1. ## Uniformly continuous functions

The question:
Suppose $g(a,b] \rightarrow \mathbb{R}$ is uniformly continuous. Show that if { $x_n$} is a Cauchy sequence in (a,b], then { $g(x_n)$} is always a Cauchy sequence.

So I need to show that $\forall \epsilon >0 \exists N \in \mathbb{N} s.t. \forall m,n>N |g(x_m)-g(x_n)|<\epsilon$. I have then written out the definition for g being uniformly continuous on (a,b] but I can't see where to go from this definition to show the Cauchy sequence. Help?

2. Applying the definition of uniformly continuous, for $\varepsilon >0$ given we can find a $\delta$ such that if $x,y\in\left(a,b\right]$ and $|x-y|<\delta$ then $|g(x)-g(y)|<\delta$.
We can find $N_0$ such that if $m,n\geq N_0$, then $|x_m-x_n|< \delta$.

3. Ok, so then I guess my question is how do we go about finding an $N_0$ that satisfies this?

4. What about $|g(x_n)-g(x_m)|$ if $n,m\geq N_0$?

5. Ok, I think i've got it thanks.