Metric Spaces, The Product of Metric Spaces

The Product of Metric Spaces

Given two metric spaces, let the set S be the direct product of the two spaces, i.e. points in the first cross points in the second. Given two points in the product space, compute their distance as follows. Take the distance between the corresponding points in the first metric space and square it, then add the square of the distance between the corresponding points in the second metric space, then take the square root.

The obvious example is the xy plane. The x and y axes are metric spaces, and their product gives the xy plane, as a set, with a metric that is the same as euclidean distance. The distance = the square root of the change in x squared + the change in y squared. If we take the product of a dozen axes, we have 12 dimensional space, with the usual distance metric. If we take the product of various dimensional spaces, some lines, some planes, etc, the result is n space, with the usual distance metric, where n is the sum of all the dimensions.

There are still a couple of issues to deal with. Is the new distance function a valid metric? Realize that the distance function is the same, whether we take a finite product of metric spaces all at once, or whether we build the product sequentially, bringing in one space at a time. Either way you get the square root of the sum of the squares of the individual distances. Therefore we only need prove our results for the product of two spaces, and it will generalize to all finite products.

The only tricky part is the triangular inequality. Assume a+b ≥ c, three distances in the first space that obey the triangular inequality. Similarly, x+y ≥ z in the second space. We would like to prove the following.

sqrt(a²+x²) + sqrt(b²+y²) ≥ sqrt(c²+z²)

The right side can only increase if we replace c with a+b, and z with x+y. Let's prove this more ambitious inequality. Square both sides; this preserves the direction of inequality. The terms a², b², x², and y² cancel out, leaving the following.

sqrt(a²b² + a²y² + x²b² + x²y²) ≥ ab + xy

Square again to get this.

a²y² + b²x² ≥ 2abxy

Move everything to the left and we have a perfect square ≥ 0, which is always true. The distance function is a metric, and the product is a metric space.

But there's another problem; we've already defined the topology for a product space. Open sets in the product are the product of open sets in the individual spaces. Does this agree with the topology induced by our metric? It does. Let's illustrate with the xy plane. This is the product of the x and y axes, and the standard product topology has a base of open rectangles, all sizes, all locations. This is what you get when you cross x intervals with y intervals, the base sets for the x and y axes. When we take the product of the x and y axes as metric spaces, the base becomes open disks, all sizes, all locations. In the last section we proved these two topologies are the same. Disks can be covered with rectangles and vice versa. This is harder to visualize when the product is not as simple as the xy plane, but the idea is the same. I'll leave the algebra to you.