Which changes of metric fix all open balls of a metric space?

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In an earlier question, I was interested in counting the number of metric spaces on N points, where I considered two metric spaces to be the same if they had the same collection of open balls. Two questions:

1. What are the usual notions of metric space equivalence? Are any of them nontrivial for finite metric spaces? (For instance, the obvious one that two metric spaces are equivalent if their topology is the same is trivial for finite vector spaces).
2. If we say that two (labeled) metric spaces are equivalent if they have the same collection of open balls, in what ways can we operate on the metric d such that we get the same collection of open balls?

For example, any two metric spaces on 2 points are equivalent in this way, so any allowable operation on metrics yields an equivalent metric space. Clearly we can always scale d by any positive real without changing the equivalence class. Now consider the two metric spaces

````x ---3--- y        x ---3--- y
 \       /          \       /
  3     4            4     5
   \   /              \   /
     z                  z
````

In both cases, the nontrivial open balls are {x, z} and {x, y}, so these metric spaces are equivalent. How can I describe the operation I'm performing on d in general terms? What other operations on d will yield metric spaces that are equivalent in this sense?

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You might be interested in the discussion here: mathoverflow.net/questions/5957/… – Qiaochu Yuan Nov 23 2009 at 17:25

1 Answer

If you take the set of distances between $n$ points, then you can define a function on the real numbers that when applied to the distances preserves the open balls on $n$ points and defines a new metric as follows. Let $m$ be the number of distinct distances. First note that zero must be fixed second the set of new distances must preserve the original open balls and also preserve the triangle inequality and be positive then we can take the sum of the polynomials $\sum_1^k x(x-d_1)(x-d_2)$...(goes through all distances but d_k)$(d_k'/d_kf_k(d_k))(f_k(x)$ summing over all such terms for all $k$ where $f_k$ are arbitrary smooth functions not zero at $d_k$ and $d_k'$ is the desired new distance we can also add a term $x(x-d_1)$...$f(x)$ where $f(x)$ is an arbitrary smooth function for all distances in this way we preserve zero and changes all the distances to new distances satisfying the triangle inequality and having positive distances and preserving the open balls on the n points and alao preserving symmetry since if $x,y$ and $y,x$ have the same distance it will be sent to a new distance which will be the distance between $x$ and $y$ as well as $y$ and $x$. Thus the new set of distances will be a metric on the $n$ points preserving the open balls on the $n$ points.

This provides a real function operating on the reals that maps one metric space to another by transforming the distances. It does not send metrics with two equal distances to metrics with two different distances so for the two equivalent metrics given it sends one to the other but that transformation is not reversible.

For any set of inequalities on distances there is a realization as a metric space. Just Let the distances be close enough to one to satisfy the triangle inequality and set distances from points to themselves as zero and the triangle inequality will be satisfied.

Let us look at what is happening in the case with three points mentioned in the problem. If one distance is maximum. It will determine the whole structure. Say it is the distance from $A$ to $B$ then at point $A$ it will be greater than the distance from $A$ to $C$ so at this point we will get the set containing $A$ and $C$ in the topology similarly we will get the set containing $B$ and $C$ in the topology. The set containing $A$ and $B$ will not be in the toplogy. In all cases the set of all elements and the set of individual elements will be in the topology so at this point we have determined the entire topology. Now we have done this without specifying distances $AC$ and $BC$ and the can be set arbitrarily as long as they are less than $AB$.

In fact we can determine the entire set of topologies for three points by looking at the number of maxima. If there are two distances which are equal and larger than the third then the topology contains the entire set, single points and the two points of the third edges. If there all three distances are equal we have the topology consists of the three pionts and the entire set.

Something similar happens in three dimensions if one distance $AB$ is greater than all the others. It forbids the three points sets containing the two points with this distance and it forces the two three point sets not containing it to be in the topology and that determines the entire structure of three points sets in the topology.

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2
I like this answer, but if you have the time, would you mind cutting the explanation into more sentences and writing the summation formula for the function? I find it somewhat hard to follow as is. – Elizabeth S. Q. Goodman Nov 24 2009 at 8:06
Seconded; I'm having trouble following exactly what your construction is, and why it works. Could you clarify? – Gabe Cunningham Nov 25 2009 at 1:32
I made some changes including adding a summation function and adding material. – Kristal Cantwell Nov 25 2009 at 18:25
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