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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, $n,m\in N$

Yiorgos S. Smyrlis
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Yusuf
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  • Isn't this just applying triangle inequality a few times? – IAmNoOne Dec 22 '14 at 09:17
  • How? Pls. can you show it? – Yusuf Dec 22 '14 at 09:18
  • What are $x_n$ and $y_n$? – Suzu Hirose Dec 22 '14 at 09:18
  • Are real sequences – Yusuf Dec 22 '14 at 09:22
  • So what is the role of $(b_n)$ in your question? – IAmNoOne Dec 22 '14 at 09:24
  • A cauchy sequence – Yusuf Dec 22 '14 at 09:25
  • This looks like a very fundamental (homework?) question. What did you try in order to solve this? where exactly did you get stuck? To use this site effectively, you must first give your question a serious attempt and provide context for your question. Then you would have picked up the obvious mismatch in the role/definition of a_n.... – Ittay Weiss Dec 22 '14 at 09:30
  • There seems to be a little confusion as to the connection of $(x_n)$, $(y_n)$, $(a_n)$ and $(b_n)$ as the first two only appear implicitly (in the "definition" of $(a_n)$) and $b_n$ is not used at all. Moreover the initialization of $(a_n)$ as an arbitrary Cauchy sequence is a little bit conflicting with defining $a_n = d(x_n,y_n)$. Could you clarify the situation and give the precise relation between the four sequences? – Matthias Klupsch Dec 22 '14 at 09:30
  • I am sorry, the question should be... $(x_n)$ and $(y_n)$ are Cauchy sequences... – Yusuf Dec 23 '14 at 08:54

1 Answers1

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

Yiorgos S. Smyrlis
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