Cartesian-Metric Space Let $(X_1,d_1),\ldots,(X_n,d_n)$ be metric spaces and let $X=X_1\times\cdots\times X_n$ be their Cartesian product. For $x=(x_1,\ldots,x_n),\ y=(y_1,\ldots,y_n)\in X$, define $\sigma(x,y)=\sum_{k=1}^nd_k(x_k,y_k)$. Show that  $\sigma$ is a metric on $X$. Show that $\sigma$ is complete if and only if $(X_i,d_i)$ is complete for each $i = 1,\ldots,n$.
 A: As a first prototype, think of each $X_i$ as it was a real line with the usual distance $d_i(x,y)=|x-y|$, then consider the more complicated distance on the Cartesian product: $\sqrt{\sum_i(d_k(x,y)^2)}$. Does it remind anything to you? 
Now, the given exersize is even simpler.
You have to verify that $\sigma$ is indeed a metric:


*

*$\sigma(x,y)=0 \iff x=y$

*$\sigma(x,y)+\sigma(y,z)\ge \sigma(x,z)$ for all $x,y,z$ points (now $\in X$).


If it makes it more comfortable, try with $n=2$, it is basically the same.
For the completeness: you have to show that any Cauchy sequence in $X$ (of course, according to the distance in $X$, that is, $\sigma$) converges, assumed that it happens in all coordinates. 
Finally, you have to investigate the case when at least one $X_k$ is not complete, i.e. there is a Cauchy sequence (i.e. a sequence that 'wants to converge') without limit available in $X_k$. Then, using this particular sequence (and, say, constants in the other coordinates) you will get that neither $X$ is complete.
