Showing that a family of metrics induce all the same topology on special sequence space Let $X = \{0,1\}$ and consider the discrete metric 
$$
 d(x,y) := \left\{ \begin{array}{ll} 0 & x = y \\ 1 & x \ne y. \end{array}\right.
$$
Now consider $X^{\mathbb N_0}$, the set of all sequences $(s_k)_{k\in \mathbb N}$ with $s_k \in X$. For $q > 1$ define the metric
$$
 d((s_k), (t_k)) := \sum_{i=0}^{\infty} \frac{d(s_i, t_i)}{q^i}.
$$
Then I want to show that for every $q > 1$, these metrics induce the same topology on $X^{\mathbb N_0}$.
If $q > 2$, then I can show that the open balls have the form
$$
 B_d\left((s_k), \frac{1}{q^n}\right) = \{ (t_k) : t_0 = s_0, t_1 = s_1, \ldots, s_n  = t_n \}
$$
As if for $(t_k)$ we have $t_0 = s_0, \ldots, t_n = s_n$, then
$$
 d((t_k), (s_k)) \le \sum_{i=n+1}^{\infty} \frac{1}{q^i} = \frac{1}{q^n} \sum_{i=1}^{\infty} \frac{1}{q^i} = \frac{1}{q^n}\cdot \frac{1}{q-1} < \frac{1}{q^n}
$$
for $q > 2$, and conversely if $d((s_k), (t_k)) < \frac{1}{q^n}$ then $d(s_0, t_0) = d(s_1, t_1) = \ldots = d(s_{n}, t_{n}) = 0$, or said differently $s_0 = t_0, \ldots, s_n = t_n$. Now as every open ball could be written as an union of balls with radii $1/q^n$ for some $n$, and these are all the same regardless of the $q > 2$, it follows that they all induce the same topology.
But here $q > 2$ is essential in the estimation
$$
 \frac{1}{q^n} \frac{1}{q-1} < \frac{1}{q^n}
$$
and I have no idea how to handle the case $q \in [2, 1)$?
EDIT: Also for $q = 2$ something odd happens in the open balls. For example consider $s_k = 0$ for all $k \ge 0$, then for $(t_k)$ with $t_0 = 0, t_i = 1$ for $i \ge 1$ we have $d((s_k), (t_k)) = \sum_{i=1}^{\infty} 1/2^i = 1$, hence
$$
 (t_k) \notin B_{d}((s_k), 1)
$$
despite $s_0 = t_0$, so we could not write the open balls as above. Instead I guess we have
$$
 B_d((s_k), 1/2^n) = \{ (t_k) : t_0 = s_0,\ldots, t_n = s_n \} \setminus \{ (t_k) : t_i = 1 - s_i, i \ge n+1 \}.
$$

For $n\in\Bbb N$ and $x=\langle x_k:k\in\Bbb N\rangle\in X^{\Bbb N}$ let 
$$B(x,n)=\left\{y\in X^{\Bbb N}:y_k=x_k\text{ for all }k<n\right\}\;;$$
the set of all such $B(x,n)$ is a base for the product topology on $X^{\Bbb N}$.
Suppose that $y\in B(x,n)$ and $q>1$. Then for $z\in B\left(y,q^{-n}\right)$ we must have $z_k=y_k$ for $k\le n$, so $B_q(y,q^{-n})\subseteq B(x,n)$, and $B(x,n)$ is open in the topology of the metric $d_q$.
Now suppose that $\epsilon>0$, and $y\in B_q(x,\epsilon)$, so that
$$\sum_{k\ge 0}\frac{d(x_k,y_k)}{q^k}<\epsilon\;.$$
Choose $n\in\Bbb N$ so that
$$\frac1{q^{n-1}(q-1)}<\epsilon-d_q(x,y)\;.$$
If $z\in B(y,n)$, then 
$$d_q(z,y)\le\sum_{k\ge n}\frac1{q^k}=\frac1{q^{n-1}(q-1)}<\epsilon-d_q(x,y)\;,$$
so $d_q(x,z)\le d_q(x,y)+d_q(y,z)<\epsilon$, and hence $B(y,n)\subseteq B_q(x,\epsilon)$. Thus, $B_q(x,\epsilon)$ is open in the product topology, and it follows that the topology of the metric $d_q$ is the product topology.
