1
$\begingroup$

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 $Y$?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Mees de Vries Mar 8 '17 at 13:40
  • $\begingroup$ $\rho(x,y)$ should have $d_n(x_n, y_n)$ in the nominator too, I think. $\endgroup$ – Henno Brandsma Mar 8 '17 at 18:34
  • $\begingroup$ @HennoBrandsma You are absolutely right! Thanks $\endgroup$ – Eduardo Mar 8 '17 at 19:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.