Metric Spaces, Distance, Isometry, Equivalent

Distance to a Set

Let S be a metric space. Define the distance between a point x and a set U as the greatest lower bound of |x,c| for c in U. Note that this distance is positive iff an open ball containing x misses U, iff x is not in the closure of U. In other words, the distance between x and U is 0 iff x is in the closure of U.

Let the "distance to U" be a function from S into the nonnegative reals. Moving x by δ cannot change the distance to U by more than δ. This is a consequence of the triangular inequality, and it can be used to show the distance function is continuous. In fact it is uniformly continuous; just set δ = ε.

Isometry

A function f between metric spaces is an isometry if it preserves distance. Since a positive distance cannot become a 0 distance, f is 1-1. For convenience, restrict attention to the image of f, whence f is onto.

Set δ = ε, and f becomes continuous. In fact it is bicontinuous. Combine this with 1-1 and onto, and f becomes a homeomorphism. The spaces are essentially the same; we are merely relabeling the points.

Equivalent

Two distance metrics f and g on a common space S are equivalent if they define the same topology. An open set under one metric must be covered by open balls in the other. Select any x and imagine it is part of an open set W under f. Select a value of ε, which could be arbitrarily small. There must be a δ such that g(x,y) < δ implies f(x,y) < ε. And this must hold in the reverse direction: f(x,y) < δ implies g(x,y) < ε. The metrics f and g are equivalent iff these relationships hold for every x in S and for every ε > 0.

If δ does not depend on x, but only on ε, then the metrics are uniformly equivalent.

Let h be a strictly increasing continuous function, with h(0) = 0. Furthermore, let h be twice differentiable, with h′′ ≤ 0. Thus h(f) is a valid distance metric. Since h is increasing, it is invertible, so let j be the inverse of h. Use h and j to verify the δ ε test above. Thus f and h(f) are equivalent metrics.

if h′ and j′ are bounded below some constant b, then f and h(f) are uniformly equivalent.

An important example is h(x) = x/(1+x). Note that h′ is positive, and h′′ is negative. Also, h is bounded by 1. Thus any metric space can be turned into a bounded metric space. The entire universe is pulled into an open ball, and the topology, i.e. the open sets, are exactly the same.

Let S be a space with two equivalent metrics f and g. The identity map that carries S onto itself maps open sets to open sets, and is bicontinuous. Thus the identity map is a homeomorphism. If f and g are uniformly equivalent then the identity map is a uniform homeomorphism.