Uniform Continuity of Cts. Functions with Comp. Domains on Met. Sps.

# Uniform Continuity of Continuous Functions with Compact Domains on Metric Spaces

Recall from the Uniform Continuity of Continuous Functions with Compact Domains page that if $(S, d_S)$ and $(T, d_T)$ are metric spaces and $A \subseteq S$, then a function $f : A \to T$ is said to be uniformly continuous on $A$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that if $x, y \in A$ and $d_S(x, y) < \delta$ then $d_T(f(x), f(y)) < \epsilon$.

We noted that a function may be continuous on some set but may fail to be uniformly continuous on that same set.

We will now look at a fundamentally important theorem which will tell us that if $f$ is a continuous function and the domain of $f$ is compact then $f$ will be uniformly continuous.

Theorem 1: Let $(S, d_S)$ and $(T, d_T)$ be metric spaces and let $A \subseteq S$. If $A$ is compact and $f : A \to T$ is continuous on $A$ then $f$ is uniformly continuous on $A$. |

**Proof:**Let $\epsilon > 0$ be given. Since $f$ is continuous on $A$ we have that for every $a \in A$ and for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists a $\delta_a > 0$ such that if $d_S(x, a) < \delta_a$ then:

\begin{align} \quad d_T(f(x), f(a)) < \epsilon_1 = \frac{\epsilon}{2} \quad (*) \end{align}

- Consider the following collection of open balls:

\begin{align} \quad \mathcal F = \left \{ B \left (a, \frac{\delta_a}{2} \right ) : a \in A \right \} \end{align}

- Clearly $\mathcal F$ is an open covering on $A$ since each $a \in A$ is covered by a ball centered at $a$. Since $A$ is compact in $S$ we have that there exists a finite open subcovering, call it $\mathcal A$ that also covers $S$ and:

\begin{align} \quad \mathcal A = \left \{ B \left ( a_1, \frac{\delta_{a_1}}{2} \right ), B \left ( a_2, \frac{\delta_{a_2}}{2} \right ), …, B \left ( a_n, \frac{\delta_{a_n}}{2} \right ) \right \} \end{align}

- Now let $\delta = \min_{k \in \{1, 2, …, n \}} \left \{ \frac{\delta_{a_k}}{2} \right \}$. Let $x, y \in A$ be such that $d_S(x, y) < \delta$. Since $x \in A$ and since $\mathcal A$ is an open covering of $A$ there exists an $i \in \{1, 2, …, n \}$ such that:

\begin{align} \quad x \in B \left ( a_i, \frac{\delta_{a_i}}{2} \right) \subset B \left ( a_i, \delta_{a_i} \right ) \end{align}

- Furthermore, for $y \in A$ we have by the triangle inequality that:

\begin{align} \quad d_S(a_i, y) \leq d_S(a_i, x) + d_S(x, y) < \frac{\delta_{a_i}}{2} + \delta < \frac{\delta_{a_i}}{2} + \frac{\delta_{a_i}}{2} = \delta_{a_i} \end{align}

- Since $x, y \in B(a_i, \delta_{a_i})$ we have that $d_S(x, a_i) < \delta_{a_i}$ and $d_S(y, a_i) < \delta_{a_i}$ and so by $(*)$ we have that:

\begin{align} \quad d_T(f(x), f(a_i)) < \epsilon_1 = \frac{\epsilon}{2} \quad \mathrm{and} \quad d_T(f(y), f(a_i)) < \epsilon_1 = \frac{\epsilon}{2} \end{align}

- Using the triangle inequality again and we see that:

\begin{align} \quad d_T(f(x), f(y)) \leq d_T(f(x) + f(a_i)) + d_T(f(a_i), f(y)) < \epsilon_1 + \epsilon_2 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

- So for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $x, y \in A$ and $d_S(x, y) < \delta$ then $d_T(f(x), f(y)) < \epsilon$ and so $f$ is uniformly continuous on $A$. $\blacksquare$