# Prove that there is a metric on the product space $\mathbb R^{\mathbb N}$ relative to which $\mathbb R^{\mathbb N}$ is complete.

Prove that there is a metric on the product space $\mathbb R^{\mathbb N}$ relative to which $\mathbb R^{\mathbb N}$ is complete.

A hint has been given to take the metric D(x,y)=

$=\sup \{\dfrac{\overline {d}(x_i,y_i)}{i}\}$.where $\overline {d(x,y)}=\min\{1,d(x,y)\}$.

I have to show that any basic open set is of the form $U_1\times U_2...\times U_n\times \mathbb R\times \mathbb R\times \mathbb R \rightarrow (1)$.

Any open ball with respect to the given metric is of the form B(x$,\epsilon)=\{$y$:D($x,y$)<\epsilon\}$ which reduces to of the form $\{y_i:d(y_i,x_i)<i\epsilon\}$ i.e of the form $B(y_i,i\epsilon)$ for each $i\in \mathbb N$.

But how to show that it is of the form $1$.

Any help on how to proceed

• What happens when $i$ is so large that $i\epsilon > 1$? – Arthur Sep 29 '15 at 7:37
• Presumably you want your metric to have some other nice properties as well; otherwise, take the discrete metric. :P – Noah Schweber Sep 29 '15 at 7:55
• A cauchy sequence in $R^N$ will be also pointwise cauchy sequence and therefore pointwise convergent. The pointwise limit is a good candidate for the limit in $R^N$ metric. To prove this it's useful to note that $\overline{d(x_i, y_i)}/i \le i$ and you only have to consider $1/\epsilon$ elements when applying the definition for convergence. – skyking Sep 29 '15 at 9:18
• @Noah: The phraseology metric on the product space makes it pretty clear that the metric is to generate the product topology. – Brian M. Scott Sep 29 '15 at 16:03
• @BrianM.Scott;Will you please help me out how to solve the above one .Really got stuck on it – Learnmore Sep 29 '15 at 16:08