Metric Case
Metric spaces - can you help with the closure and the interior of balls?
In "nice" metric spaces, like R^n, the closure of an open ball is the closed ball of same center and radius and the interior of a closed ball is the corresponding open ball. But this is not always true, an example is R with the discrete metric.
If (X,d) is a metric space, is there any condition which ensures that the closure of every open ball is the corresponding closed ball? And is there any condition (probably not the same as the previous, in case it exists) that ensures the interior of every closed ball is the corresponding closed ball?
Thank you
I answered the question about the closure of open balls more than a year ago. I didn't find the link, but I saved my answer, which is not that difficult once you know the condition.
1) 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.
2) 1) Let a belong to the metric space X. Then, a necessary and sufficient condition so that the interior of every closed ball centered at a be the corresponding open ball is that the function defined on X by x → d(x, a) has no relative maximum. 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 (1) (I'm just copying and paste from my file and when I wrote this proof the editor didn't allow mathematical symbols. But I think it's readable)
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.
The proof of (2) is not that hard, I think it's even simpler than the proof of (1). I leave it to you or to somebody else who can collaborate, I don't have time now.
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