Proof: Cauchy sequences and uniform continuity

I'm working on a proof and I'm having trouble relating definitions

I want to prove that if f is uniformly continuous, then if a sequence ${a_n}$ is Cauchy, ${f(a_n)}$ is Cauchy.

So if $f$ is uniformly continuous, then for all $\epsilon>0$ there exists a $\delta>0$ such that $d(x,y)<\delta$ implies $d(f(x),f(y))<\epsilon$

And by definition, a sequence $a_n$ is Cauchy if for $\epsilon>0,$ there exists a natural number $N_1$ such that for all $m,n>N_1, d(a_m,a_n)<\epsilon.$

How can I show that $f(a_n)$ must be Cauchy?

• I don't understand the changes. You assume $f$ is uniformly continuous, but at the same time, you're trying to prove $f$ is uniformly continuous. – kobe Jan 26 '15 at 1:25
• you're right,sorry – jestina Jan 26 '15 at 1:36
• No problem, jestina. – kobe Jan 26 '15 at 1:40

Assume $f$ is uniformly continuous, and let $(a_n)$ be a Cauchy sequence in your space. Given $\varepsilon > 0$, we may choose, by uniform continuity of $f$, a $\delta > 0$ such that for all $x, y$, $d(x,y) < \delta$ implies $d(f(x),f(y)) < \varepsilon$. Since $(a_n)$ is Cauchy, there exists a positive integer $N$ such that $d(a_m,a_n) < \delta$ for all $m,n \ge N$. Hence, if $m, n \ge N$, then $d(a_m,a_n) < \delta$, which implies $d(f(a_m),f(a_n)) < \varepsilon$. Since $\varepsilon$ was arbitrary, $(f(a_n))$ is Cauchy.