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I was trying to prove that, in general, the diameter of an open ball $B_d(x, \epsilon)$ in a metric space $(X, d)$ is equal to $2\epsilon$. It then occurred to me that this is not the case if the metric induces the discrete topology, so one necessary (but probably not sufficient) condition is that $d$ doesn't induce the discrete topology. A trivial example of where this is true is $\mathbb{R}^n$ with the euclidean metric. So, what can we impose to make sure that the diameters are exactly twice the radius?

Observation: if $A\subset X$ is bounded, the diameter of $A$ is $:= \displaystyle{\sup_{a_1, a_2 \in A} d(a_1, a_2)}$

EDIT: to the people voting to close the question - could you at least mention your reasons in a comment? It's hard to know just what I'm doing wrong otherwise...

EDIT 2: It's been brought to my attention that maybe the initial question isn't that interesting so I've edited the title. Still, if we could find a necessary and sufficient condition, that would be super nice!

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There are several nice classes of spaces for which the property in the title of your question holds:

  1. Compact (connected) Riemannian manifolds or, more generally, complete Riemannian manifolds with injectivity radius bounded below. Recall that a Riemannian metric on a connected manifold $M$ defines a distance function $d$ on $M$ ($d(p,q)$ is equals the infimum of lengths of paths connecting $p$ to $q$). Riemannian geodesics do not necessarily minimize the length, but they locally do: If two points $p, q$ are on a geodesic $c$ in $M$ and the length $L$ of the segment $pq$ of $c$ between these points is sufficiently small then $d(p,q)=L$. If $M$ is compact then one can make a bound in this statement uniform: There exists $D>0$ such that if $L\le D$ then $pq$ is a (unique) minimizing geodesic between $p$ and $q$. Now take any $r< D/2$. For $x\in M$ the metric ball $B(x,r)$ centered at $x$ and of radius $r$ (called a geodesic ball) has the property that every geodesic $c$ through $x$ and contained in $B(x,r)$ is a minimizer. In particular, if $c=pq$ where $p, q$ are on the boundary of $B(x,r)$ then $d(p,q)=2r$. Hence, the diameter of $B(x,r)$ is at least $2r$. The opposite inequality, $diam(B(x,r))\le 2r$ is immediate from the triangle inequality.

My favorite reference for this is do Carmo's "Riemannian Geometry".

  1. The same argument works for Finsler manifolds, provided that the Finsler norm is strongly convex.

  2. More generally, there is a class of metric spaces called geodesic spaces with geodesic extension property which are not Riemannian manifolds but for which the same argument applies. An example of such a space is a complete tree: A connected graph without cycles and without valence 1 vertices, which we equip with the graph-metric: declare every edge to be isometric to the unit interval and define the distance between points by minimizing lengths of paths in the graph connecting these points.

Note that geodesics in general metric spaces are defined differently from the Riemannian case: A geodesic is defined as a length-minimizing path parameterized by its arc-length, equivalently, as an isometric embedding of an interval into the metric space. A good reference is

Burago and Ivanov "A course in metric geometry", especially section 9.1.7.

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  • $\begingroup$ Great answer! Thanks for the references as well. $\endgroup$ – Matheus Andrade Feb 8 '19 at 20:06
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Not a complete answer

The discrete topology is a good example to show things aren't "nice", but there are far nicer examples. Look, for instance, at the unit sphere.

A disk of radius $\pi/2$ centered at the north pole has the equator as its boundary, so everything is nice, and the diameter is $\pi = 2r$. But if you look at a disk with a larger radius --- say $3\pi/4$ --- then you find that the diameter is still $\pi$ (achieved by a pair of antipodal points on the equator, for instance).

I suspect that for many nice spaces you'll find that for large values of $\epsilon$ your condition fails.

I guess what I'm saying is that it might be interesting to ask a modified version of your question:

For what metric spaces $(X, d)$ is there some $s > 0$ with the property that for all $\epsilon < s$ and all $x \in X$, we have $\text{diam}(B_d(x, \epsilon)) = 2\epsilon$?

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  • $\begingroup$ That's a great point, actually! I'll edit the title. $\endgroup$ – Matheus Andrade Feb 8 '19 at 15:18

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