In the last section we proved that any metric space can be given a new metric, that preserves the topology, and yields a bounded metric space with a diameter no larger than 1. Do this for each metric space Mi. Since the topologies have not changed, the topology of P has not changed either.
Given two points in P, project them into each Mi, find the distance within each Mi, multiply by ½i, and add up the results. Distance in each component counts half as much as distance in the previous component. The maximum distance between two points is ½ + ¼ + … = 1.
The triangular inequality holds in each component space Mi, hence it holds across each partial sum, and finally, it holds for the infinite sum. Verify the other properties, (straightforward), and we have a valid metric.
If the topologies are the same, base open sets under our metric must cover each base open set in the product, and conversely. Let's begin with a base open set in P, as defined by the product topology. Some of the components are restricted to open balls; most are left unconstrained. Let z be a point in this set. Let d be the shortest distance that zi might stray, before running into the edge of the open ball in Mi. If all the components of z are restricted to a ball of radius d, we're ok. Let Mn be the last constrained component; everything after that is unconstrained. Let c = d/2n. Using our new metric, build an open ball about z with radius c. If the nth component strays by more than d, then we have already move through a distance of c, regardless of the other components. This lies outside the ball of radius c. Thus the nth component cannot stray by more than d, and neither can the earlier components. The later components don't matter, so we have found an open ball covering the arbitrary point z, and the metric topology implies the product topology.
Now for the converse. Let z be a point inside an open ball, using our new metric. Let z be a distance d from the edge of the open ball. Find n such that ½n is less than half of d. If the components beyond n are allowed to run freely, the result will stray from z by a distance that does not exceed half of d. For the components 1 through n, restrict Mi to a ball about zi with radius ½d. The sum of these deviations is at most ½d. Put this all together, and the points in a base open set, according to the product topology, are never farther than d away from z, and are wholly contained in our open ball.
Each topology implies the other, and P is metrizable.