I have ,
In $X$ $\lim_{n\to \infty} f_n=f_0 $ and similarly $g_n$ converges to $g_0$, $d$ is a metric defined in $X$ . I have to show that $\lim_{n\to \infty} d(f_n, g_n)=(f_0, g_0)$ This is what i have done , Please let me know if i am right or wrong and where exactly etc..
Since $f_n$ converges $\exists$ $n\ge m$ such that $|f_n-f_0| \le\epsilon $ Similarly for $g_n$ $\exists$ $l\ge m$ such that $|g_l-g_0|\le \epsilon$ So now , $|f_n-g_l|\le|f_n-f_{n+1}|+|f_{n+1}-g_{k+1}|+|g_k-g_l|$ where , $m\ge n$ and $k\ge l$ Choose $|f_n-f_{n+1}|$ and $|g_k-g_{k+1}|$ to be $\frac {\epsilon}{2^{n+1}}$. Now repeatedly applying the triangle inequality we get $|f_n-g_l|-|f_0-g_0|\le\epsilon $ as $n\to\infty$
Another one : If $f_n$ and $g_n$ are cauchy sequence in metric space $(X,d)$ then $d(f_n,g_n)$ is also a cauchy sequence in $\mathbb R$with $d=|x-y|$, Can i argue here by saying that $d$ is a continuous map ??