How can the little l2 topology be finer than the uniform topology? How can the $\ell_2$ topology be finer than the uniform topology on the set $X$ of square-summable sequences?
If the  $\ell_2$ metric is always greater than or equal to the uniform metric. Wouldn't this mean that the epsilon balls in  $\ell_2$ are always bigger than the epsilon balls under the uniform metric? I'm trying to show that the  $\ell_2$ balls are contained inside the uniform balls, right?
 A: This is a misleading concept. Consider, for example, the real line. If you define the $\epsilon$-ball to be $B(x,\epsilon)=\{(y\space s.t.\space 2|y-x|<\epsilon\}$ this would make balls "2 times smaller", yet the topology is the same.
Remember, that even if for a fixed $\epsilon$ $d_1$-$\epsilon$-ball is smaller than the $d_2$-$\epsilon$-ball, there may still be a smaller $\delta$ such that $d_2$-$\delta$-ball is smaller than $d_1$-$\epsilon$-ball.
Now hint: to show that $\ell_2$ is finer than the uniform but coarser than the box topology consider $M(x)=\{y:|x_i-y_i|^2<\epsilon^2/2^i\}$ and show that $M(x)\subset B_{\ell_2}(x,\epsilon)\subset B_{\rho}(x,\epsilon)$ where $\rho$ is uniform metric.
A: It suffices to show that


*

*Every open uniform ball contains an $\ell_2$ ball.

*There exists an open uniform ball that is not contained in any $\ell_2$ balls.


By knowing that the $\ell_2$ metric is greater than or equal to the uniform metric, you only prove 1.
To prove 2,
suppose $B$ is an open ball of radius $\sqrt{r/2}$ with respect to the uniform metric centered at $0$. $B$ consists  of sequences $(x_1, x_2, \ldots)$ such that $\sum_{n=1}^\infty x_n^2 < \infty$ and $|x_n| < r$. It is easy to see that $\sum_{n=1}^\infty x_n^2$ is not bounded even when $x_n$ are all bounded by $r$, so $B$ is not contained in any $\ell_2$ balls. (More rigorously, for any given $L > 0$, there exists an integer $m$ such that $mr/2 > L$. The sequence defined by $x_n = \sqrt{r/2}$ for $n \le m$ and $x_n = 0$ for $n > m$ is a member of $B$, but not a member of an $\ell_2$ ball of radius $L$ because $\sum_{n=1}^\infty x_n^2 = \sum_{n=1}^m (r/2) = mr/2 > L$. Therefore, $B$ is not contained in $\ell_2$ balls of any radii.)
