Could you help me with distance between sets in metric spaces?

Question

In a metric space S, with distance function, d, the distance between 2 subsets A and B is defined as d(A, B) = infimum {d(a,b) | a in A and b is B}. Suppose A and B are disjoint. Show that, if A is compact and B is closed, then d(A,B) > 0. Show that this does not follow if, instead of compact, A is only closed

Thank you

Answer 1

I'll assume as known some basic facts about metric spaces, otherwise the proof becomes kinda long.
The distance from a point a of S to a set B is defined by d(a, B) = infimum {d(a,b) | b is in B}. It's well known that the function a --> d(a, B) is Lipschitz, so uniformly continuous on S. From the definition and properties of the infimum of a set, d(A,B) can be written as d(A,B) = infimum {d(a, B) | a is in A} (1).

We also know that d(a,B) = 0 if, and only if, a is in the closure of B. Proof: if a is in the closure of B, then, for every eps >0, there exists b in B with d(a, b) < eps, which implies, by the definition of d(a, B), that d(a, B) <= d(a,b) < eps. Since eps is arbitrary, it follows that d(a, B) = 0. If, on the other hand, d(a, B) =0, then the definition of d(a, B) implies that, for every eps >0, eps is not a lower bound of {d(a,b) | b is in B}, so that there exists b in B with d(a, b) < eps. Since eps is arbitrary, a is in the closure of B.

Since A is compact, A is closed (property of every metric space) and is its closure. Since B is closed, B is its closure. Since A and B are disjoint, then their closures don't intersect each other. No element of A is in the closure of B, which implies that d(a,B) > 0 for every a of A.

Since the function a --> d(a, B) is continuous and A is compact, this function attains a global minimum at some a* of A. Therefore, d(a, B) >= d(a*, B) >0 for every a in A . From (1), it then follows that d(A,B) = d(a*, B) >0, proving the assertion.

To see that this may not be the case if A and B are closed but none of them is compact, take S = R^2 and let A be the non negative X axis and B the graph of the function f(x) = 1/x, x> 0. A and B are closed and disjoint in R^2, and none of them is compact (both are unbounded - by Heine Borel theorem, can't be compact). But letting x -> oo, we see we can find points in A as close as desired to points in B. So, d(A,B) = 0

Now we are done.
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