coarser/finer comparison for relative topology and norm topology 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 strictly 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 strictly finer than the $p$-norm topology on $l^p(\mathbb{R})$.
Attempt:
We know that $l^p(\mathbb{R}) \subseteq \mathbb{R}^{\omega}$.
By definition the relative topology is $S = \{ V \cap l^p(
\mathbb{R}): V \in T \}$, where $T$ is the product topology on $\mathbb{R}^{\omega}$.
To show it's coarser than the $p$-norm topology, we want to show that there exists an open ball $B$ in $l^p(\mathbb{R})$ such that $B \notin S$. But I'm not sure how to find such open ball.
Now let $S' = \{ V \cap l^p(
\mathbb{R}): V \in T \}$, where $T$ is the box topology on $\mathbb{R}^{\omega}$.
I'm not sure what to do to show the $p$-norm topology is strictly coarser than $S'$.
Thanks in advance!
 A: I will show the first part and leave pointers for the second part.
It is easy to see that any basic open set of the relative product topology contains an open $p$-ball from the $p$-norm topology (justify this yourself). But, consider the unit $p$-ball in $\ell^p(\mathbb{R})$: $B_p(0, 1) = \{x \in \ell^p(\mathbb{R}) : \|x\|_p < 1\}$. This is not open in the product topology induced on $\ell^p(\mathbb{R})$. To see this, consider the sequence $(0)_{i = 1}^\infty \in B_p(0, 1)$. And let
$$U = \Big(\prod_{i = 1}^\infty (a_i, b_i)\Big) \cap \ell^p(\mathbb{R}),\ (a_j, b_j) = (-\infty, +\infty)  \text{ for all but finitely many $j \in I_{\text{finite}}$}$$ be any basic open set of the induced product topology containing $(0)_{i = 1}^\infty$. Thus, $a_i < 0 < b_i$. Then $U$ is never contained in $B_p(0, 1)$. Indeed, choose any of the infinitely many $n \in \Bbb Z_+ - I_{\text{finite}}$ such that $(a_n, b_n) = (-\infty, +\infty)$. And consider the sequence $(\delta_{n,i})_{i = 1}^\infty$ (i.e. it is $1$ at index $n$ and $0$ elsewhere). Then $\|(\delta_{n,i})_{i = 1}^\infty\|_p = \sqrt[p]{\sum_{i = 1}^\infty|\delta_{n,i}|^p} = \sqrt[p]{|\delta_{n,n}|^p} = 1$. So $(\delta_{n,i})_{i = 1}^\infty \not\in B_p(0, 1)$. But $(\delta_{n,i})_{i = 1}^\infty \in U$ easily because obviously $(\delta_{n,i})_{i = 1}^\infty$ converges in the $p$-norm and $a_i < 0 < b_i, i \neq n$ while $-\infty = a_n < 1 < b_n = +\infty$.
The situation changes with the induced box topology. Again, it is not hard to see the induced box topology contains the $p$-norm topology (justify this yourself). But, consider the following basic open set from the induced box topology:
$$
U = \Big(\prod_{i = 1}^\infty (-4^{-i/p}, 4^{-i/p})\Big) \cap \ell^p(\mathbb{R})
$$ Again, $(0)_{i = 1}^\infty \in U$ but no open $p$-ball $B_p(a, R), a \in \ell^p(\mathbb{R}), R > 0$ containing $(0)_{i = 1}^\infty$ is in $U$. To see this, first simplify: any such $B_p(a, R)$ also contains another ball centered at $(0)_{i = 1}^\infty$: $B_p(0, r), R > r > 0$ (why?). Now, the sequence $(\frac{1}{2}r\ 2^{-i/p})_{i = 1}^\infty \in B_p(0, r)$ (why does this converge in the $p$-norm? why is it in the ball?). However, $(\frac{1}{2}r\ 2^{-i/p})_{i = 1}^\infty \not\in U$ (find some $j \in \Bbb Z_+$ large enough such that $4^{-j/p} < \frac{1}{2}r\ 2^{-j/p}$; why is this enough?).
