# Uniformly continuous functions

• Mar 18th 2011, 06:30 AM
worc3247
Uniformly continuous functions
The question:
Suppose $\displaystyle g(a,b] \rightarrow \mathbb{R}$ is uniformly continuous. Show that if {$\displaystyle x_n$} is a Cauchy sequence in (a,b], then {$\displaystyle g(x_n)$} is always a Cauchy sequence.

So I need to show that $\displaystyle \forall \epsilon >0 \exists N \in \mathbb{N} s.t. \forall m,n>N |g(x_m)-g(x_n)|<\epsilon$. 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?
• Mar 18th 2011, 06:55 AM
girdav
Applying the definition of uniformly continuous, for $\displaystyle \varepsilon >0$ given we can find a $\displaystyle \delta$ such that if $\displaystyle x,y\in\left(a,b\right]$ and $\displaystyle |x-y|<\delta$ then $\displaystyle |g(x)-g(y)|<\delta$.
We can find $\displaystyle N_0$ such that if $\displaystyle m,n\geq N_0$, then $\displaystyle |x_m-x_n|< \delta$.
• Mar 18th 2011, 07:26 AM
worc3247
Ok, so then I guess my question is how do we go about finding an $\displaystyle N_0$ that satisfies this?
• Mar 18th 2011, 07:32 AM
girdav
What about $\displaystyle |g(x_n)-g(x_m)|$ if $\displaystyle n,m\geq N_0$?
• Mar 18th 2011, 09:04 AM
worc3247
Ok, I think i've got it thanks. :)