Completeness and Compactness of Cartesian Product of Metric Spaces Suppose $\{(X_i,d_i)\}_{i\in \mathbb{N}}$ is a collection of metric spaces with the cartesian product defined by $A=\prod_{i=1}^{\infty}X_i$. Let the metric on $A$ be given by $d(x,y)=\sum_{i=1}^{\infty}\frac{1}{2^i}\frac{d_i(x_i,y_i)}{d_i(x_1,y_i)+1}$. How would one go about proving that $(A,d)$ is complete iff each $(X_i,d_i)$ is complete and $(A,d)$ is compact iff each $(X_i,d_i)$ is compact?
 A: The proof of this is rather long, but straightforward. To show this space is complete, take a Cauchy sequence. Show that it projects to a Cauchy sequence in each coordinate. Since each factor space is complete, you have a natural candidate $x$ for convergence. You can find an $N$ large enough that the $i$th coordinate of the $n$th term of the Cauchy sequence is within $\epsilon/2$ of the $i$th coordinate of $x$ whenever $i$ and $n$ are less than $N$, and so that $\sum_{i=N}^\infty 2^{-i}<\epsilon/2$. Then apply a standard $\epsilon/2$ argument by breaking the series at $i=N$. Compactness is proved via sequential compactness with essentially the same argument. Start with an arbitrary sequence. It has subsequences which converge in each coordinate. Carefully build a subsequence which converges in every coordinate by first taking a subsequence that converges in the first coordinate, then taking a subsubsequence which converges in the first two coordinates, and so on. By the above argument, this entire subsequence converges.
To prove the converse of each part, you have one $X_i$ that is not complete(compact), so there exists a bad Cauchy sequence which (sequence such that every subsequence) doesn't converge. Choose a sequence in the product space which matches your bad sequence in the $i$th coordinate, and is constant in the other coordinates. This sequence will behave poorly in the product space too.
