Suppose $p\in [1, \infty]$. Show the relative topology on $l^p(\mathbb{R})$ induced from the product topology on $\mathbb{R}^{\omega}= \Pi_{l=1}^{\infty} \mathbb{R}$ is coarser than the $p$-norm topology on $l^p(\mathbb{R})$, but the relative topology on $l^p(\mathbb{R})$induced by the box topology on $\mathbb{R}^{\omega}$ is finer than the $p$-norm topology on $l^p(\mathbb{R})$.
Attempt:
I know that any basic open set $U$ in the relative product space is an union of the form $\Pi_{i=1}^{\infty} O_i$, where $O_i \in \mathbb{R}$. So it's like a box. And it's clear geometrically that we can always find a $p$-ball $B(x,r)$ such that it's contained in $U$ for every $x\in U$. But I'm not sure how to construct such $r$.
Thanks!
Best Answer
A basic open set containing a point $x_0$ in the product topology is of the form $\{x: |p_1(x)-p_1(x_0)| <r_1, ...., |p_N(x)-p_N(x_0)|<r_N\}$ where $p_1,p_2,..$ are the projection maps. If $r=\min \{r_1,r_2,..,r_N\}$ then the ball of radius $r$ in $\ell^{p}$ around $x_0$ is contained in this basic neighborhood. This proves the first part.
Consider any ball $B(x_0,r)$ in $\ell^{p}$. $\{x: |p_i(x)-p_i(x_0)|<\frac r {2^{i/p}} \forall i\}$ is an open set containing $x_0$ in the Box topology which is contained in $B(x_0,r)$ .