For $l^2 (\mathbb{N})$ is $(l^2, d_2)= (l^2, d)$ topologically, where $d$ forms the usual product topology on $\mathbb{R}^{\mathbb{N}}$? If we define $d(x,y)= \max_{n \in \mathbb{N}}( \min \{2^{-n}, |x_n - y_n| \})$ then does this distance form the same topology on $l^2 ( \mathbb{N})$ (the set of all square-summable real sequences) as the distance $d_2(x,y)=\sqrt{\sum |x_i-y_i|^2}$? 
 A: No, certainly not. Convergence in $\mathbb {R}^{\mathbb {N}}$ is equivalent to convergence is each slot. Thus $e_n$ converges to the $0$ sequence in $\mathbb {R}^{\mathbb {N}}$ (here $e_n$ is the sequence with $1$ in the $n$th slot, $0$'s elsewhere), while $e_n$ does not converge in $l^2(\mathbb N ).$
A: Claim 1: The topology induced by $d_2$ is finer than the topology induced by $d$.
Proof of Claim 1: It is enough to show that for every $\epsilon > 0$ we can find $\delta$ such that $$B_{d_2}(0,\delta) \subset B_{d}(0,\epsilon).$$
This turns out to be easy: let $\delta := \frac{\epsilon}{2}.$ Then if $x \in B_{d_2}(0,\delta)$, we have $$\sum_{n = 1}^{\infty}x_n^2 < \delta^2.$$
Since all the terms in the sum are positive, it follows that for every $n$, $x_n^2 < \delta^2$, or equivalently $|x_n| < \delta.$ To conclude notice that $$\max_n\{\min\{2^{-n}, |x_n|\}\} \le \max_n\{|x_n|\} \le \max_n\{\delta\} = \delta < \epsilon.$$
This shows that $x \in B_{d}(0,\epsilon).$
Claim 2: The other inclusion is false.
Proof of Claim 2: It is enough to show we cannot find $\delta$ such that $$B_{d}(0,\delta) \subset B_{d_2}(0,1).$$ 
Let $N$ be such that $2^{-N} \le \delta$ and let $$x = (\underbrace{0,\dots,0}_{N},1,0,0,\dots) = e_{N+1}.$$ Then it is easy to check that $x \in B_{d}(0,\delta)$ but $x \notin B_{d_2}(0,1).$
