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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?

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goblin GONE
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1 Answers1

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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.