Infinite Cartesian Product, Metric Triangle Inequality

304 Views Asked by At

For $\{X_j : j\in \mathbb{Z}^+\},$ each compact metric spaces, the infinite Cartesian product metric space is defined as $$X = \prod_{j=1}^{\infty} X_j$$

We make X a metric space by setting $$d(x,y) = \sum_{j=1}^{\infty} 2^{-j} \frac{d_j(p_j(x),p_j(y))}{1+d_j(p_j(x),p_j(y))}$$

where $p_j(x) = x_j$, the j'th component of x, and $d_j$ is just the standard distance function for the corresponding component in x and y.

I am trying to demonstrate that the triangle inequality holds here, namely that $$d(x,y) \leq d(x,z) + d(y,z)$$

I'm given the hint to note that $\phi(r) = \frac{r}{1+r}$, but I'm unsure of how to put it to use.