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The problem statement

Let $(X_n,d_n)_{n \in \mathbb N}$ be a sequence of metric spaces. Consider the product space $X=\prod_{n \in \mathbb N} X_n$ with the distance $d((x_n),(y_n))=\sum_{n \in \mathbb N} \dfrac{d_n(x_n,y_n)}{n^2[1+d_n(x_n,y_n)]}$

$a)$ Prove that $(X,d)$ is complete if and only if each $(X_n,d_n)$ is complete.

$b)$ Prove that $(X,d)$ is compact if and only if each $(X_n,d_n)$ is compact.

My attempt at a solution:

For $a)$, after Willie's answer I could do the following:

$\implies$ Let $n$ fixed, call it $n=n_0$ and let $\{y^{j}\}_{j \in \mathbb N}$ be a Cauchy sequence in $(X_{n_0},d_{n_0})$. I fix an arbitrary $x_i \in (X_i,d_i)$ for $i \neq n_0$ and now I consider the sequence $\{\vec{x}^{j}\}_{j \in \mathbb N}$ defined as $\vec{x}^{(j)} = (x_1, x_2, \ldots, x_{n_0-1}, y^{(j)}, x_{n_0+1} , \ldots) \in X$.

Lets prove $\{\vec{x}^{j}\}_{j \in \mathbb N}$ is a Cauchy sequence: let $\epsilon>0$, by hypothesis, there is $N \in \mathbb n$ : $\space \forall \space m,n \geq N$, $d_{n_0}(y^n,y^m)< \epsilon$. For $k \neq n_0$, $d_k(x_{k}^m,x_{k}^n)=d_k(x_k,x_k)=0$.

Then, $d(\vec{x}^{m},\vec{x}^{n})=\sum_{k \in \mathbb N} \dfrac{d_k(x_{k}^m,x_{k}^n)}{k^2[1+d_k(x_{k}^m,x_{k}^n)]}=\dfrac {d_{n_0}(x_{n_o}^m,x_{n_0}^n)}{{n_0}^2[1+d_{n_0}(x_{n_o}^m,x_{n_0}^n)]}\leq d_{n_0}(y^m,y^n)<\epsilon \space \forall n,m \geq N$.

This proves $\{\vec{x}^{j}\}_{j \in \mathbb N}$ is a Cauchy sequence in $X$. which means $\vec{x}^{j} \to \vec{a}^{\infty}$, $\vec{a}^{\infty}=(a_1,a_2,...,,a_{n_0-1},a_{n_0},a_{n_0+1},...)$. Lets prove that $y^j \to a_{n_0}$ in $(X_{n_0},d_{n_0})$. Given $0<\epsilon<1$, there is $N \in \mathbb N$ : $\space \forall \space n\geq N$, $d(\vec{x}^n,\vec{a}^{\infty})<\dfrac{\epsilon}{n_{0}^2}$.

So, $\dfrac{d_{n_0}(y^j,a_{n_0})}{n_{0}^2[1+d_{n_0}(y^j,a_{n_0})]}\leq \sum_{k \in \mathbb N} \dfrac{d_k(x_{k}^n,a_k)}{k^2[1+d_k(x_{k}^n,a_k)]}=d(\vec{x}^n,\vec{a})< \dfrac{\epsilon}{n_{0}^2}$ for $n \geq N$.

From here, it follows that $d_{n_0}(y^j,a_{n_0})<\dfrac{\epsilon}{1-\epsilon}<\epsilon$ for all $n\geq N$. We've proved that $\{y^{j}\}_{j \in \mathbb N}$ is a convergent sequence in $(X_{n_0},d_{n_0})$, since the Cauchy sequence and $n=n_0$ were arbitrary, one can conclude that for every $n \in \mathbb N$, $(X_n,d_n)$ is complete.

Now it remains to prove the other implication (I had problems with this one):

Let $\{\vec{x}^n\}_{n \in \mathbb N}$ be a Cauchy sequence in $(X,d)$. For a fixed $n=n_0$, lets prove that $\{x_{n_0}^{n}\}_{n \in \mathbb N}$ is a Cauchy sequence in $(X_{n_0},d_{n_0})$. Let $0<\epsilon<1$, by hypothesis, there is $N \in \mathbb N: n,m\geq N \implies d(\vec{x}^n,\vec{x}^m)<\dfrac{\epsilon}{{n_0}^2}$.

But then, $\dfrac{d_{n_0}(x_{n_0}^n,x_{n_0}^m)}{n_{0}^2[1+d_{n_0}((x_{n_0}^n,x_{n_0}^m)]}\leq \sum_{k \in \mathbb N} \dfrac{d_k(x_{k}^n,x_{k}^m)}{k^2[1+d_k((x_{k}^n,x_{k}^m)]}=d(\vec{x}^n,\vec{x}^m)<\dfrac{\epsilon}{{n_0}^2}$.

We have that $d_{n_0}(x_{n_0}^n,x_{n_0}^m)<\dfrac{\epsilon}{1-\epsilon}<\epsilon$. So the sequence $\{x_{n_0}^n\}_{n \in \mathbb N}$ in $(X_{n_0},d_{n_0})$, which means there is $y^{n_0}=\lim_{n \to \infty} x_{n_0}^n$.

If we call $\vec{y}^{\infty}=(y_1,y_2,...,y_n,...)$, lets show that $\vec{y}^{\infty}=\lim_{n \to \infty}\vec{x}^n$.

Here I got stuck, given $\epsilon>0$, I don't know which $N$ to choose such that for $n\geq N$, $d(\vec{x}^n,\vec{y}^{\infty})<\epsilon$, for every term of the sequence $\{\vec{x}^n\}_{n \in \mathbb N}$, I will have a different $N_k$ that will depende on the sequence of the space $(X_k,d_k)$.

A secondt attempt for this last part:

Note that $0\leq d(\vec{x}^n,\vec{y}^{\infty})=\sum_{k \in \mathbb N} \dfrac{d_k(x_{k}^n,y_k)}{k^2[1+d_k(x_{k}^n,y_k)]}\leq \sum_{k \in \mathbb N} \dfrac{1}{k^2}$. Then, $\sum_{k \in \mathbb N} \dfrac{d_k(x_{k}^n,y_k)}{k^2[1+d_k(x_{k}^n,y_k)]}$ is convergent, which means that given $\epsilon>0$, there is $N \in \mathbb N$: $\sum_{N+1}^{\infty} \dfrac{d_k(x_{k}^n,y_k)}{k^2[1+d_k(x_{k}^n,y_k)]}<\dfrac{\epsilon}{2}$.

For a fixed $k$, $\{x_{k}^n\}_{n \in \mathbb N}$ converges to $y_k$ in $(X_k,d_k)$, so there is $n_k : \space \forall \space n_k\geq n$, $d_k(x_{k}^n,y_k)<\dfrac{\epsilon}{2N}$. Consider $M=\max\{n_1,n_2,...,n_N\}$,then

$\sum_{k=1}^N \dfrac{d_k(x_{k}^n,y_k)}{k^2[1+d_k(x_{k}^n,y_k)]}\leq \sum_{k=1}^N d_k(x_{k}^n,y_k)<N\dfrac{\epsilon}{2N}=\dfrac{\epsilon}{2}$.

So we have that for $n\geq M$,

$d(\vec{x}^n,\vec{y}^{\infty})=\sum_{k=1}^N \dfrac{d_k(x_{k}^n,y_k)}{k^2[1+d_k(x_{k}^n,y_k)]}+\sum_{k=N+1}^{\infty} \dfrac{d_k(x_{k}^n,y_k)}{k^2[1+d_k(x_{k}^n,y_k)]}<\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon$

This proves that $\vec{x}^n \to \vec{y}^{\infty}$; since the sequence $\{\vec{x}^n\}_{n \in \mathbb N}$ was an arbitrary Cauchy sequence in $(X,d)$, then $(X,d)$ is complete.

Sorry if my notation is confusing, I did the best I could but with so many sequences and indexes I easily get lost.

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  • $\begingroup$ I removed the (completeness) tag because it is vague. See this meta discussion. $\endgroup$ – Willie Wong Dec 9 '13 at 10:23
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    $\begingroup$ On a quick read through, your second attempt looks fine. Good job! $\endgroup$ – Willie Wong Dec 10 '13 at 8:39
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I think your problem is mostly that you don't understand what the product space is. The product space, by definition, have as its elements $\vec{x} \in X$ vectors of the form $$ \vec{x} = (x_1,x_2, \ldots, x_n, \ldots) $$ where $x_i \in X_i$. The metric that is chosen on $X$ is given in your question.

So a sequence $\vec{x}^{(n)}$ in $X$ can be interpreted as a double sequence $(x_i^{(j)})$ where both $i,j\in \mathbb{N}$: but note that $i$ indexes the "coordinate" while $j$ indexes the "term".

Let me sketch to you how I would prove (a, $\implies$), and maybe you can figure out how to do the other problems.

Let $y^{(j)}$ be a sequence in $X_n$, for some $n$ fixed. Assume that $y^{(j)}$ is a Cauchy sequence. Now fix an arbitrary $x_i \in X_i$ for $i\neq n$. Consider the sequence given by

$$ \vec{x}^{(j)} = (x_1, x_2, \ldots, x_{n-1}, y^{(j)}, x_{n+1} , \ldots) \in X $$

  1. First you show that $\vec{x}^{(j)}$ is Cauchy.
  2. This implies that it converges to $\vec{x}^{(\infty)}$.
  3. Show that the $n$th coordinate of $\vec{x}^{(\infty)}$, $x_n^{(\infty)}$, is in fact the limit of $y^{(j)}$ as $j\to \infty$.
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  • $\begingroup$ Thanks, now I understand much better, I think I could prove the forward implication in $a)$ but I had some problems with the other implication, I'll write it now in my original post. $\endgroup$ – user100106 Dec 9 '13 at 18:24

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