Let $X$ be the subset of $\Bbb{R}^\omega$ all of sequences $y = (y_i)$ for which $\sum y_i^2 < \infty$. Then $d(x,y) = \left[ \sum_{i=1}^\infty (x_i-y_i)^2 \right]^{1/2}$ defines a metric on $X$, and the topology induced by it is called the $\ell^2$ topology. Let $\Bbb{R}^\infty$ be the set of all sequences which are eventually zero, which is obviously contained in $X$. Show that the for topologies $\Bbb{R}^\infty$ inherits as a subspace of $X$ are all distinct.
I am working on showing that the last two topologies, the box and $\ell^2$ topologies, are distinct. Here is my scratch work. By way of contradiction, suppose that they aren't distinct; and let $0$ denote the zero sequence. Then $\prod (-\frac{1}{i},\frac{1}{i})$ is some open nhbd of $0$ that's open in the box topology, which means that there exists a $\delta > 0$ such that $B(0,\delta) \subseteq \prod (-\frac{1}{i},\frac{1}{i})$. First, imagine that all of these open sets are intersected with $\Bbb{R}^\infty$ (I don't feel like typing it). So, obviously, the goal is to construct a point $y=(y_i)$ that's in $B(0, \delta)$ but not in $\prod (-\frac{1}{i},\frac{1}{i})$, or that $\sqrt{\sum y_i^2} < \delta$ but $|y_n| \ge \frac{1}{n}$ for some $n \in \Bbb{N}$. The sum will be finite, since $y \in \Bbb{R}^\infty$ (i.e., infinitely many terms are zero). For simplicity, let the first $n^2$ terms be nonzero, where $n$ is to be determined. Again, for simplicity, let these $n^2$ nonzero terms be the same, which is to be determined. Then $\sqrt{\sum y_i^2} < \delta$ becomes $ny_1 < \delta$, where all the nonzero terms equal $y_1$, which is to be determined.
I figured choosing $n$ and $y$ to satisfy the inequality would work, but it isn't clear that it will. I thought that by choosing the same number $y_1$, it would "eventually" exceed $\frac{1}{i}$. But it isn't clear that this will happen, since the longer I make that sum, the smaller $y$ will have to be, which means that it "takes" longer to exceed $\frac{1}{i}$. But if it takes "too long," I might run out of nonzero terms and so I wouldn't in fact have a point $y$ that isn't in $\prod (-\frac{1}{i},\frac{1}{i})$.
I could use help constructing a point $y$.
Your idea is correct. Just pick one index $i$ such that $1/i < \delta$ and define $y = \frac{1}{i}e_i$ , meaning the sequence that is 0 for all indices except $i$, and is equal to $1/i$ at $i$. Then $y\in B(0, \delta)$ but $y$ is not in the product of the intervals $(-1/i, 1/i)$.
Intuitively, the box $\prod (-\frac{1}{i},\frac{1}{i})$ gets thinner and thinner as the coordinate index increases, while a ball $B(0, \delta)$ is of the same size in each direction. So if we just look at some direction with large coordinate index, it will be clear than the ball does not fit inside the box.