Let $A \subset \mathbb{R}$ be a bounded interval. Assume a function $f:A \longrightarrow \mathbb{R}$ satisfies that for every cauchy sequence ${x_n}$ where $x_n \in A$ the sequence ${f(x_n)}$ is also a cauchy sequence. How can I prove that $f$ is uniformly continuous in $A$?
I have tried extending $A$ to be a closed interval $A'=[a,b]$ where $a=\inf A ,b=\sup A$, and proving that for every convergent series $x_n \in A'$ the series $f(x_n)$ also converges. This would allow me to use the series definition of continuity and thus because this is a closed interval it would also be uniformly continuous. But because the function is not necessarily defined at $f(a)$ I haven't been able to do that.
$A=(a,b)$. Let $(x_n)$ be any sequence in $A$ converging to $b$, for example $x_n := b-(b-a)/n$. Then $(x_n)$ is Cauchy. By our hypothesis $f(x_n)$ is Cauchy, hence converging to say $\kappa$ as $\mathbb{R}$ is complete. Similarly, we can find that for any sequence in $A$ that converges to $a$ converges to say $\lambda$ when $f$ is applied. Extend $f$ to $\tilde{f}$ as $\tilde{f}(x)=f(x)$ when $x \in A$, $\tilde{f}(b)=\kappa$ and $\tilde{f}(a)=\lambda$. Continuity of $\tilde{f}$ follows from our construction and the rest follows from your argument.
I hope this helps.