We know that if a function $f$ is uniformly continuous on an interval $I$ and $(x_n)$ is a Cauchy sequence in $I$, then $f(x_n)$ is a Cauchy sequence as well.
Now, I would like to ask the following question:
The function $g:(0,1) \rightarrow \mathbb{R}$ has the following property: for every Cauchy sequence $(x_n)$ in $(0,1)$, $(g(x_n))$ is also a Cauchy sequence. Prove that g is uniformly continuous on $(0,1)$.
How do we go about doing it?