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 $(0, 1)$.

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