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:

(1)

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

• Consider the following collection of open balls:

(2)

\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:

(3)

\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:

(4)

\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:

(5)

\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:

(6)

\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:

(7)

\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 $A$. $\blacksquare$