Uniformly Continuity Implies Continuity of Functions on Metric Spaces

# Uniformly Continuity Implies Continuity of Functions on Metric Spaces

Recall from the Uniform Continuity of Functions on Metric Spaces page that that if $(S, d_S)$ and $(T, d_T)$ are two metric spaces, $A \subseteq S$, and $f : A \to T$, then $f$ is said to be uniformly continuous on $A$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that for all $x, y \in A$ where $d_S(x, y) < \delta$ we have that $d_T(f(x), f(y)) < \epsilon$.

We will now look at a very important theorem which says that if $f$ is uniformly continuous on $A$ then $f$ is also continuous on $A$.

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

Note that the converse of Theorem 1 is not true in general. A function may be continuous but may fail to be uniformly continuous.

• Proof: Let $\epsilon > 0$ be given and let $a \in A$. We want to find a $\delta > 0$ such that if $d_S(x, a) < \delta$ then $d_T(f(x), f(a)) < \epsilon$.
• Since $f$ is uniformly continuous on $A$ there exists a $\delta^* > 0$ such that if for all $x \in X$ such that $d_S(x, a) < \delta^*$ we have that $d_T(f(x), f(a)) < \epsilon$.
• So set $\delta = \delta^*$. Then $f$ is continuous at $a$. So $f$ is continuous on all of $A$. $\blacksquare$

One of the most common examples that shows the converse of Theorem 1 is not true in general is the following function $f : (0, 1) \to \mathbb{R}$ (where $(\mathbb{R}, d)$ is the set of real numbers with the usual Euclidean metric and $((0, 1), d)$ is a metric subspace) defined by:

(1)
\begin{align} \quad f(x) = \frac{1}{x} \end{align}

It's not hard to show that $f$ is continuous on $(0, 1)$ but is not uniformly continuous on $(0, 1)$.