Metric spaces - when does the interior of a closed ball coincide with the coresponding open ball?

Question

Yesterday I came across a very interesting question at

http://answers.yahoo.com/question/index;_ylt=AiZ_9436.iY9WMwEiVbaSGTsy6IX?qid=20070726145922AAp4oEw&show=7#profile-info-9ea5b56c0000f60ba86311a887f43581aa involving closures of open balls. Then, I started wondering about a related prolem. Let X be a metric space with metric d. Is there any necessary or sufficient or both condition so that the interior of a closed ball be the open ball of same center and radius?

In the euclidean spaces R^n, this is always true. But if you take R^n with the discrete metric, than this doesn't hold. The closed ball of center at 0 and radius 1 is the whole space and coincides with its interior. But the open ball of center 0 and radius 1 is {0}.

Is anyone interested in thinking about this?
Thank you
Steiner

Answer 1

Mimicking the proof for the similar case that inspired your question, we can prove that a necessary and sufficient condition is:

For every p in X, the function x --> d(p,x) defined on X\{p} has no local maximum.

Proof. Let's call

B(p,r) the open ball with radius r centered at p
B[p,r] the corresponding closed ball
B°[p,r] the interior of B[p,r]
X\S the complement of a set S

Now suppose y in X\{p} is a local maximum, and set r=d(p,y). Then there is a neighbourhood U of y such that, for every z in U, d(p,z) =< d(p,y), so U is contained in B[p,r] and thus, since y is contained in a neighbourhood contained in B[p,r], y belongs to B°[p,r]. But y does not belong to B(p,r), so the two sets do not coincide.

For the converse, let y be in X with d(p,y)=r. Since y is not a local maximum, for every neighbourhood of y there is at least one point z such that d(p,z) > d(p,y), so we can build a sequence converging to y and made up of points in X\B[p,r]. Thus y belongs to the closure of X\B[p,r], which in turn is X\B°[p,r] (this is a general fact of closure and interior). So we conclude that y does not belong to B°[p,r]. But y was an arbitrary point of the boundary of B[p,r] and, since B[p,r] is the disjoint union of its boundary (i.e. the points with distance r from p) and B(p,r), then we have just showed that B°[p,r] contains at most B(p,r). In fact, as B(p,r) is open and contained in B[p,r], it follows at once that B°[p,r]=B(p,r). End of proof.

Answer 2

Well, the open ball is always in the interios of the closed ball. So, a necessary and sufficient condition is that no point of the boundary of the closed ball is an interior point.

So, a necessary and sufficient conditioncon is: if x is in X, then, for every a in X distinct from a, every neighborhood of x contains an y with d(y,a) > d(x,a).

Answer 3

I think this may be necessary and sufficient, but I haven't proved it:

For every x, if B(s,x) is a subset of F(r,c) for some s>0, then x is in B(r,c).

Here B(a,b) is the open ball of radius a centered at b, and F(a,b) is the closed ball of radius a centered at b.