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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, $n,m\in N$

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  • $\begingroup$ 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
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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 $n_0=\max\{n_1,n_2\}$, $$ m,n\ge n_0\quad\Longrightarrow\quad \lvert a_m-a_n\rvert\le d(x_m,x_n)+d(y_m,y_n)<\varepsilon. $$

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  • $\begingroup$ 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

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