Hi Ilya.

Let $D(x,y) = \sup_{n \in \mathbb N} \frac{d(x_n , y_n)}{1 + d(x_n , y_n)}$ where $d(.,.)$ is the Euclidean metric on $\mathbb R$.

Then $D$ is a metric on $\mathbb R^{\mathbb N}:

$0 \le D(x,y) \le 1$ follows from the definition. So does $D(x,y) = D(y,x)$.

As for the $\Delta$-inequality, we use that $\sup (f(x) + g(x)) \le \sup f(x) + \sup g(x)$:

$\begin{align}D(x,z) &= \sup_{n \in \mathbb N} \frac{d(x_n , z_n)}{1 + d(x_n , z_n)} \le \sup_{n \in \mathbb N} \left ( \frac{d(x_n , y_n)}{1 + d(x_n , y_n)} + \frac{d(y_n , z_n)}{1 + d(y_n , z_n)} \right ) \\ &= \sup_{n \in \mathbb N} (f(y) + g(y)) \le \sup_{n \in \mathbb N} f(y) + \sup_{n \in \mathbb N} g(y) \\ &= \sup_{n \in \mathbb N} \frac{d(x_n , y_n)}{1 + d(x_n , y_n)} + \sup_{n \in \mathbb N} \frac{d(y_n , z_n)}{1 + d(y_n , z_n)} = D(x,y) + D(y,z)\end{align}$.

Then $D$ induces the product topology on $\mathbb R^{\mathbb N}$:

Let $B'_\varepsilon (x) = \{y \mid D(x,y) \le \varepsilon \}$ denote the closed epsilon ball around $x$.

Note that $D(x,y) \le \varepsilon \iff d(x_n, y_n) \le \frac{\varepsilon}{1 - \varepsilon}$ for all $n$.

Hence $B'_{\varepsilon, D} (x) = \prod_{n \in \mathbb N} B'_{\frac{\varepsilon}{1- \varepsilon}, d} (x_n)$.

That is, a closed ball in the $D$ metric corresponds to a product of closed balls in the Euclidean metric $d$.

The topology generated by the closed balls $B'_D$ is the topology generated by $\{\prod_n B'_{\varepsilon, d}(x_n) \}$ and the topology generated by these products of closed balls is the same as the topology generated by the products of closed sets (I did not prove this, one will have to prove it too). But this is the product topology since it is the coarsest topology such that the projections $\pi_i$ are continuous.

wait-wait-wait

I'm not sure, but do the closed balls give the complete characterization of the product topology?

And for example, both the product topology and the D-topology on $\Bbb R^\Bbb N$ are metrizable and hence sequential

now, the sequence $(1,1,1,\dots)$, $(0,1,1,\dots)$, $(0,0,1,\dots)$ converges to the zero vector in the product topology but not in D, thus they are different

so don't look for the proof that they are the same :)

and the box topology is not metrizable

so D-topology is neither of two

however, it induces the same Borel $\sigma$-algebra as the product topology does - I don't know how it can help :)

Thanks for the example sequence. I will have to think more. What I wrote above is stupidity.

But this is fun : )

BBL