# Limit in metric space.

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 ??

-
Just some general comments: 1. If X is a metric space, there is no $|\;|$, only a metric $d$. There is quite a difference between a norm and a metric and if you will use the norm notation for distance, you might find yourself using in your proof properties of a norm that don't hold in general metric spaces. 2. Don't use $f_n$ and $g_n$ for general elements of a metric space. Reserve those letters for functions. Use $x$, $y$, $p$, etc. – levap Oct 25 '12 at 20:55

I couldn't really follow your proof, but I think this is cleaner: We have converging sequences $f_n\rightarrow f_0$ and $g_n\rightarrow g_0$. Then for each $\epsilon > 0$, we can find an $N>0$ such that for all $n,m>N$, we have $d(f_n,f_0)<\epsilon$ and $d(g_m,g_0)<\epsilon$. Choose $k\geq \text{max}(n,m)$. Then we have $d(f_k,g_k)\leq d(f_k,f_0)+d(f_0,g_0)+d(g_0,g_k)<d(f_0,g_0)+2\epsilon$. From this it is not hard to show that for each $\epsilon >0$, there exists a $k>0$ such that $|d(f_k,g_k)-d(f_0,g_0)|<\epsilon$.

As for the other one, yes. This one is much of the same. Use the Cauchy property along with the inequality $d(f_n,g_n)\leq d(f_n,f_m)+d(f_m,g_m)+d(g_m,g_n)$.

-

Both follow from the same triangular inequality :

• In the first case, write $|d(f_n,g_n)-d(f_0,g_0)|\leqslant d(f_n,f_0)+d(g_n,g_0)$, which converges to zero when $n\to\infty$.

• In the second case, write $|d(f_n,g_n)-d(f_m,g_m)|\leqslant d(f_n,f_m)+d(g_n,g_m)$, which converges to zero when $n,m\to\infty$.

-