# Restricted product metric.

Some Facts

Let $\{(X_n,d_n)\}$ be a sequence of metric spaces. If $\sum diam(X_i)<\infty,$ then the metric $\gamma$ given by $$\gamma(x,y)=\sum \dfrac{1}{2^n}d_n(x_n,y_n)$$ provides a metric to the product space $\prod X_n$ which is equivalent to the product topology.

On the other hand if the metrics spaces $X_n$ are not bounded we just consider the metric $$\rho(x,y)=\sum \dfrac{1}{2^n}\dfrac{d_n(x_n,y_n)}{1+d_n(x_n,y_n)}$$ which is also equivalent to the product topology of $X$.

Question

Consider the family of metric spaces $(X_n,d_n)=(\mathbb{R}, |\cdot|)$ and $(Y_n,d_n)=([-k,k],|\cdot|)$ where $|\cdot|$ denote the absolute value. We will consider in $X=\prod_{n} \mathbb{R}$ the metric given by product metric given above by $\rho$, and we will consider in $Y=\prod_{n}[-k,k]$ the metric given above by $\gamma$.

My question is the following: Is the metric $\gamma$ generate the same topology as the metric $\overline{\rho},$ which consists of the restriction of $\rho$ to the subspace $Y$?

• Perhaps I am totally mistaken, but that metric does not seem to give the product topology; it seems to give something more akin to the box topology. – Mees de Vries Mar 8 '17 at 13:40
• $\rho(x,y)$ should have $d_n(x_n, y_n)$ in the nominator too, I think. – Henno Brandsma Mar 8 '17 at 18:34
• @HennoBrandsma You are absolutely right! Thanks – Eduardo Mar 8 '17 at 19:51