I need a little help with the following problem:
Suppose $\left\{ \left(X_{n},d_{n}\right)\right\} _{n=1}^{\infty}$ is a family of metric spaces such that $d_{n}$ is upper-bounded by $1$ for all $n$. I'm trying to show the metric defined on $\prod X_{n}$ by $$d\left(\left(x_{n}\right)_{n=1}^{\infty},\left(y_{n}\right)_{n=1}^{\infty}\right)={\displaystyle \sum_{n=1}^{\infty}\frac{1}{2^{n}}d\left(x_{n},y_{n}\right)}$$ induces the product topology on $\prod X_{n}$. What I need to in order to finish is to show that given an $\varepsilon-ball$ $B\left(\left(x_{n}\right)_{n=1}^{\infty},\varepsilon\right)$ with the given metric for each point $\left(y_{n}\right)_{n=1}^{\infty}$ in the ball there is a basis element of the product topology $V$ such that: $$\left(y_{n}\right)_{n=1}^{\infty}\in V\subseteq B\left(\left(x_{n}\right)_{n=1}^{\infty},\varepsilon\right)$$ I've messed around trying to define such a basis element but I haven't managed to do that. Help would be appreciated.
Let $n_0$ large enough so that $2^{-n_0} < \epsilon/2$. For $i=1,...,n_0$ let $U_i = B(y_i, \frac{\epsilon}{2 n_0})$ be the open ball around $y_i$ in coordinate space $X_n$ with metric $d_n$. Finally, let $$V = U_1\times\ldots\times U_{n_0}\times X_{n_0+1}\times X_{n_0+2}\times\ldots$$ be a nbhd of $(y_n)_{n=1}^\infty$ in the products space. I claim that $V \subset B((y_n)_{n=1}^\infty, \epsilon)$.
If $(x_n)\in V$, then $d((x_n),(y_n))\le\sum_{i=1}^{n_0} \frac{\epsilon}{2 n_0} + \sum_{n=n_0+1}^{\infty} 2^{-i} \le n_0\frac{\epsilon}{2n_0} + 2^{-n_0} < \epsilon$.