# Continuity and Cauchy sequences

So I'm revising for my final and I have encountered this problem that uses continuity and Cauchy sequences.

Let $f:D\rightarrow \Bbb{R}$ be continuous and let $(x_n)$ be a Cauchy sequence.

a) Give an example to show that $(f(x_n))$ isn't Cauchy

b) If $D$ is compact, then show that $(f(x_n))$ is Cauchy

For a I think I have an example, let $(x_n) = \dfrac{1}{n}$ and then let $f(x_n) = \dfrac{1}{x_n}$

But for the second part I'm a little stumped, I think I'm going to have to go from the definition of continuity to the the definition of a Cauchy sequence but can't get very far.

• Hint: A continuous function on a compact metric space is uniformly continuous. – E.Lim Dec 8 '15 at 6:45
• Adding a bit more to E.Lim's hint: write down the definition of uniform continuity and exploit the definition of Cauchy seq. to deduce something about $(f(x_n))$. – user160738 Dec 8 '15 at 6:54
• You have to specify $D$ too for question 1. – Henno Brandsma Dec 8 '15 at 10:06

We define $x:=\lim_{n\to\infty}x_n$ (a Cauchy sequence is convergent). If $D$ is compact, then $x\in D$. Let $\varepsilon>0$. Thus: $$\exists\delta>0\forall y\in D:\space |x-y|<\delta\implies |f(x)-f(y)|<\frac{\varepsilon}{2}$$ Furthermore: $$\exists N\in\mathbb{N}\forall n>N: |x-x_n|<\delta$$ So finally, if $m,n>N$: $$|f(x_n)-f(x_m)|\le |f(x_n)-f(x)|+|f(x)-f(x_m)|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\epsilon$$ Thus, $(f(x_n))$ is Cauchy.
You could also argue that if $D$ is compact, $f$ is uniformly continuous and therefore: $$\exists\delta'>0\forall x,y\in D:\space |x-y|<\delta\implies |f(x)-f(y)|<\varepsilon\\ \exists N'\in\mathbb{N}\forall m,n>N': |x_m-x_n|<\delta\implies\\ \forall m,n>N': |f(x_m)-f(x_n)|<\varepsilon$$