# Is this (countable) product space complete?

Let $((X_n, {\rm d}_n))_{n \geq 0}$ a sequence of complete metric spaces. Suppose that all the metrics are bounded by $1$. Consider $X = \prod_{n \geq 0}X_n$ with the metric given by: $${\rm d}((x_n)_{n \geq 0},(y_n)_{n \geq 0}) = \sum_{n \geq 0}\frac{1}{2^n}{\rm d}_n(x_n,y_n).$$

I'm trying to prove that $(X, {\rm d})$ is complete. I am not sure that it is, though. Let $$(\xi_n)_{n \geq 0} = ((x_k^{(n)})_{k \geq 0})_{n \geq 0}$$ be a ${\rm d}$-Cauchy sequence in $X$. Fix $k$. Let $\epsilon > 0$. There exists $n_0 \in \mathbb{N}$ such that for $m,n > n_0$, we have that: $$\sum_{r \geq 0}\frac{1}{2^r}{\rm d}_r(x_r^{(n)},x_r^{(m)}) < \frac{\epsilon}{2^k}.$$With this: $$\frac{1}{2^k}{\rm d}_k(x_k^{(n)},x_k^{(m)}) \leq \sum_{r \geq 0}\frac{1}{2^r}{\rm d}_r(x_r^{(n)},x_r^{(m)}) < \frac{\epsilon}{2^k} \implies {\rm d}_k(x_k^{(n)},x_k^{(m)})< \epsilon,$$ and so $(x_k^{(n)})_{n \geq 0}$ is a ${\rm d}_k$-Cauchy sequence in $(X_k,{\rm d}_k)$, and so converges: $x_k^{(n)}\stackrel{n \to +\infty}{\longrightarrow_{{\rm d}_k}} x_k$.

Call $\xi = (x_k)_{k \geq 0}$, naturally. I want to prove now that $\xi_n \stackrel{n \to +\infty}{\longrightarrow_{\rm d}} \xi$, but I'm unsure of how to bound each ${\rm d}_k(x_k^{(n)},x_k)$. And even if I could, I would have a special $n_0^{(k)}$ for each sequence, and nothing ensures that $\sup_{k \geq 0}n_0^{(k)} < +\infty$. Please help.

Note: I looked around a bit in other questions here, such as this one and the related ones, but it wasn't helpful to me.

With the comments, we can pick $k_0 \in \mathbb{N}$ such that $\frac{1}{2^k} < \epsilon$ for all $k \geq k_0$. Then we can write: $$\sum_{r \geq 0}\frac{1}{2^r}{\rm d}_r(x_r^{(n)},x_r) = \sum_{r = 0}^{k_0}\frac{1}{2^r}\,{\rm d}_r(x_r^{(n)},x_k)+\sum_{r \geq k_0+1}\frac{1}{2^r}\,{\rm d}_r(x_r^{(n)},x_k),$$ and we can bound the second term by $\epsilon \sum_{r \geq k_0+1}{\rm d}_r(x_r^{(n)},x_r)$, but this doesn't give anything better. I still need help.

• You don't need to consider every $k$. Since your metric is scaled component wise, for every $\epsilon$, you just have to pick $n$ for the first $k$ coordinates for which $1/2^k < \epsilon$ – TYS Jun 14 '15 at 1:51
• Makes a bit of sense. Let me try to digest that. – Ivo Terek Jun 14 '15 at 1:52
• For $k$ big enough, we have $1/2^k < \epsilon$. Maybe you meant for the $k$ first coordinates for which $1/2^k \geq \epsilon$? – Ivo Terek Jun 14 '15 at 1:54
• Yeah that's what i meant, sorry for being unclear – TYS Jun 14 '15 at 1:54
• Don't worry, it was helpful anyway. Let's see if anything comes up here, thanks. – Ivo Terek Jun 14 '15 at 1:56

Fix $\epsilon>0$. There is an $m\in\Bbb N$ such that $\sum_{n\ge m}\frac1{2^n}<\frac{\epsilon}2$. For $k<m$ you have $n_0^{(k)}\in\Bbb N$ such that $d_k\left(x_k^{(n)},x_k\right)<\frac{\epsilon}{2m}$ whenever $n\ge n_0^{(k)}$. Let $n_0=\max_{k<m}n_0^{(k)}$; then
\begin{align*} d(\xi_n,\xi)&=\sum_{k\in\Bbb N}\frac1{2^k}d_k\left(x_k^{(n)},x_k\right)\\ &=\sum_{k<m}\frac1{2^k}d_k\left(x_k^{(n)},x_k\right)+\sum_{k\ge m}\frac1{2^k}d_k\left(x_k^{(n)},x_k\right)\\ &<m\cdot\frac{\epsilon}{2m}+\sum_{k\ge m}\frac1{2^k}\\ &<\frac{\epsilon}2+\frac{\epsilon}2\\ &=\epsilon\;. \end{align*}
• Thank you, that was very clear. It didn't even cross my mind that the index $m$ would appear in one of the bounds - in my head, these positive integers were being something separate, and we would choose the maximum of them just to make sure every bound would hold at the same time. But now I see. You saved the day again, well earned +1 and accept :) – Ivo Terek Jun 14 '15 at 2:53