Let $a, b \in \mathbb{R}$ such that $a<b$ and let $f:(a,b)\to\mathbb{R}$ be a continuous function. Prove that are equivalent
i) There exists a continuous function $g:[a,b]\to\mathbb{R}$ such that $g(x) = f(x)$ for all $x \in (a,b)$.
ii) $f$ is uniformly continuous.
(i) implies (ii) comes from Heine´s Theorem, since $g$ is continuous and defined on a closed and bounded interval, it is uniformly continuous. So, in particular, the restriction of $g$ to $(a,b)$ is also uniformly continuous. Therefore, $f$ is uniformly continuous.
For (ii) implies (i), I´ve thought about using Cauchy sequences. Namely, take a sequence $(x_{n})$ of points in $I$ that converges to $a$ (this is possible because $a$ is a limit point of $(a,b)$). Since $(x_{n})$ is convergent, it is Cauchy, and because $f$ is assumed to be uniformly continuous, it preserves Cauchy sequences. So we get a Cauchy sequence $(f(x_{n}))$ of points in $f((a,b))$ that has to be convergent to a real number $L$. Similarly, we can find a sequence $(y_{n})$ converging to $b$ that gives rise to a Cauchy sequence $(f(y_{n}))$ of $f((a,b))$, that is forced to converge to a real number $M$. Now extend $f$ by defining $g(x) = f(x)$ for all $x \in (a,b)$, $g(a)=L$ and $g(b) = M$, which gives a continuous extension of $f$.
I´m not sure if my argument would be entirely correct or I should be more exhaustive (for example, giving an $\epsilon - \delta$ proof).
