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.

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

The key idea is that tails of the sequences have increasingly negligible effect on the distance between two points of the product. You can bound the effect of the tail by starting it far enough out, and the rest behaves like a finite product.

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

$$\begin{align*} d(\xi_n,\xi)&=\sum_{k\in\Bbb N}\frac1{2^k}d_k\left(x_k^{(n)},x_k\right)\\ &=\sum_{k<m}\frac1{2^k}d_k\left(x_k^{(n)},x_k\right)+\sum_{k\ge m}\frac1{2^k}d_k\left(x_k^{(n)},x_k\right)\\ &<m\cdot\frac{\epsilon}{2m}+\sum_{k\ge m}\frac1{2^k}\\ &<\frac{\epsilon}2+\frac{\epsilon}2\\ &=\epsilon\;. \end{align*}$$

  • $\begingroup$ 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 :) $\endgroup$ – Ivo Terek Jun 14 '15 at 2:53
  • $\begingroup$ @Ivo: You’re welcome! $\endgroup$ – Brian M. Scott Jun 14 '15 at 2:56

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