If $(X,d)$ is a metric space, $(x_n)$ and $(y_n)$ are Cauchy sequences in $(X,d)$. How do i show that $(a_n):=d(x_n,y_n)$ converges?

Here is what i did: Let $(x_n)$ and $(y_n)$ be Cauchy sequences, then $\lim_{n\to\infty}d(x_n,x_{n+1})=0$ and $\lim_{n\to\infty}d(y_n,y_{n+1})=0$. I tried using triangle inequality as follows: $d(a_n,a_m)=|a_n−b_m|=d((x_n,y_n),(x_m,y_m))=|(x_n-x_m) + (y_n-y_m)|\leq |x_n-x_m| + |y_n-y_m|= d(x_n,x_m)+d(y_n,y_m),$ wheren, $n,m\in N$

  • $\begingroup$ Isn't this just applying triangle inequality a few times? $\endgroup$ – IAmNoOne Dec 22 '14 at 9:17
  • $\begingroup$ How? Pls. can you show it? $\endgroup$ – Yusuf Dec 22 '14 at 9:18
  • $\begingroup$ What are $x_n$ and $y_n$? $\endgroup$ – Suzu Hirose Dec 22 '14 at 9:18
  • $\begingroup$ Are real sequences $\endgroup$ – Yusuf Dec 22 '14 at 9:22
  • $\begingroup$ So what is the role of $(b_n)$ in your question? $\endgroup$ – IAmNoOne Dec 22 '14 at 9:24

It suffices to show that $\{a_n\}_{n\in\mathbb N}=\{d(x_n,y_n)\}_{n\in\mathbb N}\subset \mathbb R$ is a Cauchy sequence, and as $\mathbb R$ is complete, then it converges.

Simply observe that $$ \lvert a_m-a_n\rvert=\lvert d(x_m,y_m)-d(x_n,y_n)\rvert\le d(x_m,x_n)+d(y_m,y_n). $$ Now, for every $\varepsilon>0$, there exist $n_1,n_2>0$, such that $$ m,n\ge n_1\quad\Longrightarrow\quad d(x_m,x_n)<\frac{\varepsilon}{2} $$ and $$ m,n\ge n_2\quad\Longrightarrow\quad d(y_m,y_n)<\frac{\varepsilon}{2}. $$ Hence, for $n_0=\max\{n_1,n_2\}$, $$ m,n\ge n_0\quad\Longrightarrow\quad \lvert a_m-a_n\rvert\le d(x_m,x_n)+d(y_m,y_n)<\varepsilon. $$

  • $\begingroup$ That is exactly what i did! But i am stuck there. $\endgroup$ – Yusuf Dec 23 '14 at 10:21
  • $\begingroup$ See my updated answer. $\endgroup$ – Yiorgos S. Smyrlis Dec 23 '14 at 10:28
  • $\begingroup$ Finally, thank you. $\endgroup$ – Yusuf Dec 23 '14 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.