# Convergence of distances in metric space

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, $n,m\in N$

• Isn't this just applying triangle inequality a few times? – IAmNoOne Dec 22 '14 at 9:17
• How? Pls. can you show it? – Yusuf Dec 22 '14 at 9:18
• What are $x_n$ and $y_n$? – Suzu Hirose Dec 22 '14 at 9:18
• Are real sequences – Yusuf Dec 22 '14 at 9:22
• So what is the role of $(b_n)$ in your question? – 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 $n_0=\max\{n_1,n_2\}$, $$m,n\ge n_0\quad\Longrightarrow\quad \lvert a_m-a_n\rvert\le d(x_m,x_n)+d(y_m,y_n)<\varepsilon.$$