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?

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.
• I'm struggling to show that a Cauchy sequence in A projects to a Cauchy sequence in each coordinate. If we suppose $x=x_n$ is a Cauchy sequence, then for all $\epsilon > 0$ we have there exists $N$ s.t. for all $k,m > N$ that $\sum_{i=1}^{\infty}\frac{1}{2^i}\frac{d_i(x_{k_i},y_{m_i})}{d_i(x_{k_i},y_{m_i})+1} < \epsilon$. Then it follows that each $\frac{1}{2^i}\frac{d_i(x_{k_i},y_{m_i})}{d_i(x_{k_i},y_{m_i})+1} < \epsilon$. Then we can rearrange to show that $d_i(x_{k_i},x_{m_i}) < \frac{\epsilon}{\epsilon^{-i}-\epsilon} = \delta$.But how does this show that either $x_k$ or $x_m$ is Cauchy?? – Cococabana Mar 27 '16 at 9:24
• You can get $\sum_{i=1}^\infty\frac{1}{2^i}\frac{d(x_i,y_i)}{1+d(x_i,y_i)}<\epsilon 2^{-k}$, so $d_k(x_k,y_k)<\frac{2^k}{2^k}\frac{d(x_k,y_k)}{1+d(x_k,y_k)}\leq 2^k\sum_{i=1}^\infty\frac{1}{2^i}\frac{d(x_i,y_i)}{1+d(x_i,y_i)}<\epsilon.$ – Alex S Mar 27 '16 at 14:37
• But $x_k$ is an element of $x$ and $y_k$ is an element of $y$. This does not show that $x$ or $y$ is cauchy in $X_k$...? – Cococabana Apr 2 '16 at 13:07
• I just used $x$ and $y$ as example symbols replace the letters with the ones you need to get the job done. – Alex S Apr 2 '16 at 15:25