Countable product metric ($D(x,y)={\sum }_{i\ge 1}{2}^{-i}\frac{d_i(x_i,y_i)}{1+d_i(x_i,y_i)}$)

The displayed metric on a countable Cartesian product of metric spaces combines the bounded metric transform with summable positive weights. It is finite even when individual factor metrics are unbounded. The triangle inequality follows from subadditivity of $t/(1+t)$ and summation. Zero distance forces equality in each coordinate.

Convergence in this metric is exactly coordinatewise convergence. One term is bounded by the sum, so $D(x^{(n)},x)\to 0$ implies each $d_i(x_i^{(n)},x_i)\to 0$. Conversely, choose a finite head whose omitted weight sum is small; convergence in those finitely many coordinates controls the head uniformly, while boundedness of the transformed distances controls the entire tail. Thus it induces the countable product topology. Arbitrary positive summable weights work by the same proof.

If every factor is a complete metric space, the countable product metric is complete. For a product-Cauchy sequence, the inequality $2^{-i}\varphi (d_i(x_i^{(m)},x_i^{(n)}))\le D(x^{(m)},x^{(n)})$, with $\varphi (t)=t/(1+t)$, makes each coordinate sequence Cauchy. Its coordinate limits define a product point. Coordinatewise convergence and the uniformly small summable tail imply convergence to that point in $D$. It matters that the original distances are bounded before they are summed: summing arbitrary unbounded metrics directly need not be finite.