# Sequence of metric spaces is compact iff each metric space is compact

1. Dec 9, 2013

### mahler1

The problem statement, all variables and given/known data.

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.

2. Dec 10, 2013

### pasmith

For any sequence on $X$, there is a subsequence such that its projection onto $X_1$ converges. Take this subsequence. It has a subsequence whose projection onto $X_2$ converges. Take this subsequence. It has a subsequence whose projection onto $X_3$ converges. ...

EDIT: One point which is worth noting is that the metrics $d_k$ and $d_k' = \displaystyle\frac{d_k}{k^2(1 + d_k)}$ give rise to exactly the same topology on $X_k$, so a sequence converges with respect to $d_k$ if and only if it converges with respect to $d_k'$.

Last edited: Dec 10, 2013
3. Dec 10, 2013

### mahler1

This last remark was particularly helpful, it was much more difficult trying to prove $d_k<\epsilon$ without using that fact.