Continuous_Extension_Theorem

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Definition

A function $f$ is uniformly continuous on a set $A$ if and only if it is possible to define a continuous function $g$ on the closure $\overline{A}$ such that $g(x) = f(x)$ for all $x \in A$.

Proof

($\Rightarrow$) Assume $f$ is uniformly continuous on $A$. For $c \in \overline{A} \setminus A$, there exists a sequence $(x_n) \subseteq A$ with $x_n \to c$.

Since $(x_n)$ is Cauchy and $f$ is uniformly continuous, $(f(x_n))$ is Cauchy, hence convergent. Define $g(c) = \lim f(x_n)$.

This is well-defined (independent of sequence choice) and $g$ is continuous on $\overline{A}$.

($\Leftarrow$) If $g$ is continuous on the closed bounded set $\overline{A}$, then $g$ is uniformly continuous by the theorem on compact sets. Hence $f = g|_A$ is uniformly continuous.