Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This relates to an earlier conjecture that was shown false.

Question: is it possible to characterize a metric space by its diameter function?

Here are my thoughts so far. Assume a diameter function that is non-negative, and which satisfies $\mathrm{diam}(A)=0$ iff $|A| < 2$. Thus defining $d(x,y)=\mathrm{diam}\{x,y\}$, we obtain that $d$ has all the properties of a metric except the triangle inequality.

Two questions remain.

  1. What needs to be assumed so that we can recover the triangle inequality?

  2. If we begin with a diameter function and define $d$ as above, is it necessarily true that $\mathrm{diam}(A)=\sup \{d(a,b) | a,b \in A\}$.

Ideas, anyone?

share|cite|improve this question
up vote 2 down vote accepted

For a diameter function to be worthy of its name, you should assume that $\operatorname{diam}(A\cup B)\le \operatorname{diam}(A)+\operatorname{diam}(B)$ whenever $A$ and $B$ are two sets with nonempty intersection. This property is satisfied by the diameter in any metric space.

Under the above assumption, the quantity $d(a,b)=\operatorname{diam}(\{a,b\})$ is indeed a metric. In general, it will not reproduce the given diameter function. For example, consider $\operatorname{diam}(A)=|A|-1$. This diameter function leads to $d(a,b)=1$ for all $a\ne b$, hence $\operatorname{diam}(A)=1$ whenever $|A|>1$.

You can additionally assume that $\operatorname{diam}(A)=\sup_B \operatorname{diam}(B)$ with the supremum taken over all two-point subsets $B\subset A$. This assumption is also satisfied by the diameter in any metric space, and with it the whole thing becomes a tautology.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.