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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)_{n \in \mathbb N},(y_n)_{n \in \mathbb N})=\sum_{n \in \mathbb N} \dfrac{d_n(x_n,y_n)}{n^2[1+d_n(x_n,y_n)]}##.

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

The attempt at a solution.

I didn't have problems to prove the forward implication. Let ##n \in \mathbb N## fixed, call it ##n=n_0## and let

Now let ##x_i## an arbitrary element in the space ##(X_i,d_i)## for ##i \neq n_0## and define the sequence ##\{\vec{x}^n\}_{n \in \mathbb N}## in ##(X,d)## as follows:

##\vec{x}^j=(x_1,x_2,...,x_{n_0-1},y^k,x_{n_0+1},...)##. By hypothesis, there is ##\{\vec{x}^{n_k}\}_{k \in \mathbb N}## convergent subsequence of ##\{\vec{x}^n\}_{n \in \mathbb N}##. Let ##\{a^{\infty}\}=(a_1,a_2,...,a_{n_0-1},a_{n_0},a_{n_0+1},...,)## be the limit of ##\{\vec{x}^{n_k}\}_{k \in \mathbb N}##. Lets prove that ##\lim_{k \to \infty} y^{n_k}=a_{n_0}##

Let ##0<\epsilon<1##, then, there is ##N: \space \forall k\geq N##, ##d(\vec{x}^{n_k},\vec{a}^{\infty})<\dfrac{\epsilon}{{n_0}^2}##.

But then, ##\dfrac{d_{n_0}(y^{n_k},a_{n_0})}{{n_0}^2[1+d_{n_0}(y^{n_k},a_{n_0})]}=\dfrac{d_{n_0}({x_{n_0}}^{n_k},a_{n_0})}{{n_0}^2[1+d_{n_0}({x_{n_0}}^{n_k},a_{n_0})]}\leq \sum_{j \in \mathbb N} \dfrac{d_j({x_j}^{n_k},a_j)}{j^2[1+d_j({x_j}^{n_k},a_j)]}=d(\vec{x}^{n_k},\vec{a}^{\infty})<\dfrac{\epsilon}{{n_0}^2}## for all ##k\geq N##.

So, ##d_{n_0}(y^{n_k},a_{n_0})<\dfrac{\epsilon}{1-\epsilon}<\epsilon## for all ##k\geq N##. This shows ##y^{n_k} \to a_{n_0}##, so the sequence ##\{y^j\}_{j \in \mathbb N}## has a convergent subsequence in ##(X_{n_0},d_{n_0})##. As ##\{y^j\}_{j \in \mathbb N}## was an arbitrary sequence, the metric space ##(X_{n_0},d_{n_0})## is compact. Since ##n_0## was an arbitrary natural number, then for each ##n \in \mathbb N##, the space ##(X_n,d_n)## is compact.

I have problems to prove the other implication. I want to show that if each ##(X_n,d_n)## is compact, then the space ##(X,d)## has to be compact. So, let ##\{\vec{x}^n\}_{n \in \mathbb N}## be a sequence in ##(X,d)## with ##\vec{x}^n=(x_{1}^n,x_{2}^n,...,x_{k}^n,...)##. Then ##\{x_{k}^n\}_{n \in \mathbb N}## is a sequence in the space ##(X_k,d_k)##. For each ##k##, I know there is a convergent subsequence Then ##\{x_{k}^{n_j}\}_{j \in \mathbb N}##. The problem is I have an infinite number of indexes ##j## dancing around, for each space ##(X_k,d_k)## I have a convergent subsequence, how can I combine the indexes of each space to choose the appropiate convergent subsequence of ##\{\vec{x}^n\}_{n \in \mathbb N}## in the space ##(X,d)##?

I hope my question and notation are clear. For me, one of the difficulties of this exercise is that one can easily be confused by all the sequences and indexes playing here.