# Continuous Function on Closed Interval is Uniformly Continuous

## Theorem

Let $\closedint a b$ be a closed real interval.

Let $f: \closedint a b \to \R$ be a continuous function.

Then $f$ is uniformly continuous on $\closedint a b$.

## Proof

We have that $\R$ is a metric space under the usual (Euclidean) metric.

We also have from the Heine-Borel Theorem that $\closedint a b$ is compact.

So the result Heine-Cantor Theorem applies.

$\blacksquare$

## Alternative Proof Using Sequences

Suppose for the sake of contradiction that $f$ is not uniformly continuous.

Then there exists an $\varepsilon>0$ such that for all $\delta>0$ there are points $x,y \in \closedint a b$ for which $|x-y|<\delta$ and $|f(x) - f(y)| \geq \varepsilon$.

We will now choose for each $k \in \N$ numbers $x_k,y_k \in \closedint a b$ such that $|x_k-y_k| < \frac{1}{k}$ and $|f(x_k)-f(y_k)| \geq \varepsilon$.

Because the sequence $x_k$ is bounded, by Bolzano-Weierstrass Theorem there exists a convergent subsequence $(x_{k_j})$, whose limit we will denote $x_0$.

Now we have $|x_0-y_{k_j}| \leq |x_0-x_{k_j}| + |x_{k_j} - y_{k_j}| \leq |x_0 - x_{k_j}| + \frac{1}{k_j}, $

Therefore the sequence $(y_{k_j})$ also converges to $x_0$. Because $f$ is sequentially continuous, $\lim_{j\to\infty} f(x_{k_j}) = f(x_0) = \lim_{j\to\infty} f(y_{k_j})$.

This is however, a contradiction, as $|f(x_k) - f(y_k)| \geq \varepsilon$ for all $k$ and thus all $k_j$. Therefore $f$ is uniformly continuous.

$\blacksquare$

## Sources

- 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $5.8$: Compactness and Uniform Continuity: Proposition $5.8.2$