(x_{n_k}) - f(y_{n_k})| \geq \epsilon_0 \end{align*} $$

Hence, \(f\) must be uniformly continuous on \(I\). \(\ \blacksquare\)

References