Distance between sets, can you help please?

Question

If A and B are subsets of a metric space (X, d), the distance between them, d(A,B), is defined as d(A, B) = infimum {d(a, b) | a is in A, b is in B}. Show that:

a) If the closures of A and B intersect, then d(A,B) = 0. Is the converse true?

b) If A is compact, B is closed and A and B are disjoint, then d(A,B) > 0. If, instead of assuming A is compact, we just assume it's closed, then is it still true that d(A,B) > 0?

Thank you

Answer 1

a) Let x be a common point of the closures of A and B. Then, for every eps >0, there are a in A and b in B such that d(a,x) < eps/2 and d(b,x) < eps/2. By the triangle inequality and the definition of d(A,B), it follows that 0 <= d(A,B) <= d(a,b) < eps. Since eps is arbitray, it follows d(A,B) =0

The converse is not true. Take R^2 with the usual (Euclidean) metric. Let A be the graph of f(x) = 1/x, x > 0, that is, A = {(x,y) | x >0,y =1/x} and let B be the horizontal axis {(x,0)}. Both are closed, so that they are their own closures, and they don't intersect. It's easy to see that, letting x --> oo, we can find a in A and b in B as close to each other as desired, so that d(A,B) = 0. So, d(A,B) = 0 but the closures of A and B are disjoint.

b) The distance from a point x in X to a subset B of X is defined as d(x, B) = inf {d(x,b) | b is in B)}. We can show the function defined on X and with values in [0, oo) given by f(x) = d(x, B) is continuous (actually, Lipschitz w/ constant 1). We can also show that d(A, B) = inf {d(a,B | a is in A} = inf {f(a) | a is in A} (I leave the proofs of these facts to you, it's a good exercise).
Since A is compact and f is continuous, f attains a global minimum at some a* of A. So, d(A,B) = f(a*). Since A is closed (in every metric space compacts set are closed), B is closed and A and B are disjoint, it follows no element of A lies in the closure of B, so that f(a) = d(a, B) >0 for every a of A. It follows that d(A,B) = f(a*) >0, proving the assertion.

But if we suppose A is only closed and not compact, then we can have d(A,B) = 0 with A and B disjoint. Exactly the same example given in (1) does the job.

Answer 2

"a) If the closures of A and B intersect, then d(A,B) = 0. Is the converse true?"

No. Take A to be the real line, and take B={(x,y);x>=0 and y=1/(1+x)}. Then d(A,B)=0 but their closures are disjoint.

"b) If A is compact, B is closed and A and B are disjoint, then d(A,B) > 0. If, instead of assuming A is compact, we just assume it's closed, then is it still true that d(A,B) > 0?"

No. See counterexample to a) above.

Answer 3

b) If, instead of assuming A is compact, we just assume it's closed, then is it still true that d(A,B) > 0?
I tried to look at R and some possible closed non-compact sets but I didn't figure out. Maybe it is needed a different topological space.

Answer 4

the area formula is often observed as "the Pythagorean theorem." d = sqrt { (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2 } that's distance in a at as quickly as line. once you're finding for distance alongside an arc on the floor of a sphere, the formula to apply is D = R ArcCos {a million - d^2 / (2 R^2)} the place R is the radius of the sphere and d is the at as quickly as line distance from earlier. make helpful that your calculator is desperate to return angles in radians once you employ the ArcCosine button. in case you're able to alter variety and longitude to geocentric oblong coordinates, you need to use those formulae: x = R cos (lat) cos (long) y = R cos (lat) sin (long) z = R sin (lat)