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?

  • $\begingroup$ 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. $\endgroup$ – kobe Jan 26 '15 at 1:25
  • $\begingroup$ you're right,sorry $\endgroup$ – jestina Jan 26 '15 at 1:36
  • $\begingroup$ No problem, jestina. $\endgroup$ – 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.


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