Is the Cartesian product of finitely many metric spaces also a metric space? If so, what about completeness? Let $n$ be a positive integer, and let  $p$ be a real number such that $p \geq 1$. Let $(X_1, d_1), \ldots, (X_n, d_n)$ be metric spaces, and let the set $X$ be given by
$$X \colon= \Pi_{k=1}^n =  X_1 \times \cdots \times X_n.$$
Let the function $d \colon X \times X \to \mathbb{R}$ be defined as follows: for any points $x \colon= (x_1, \ldots, x_n)$, $y \colon= (y_1, \ldots, y_n) \in X$, let 
$$
d(x, y) \colon= \sqrt[p]{ \sum_{k=1}^n \left[d_k (x_k, y_k) \right]^p}. 
$$
Then is $d$  a metric on $X$?  
What if $0 < p < 1$? 
For verifying the properties (M1) through (M3) for $d$, we can apply the corresponding properties of the "coordinate" metrics. 
What about (m4), the  triangle inequality? 
My feeling is that $d$ is a metric at least for $p \geq 1$. 
Moreover, if $d$ is a metric, then  what about the following assertion?
$(X,d)$ is complete if and only if $(X_k, d_k)$ is complete for each $k = 1, \ldots, n$. 
 A: Yes its an $l_p$ metric for $p \geq 1$ for $0 < p < 1$ its a pseudometric, you need to prove 2 previous inequalities before proving M4), Young's and Hölder's. As for the second question, yes its complete, I'll do the one for $n=2$ just to simplify. 
Suppose $(X_1, d_1)$ and $(X_2, d_2)$ are complete, then given a cauchy sequence in any of the spaces you know it has a limit which resides in them. Now take a cauchy sequence in the product space, $z_n = (x_n, y_n)$, you know that given $\varepsilon$ there is some $N > 0$ such that for all $n, m > N$, $d(z_n, z_m) < \varepsilon$. But this means that $$d(z_n, z_m) = \sqrt[p]{ d_1 (x_n, x_m)^p + d_2(y_n, y_m)^p } < \varepsilon$$
From here you get that the coordinates are cauchy themselves, because  $$d_1(x_n, x_m) = \sqrt[p]{ d_1 (x_n, x_m)^p} < d(z_n, z_m) = \sqrt[p]{ d_1 (x_n, x_m)^p + d_2(y_n, y_m)^p } < \varepsilon$$ and $$ d_2(y_n, y_m) = \sqrt[p]{ d_1 (y_n, y_m)^p} < d(z_n, z_m) = \sqrt[p]{ d_1 (x_n, x_m)^p + d_2(y_n, y_m)^p } < \varepsilon$$  if you pick the original limit in each for each coordinate you'll get that this new 2-tuple is the limit  in the product space. To prove the the inverse you use a similar logic but this time getting a limit in the product space (for this in the product space let all the coordinates be constant except the one that is cauchy in the space you want to prove is complete).
