Open and closed balls, can you help me (metric spaces)?

Question

Let X be a metric space with distance function d. Is there any condition that ensures that the closures of open balls of X are the closed balls of same center and radius? In R^n with the euclidean metric this is always true. But if we take R^n with the discrete metric, d(x,y) = 1 if x<>y and d(x,y) = 0 if x = y, then this is not true. For example, the open ball centered at 0 and with radius 1 is {0} and its closure is {0}. But the closed ball centered at 0 and with radius 1 is the whole R.

So, I'd like to know if there's a condition, necessary, sufficient or both, that ensures the desired condition in a general metric space.

Thank you

Answer 1

The other answers were good. But I think I have a necessary and sufficient condition:

Let a belong to the metric space X. Then, a necessary and sufficient condition so that the closure of every open ball centered at a be the corresponding closed ball is that a is the unique local minimum of the function defined on X by x --> d(x, a). So, a necessary and sufficient condition for this to happen all over X is that the condition given for a hold for every element of X. Let's prove this:

First, let's show that, in every metric space, the closure of an open ball is always contained in the corresponding closed ball. For r>0, let B(r) be the open ball of center a and radius r and let F(r) be the corresponding closed ball. Let B'(r) be the closure of B(r). If x is in B'(r), then, for every eps >0, there is y in B(r) such that d(x,y)
Now, suppose that, for every r >0, we have B'(r) = F(r) and define f(x) = d(x, a), for x in X. Since d(a,a) = 0, it's immediate that a is a global, so a local, minimum of f. Let x <> a, let r = d(x,a) >0 and let U be any neighborhood of x. Then, x is in F(r) = B'(r) and, therefore, U contains an element y of B(r). For this y we have f(y) = d(y,a) < r = d(x,a) = f(x) => f(y) < f(x). Since this holds for every neighborhood U of x, it follows f does not have a local minimum at x<>a, which implies a is the only local minimum of f in X. This proves the 1st part.

Now, for the converse, suppose a is the only local minimum of f in X. We'll show that, for every r >0, F(r) is contained in B'(r). It's immediate a is in B'(r). Let x<>a be in F(r). if d(x,a) < r, then x is in B(r) and, therefore, is automatically in B'(r). So, suppose d(x,a) = r and let U be any neighborhood of X. Since x<>a, x is not a local minimum of f and, therefore, there exists y in U such that f(y) < f(x). So, according to the definition of f, d(y,a) < d(x,a) = r, which shows d(y,a)
The proof is now complete

We can see that, in the case of R with the discrete metric, this condition is not verfied. If a is in R, then

f(a) = d(a,a) = 0
f(x) = d(x,a) = 1 if x <> a. Since f is constant in R - {a}, every element of R is a local minimum of f. a is the global one.

Best regards
Steiner


Steiner

Answer 2

First we need to clarify terminology.
"The open ball of radius r about point p" is the set of all points in the m.s. whose distance from p is less than r.
"The closed ball of radius r about point p" is the set of all points in the m.s. whose distance from p is less than or equal to r.
The first source below has a good discussion of this using pretty reasonable notation.

Usually the closure of an open ball is the corresponding closed ball, but not always, as your discrete metric example shows. The fact that you immediately jump to wondering about characterizing this with some hoped-for condition indicates you may have a point-of-view helpful for a mathematician. The "Gee, I wonder ...." p-o-v.

Another answer pointed out that for a discrete m.s. all open balls are closed. True, but this should not be allowed to mask the fact that the word "closed" in the phrase "closed ball" is there by definition, and does not imply any properties relating to closure.

I don't know what condition might characterize the property of "the closure of each open ball equals the corresponding closed ball" but it is a worthy goal to try to find such. Here's a start.

The second source below (in physicsforums for some reason) gives a completely different example. You are familiar of course with the m.s. of the real numbers with the usual distance. Now keep that same distance but change the underlying set from all the reals to being the reals with the open interval (1,2) removed. That is,
(-infinity,1] union [2,+infinity). Just to be sure you understand my notation, the numbers 1 and 2 ARE INCLUDED in the underlying set. The open ball around 0 of radius 2 is (-2,1), and the closed ball around 0 of radius 2 is [-2,1)union{2}, but the closure of the open ball is [-2,1). All 3 are different!

One tried-and-true approach for such a question is to look for examples, and ask yourself what do those examples have in common.

Another approach is to assume you have a m.s. and try to prove that the closure of every open ball is the corresponding closed ball. You will NOT be able to do that, clearly, because we have just seen 2 m.s. where that's not true. However you can look for some EXTRA assumption that will let the proof proceed.

Anyway, good luck, and have fun with it.

Answer 3

Answer edited due to misreading: I thought you wanted it enclosed in some closed ball, not a particular closed ball (of the same radius). I wasn't reading carefully enough. My apologies.

Also: PLEASE NOTE that user Steiner duplicated my answer and plagiarized the reference I gave. His answer is identical to mine, and the proof that he "came up with" on his own is actually word-for-word the proof provided by the link I gave.

The necessary and sufficient condition is that for any p the function f(x) = d(x,p) has a local minimum (nonstrict) only at x=p (where the distance is 0).

In other words, the distance has to increase continuously over the set as you move "away from" p. Distances in the discrete metric, for example, do not vary at all! The distance is always 1! (Except at x=p.)

Now then, an important necessary condition we can get from that is that the topology MUST NOT contain any isolated points. All points are isolated in the discrete metric, so that's right out the window.

Notice that this is not true if d(x,p) is constant over some subset of the space - like with the discrete topology, where it is constant over the entire space (except p) and thus has local minimum almost everywhere else.

Answer 4

Let me get home to grab my topology book and I'll get back to you on this.

Ah, so--there is no need. The question has been answered.