Let AcX, A is not the empty set.
Prove if every b in X there exists an a in A such that d(b,A)=d(b,a) then A is a closed subset of X.
So for A to be a closed subset, that means that it has to include all of its limit points. So should I assume that b is also in A?
I was thinking that I could say that A has a sequence {an} that converges to a, but how does that relate to b? with d(b,A)=d(b,a)? I am not sure how to put the two together.
So I'm assuming c is in A and is a limit point of A. Therefore, there is a ball B(c,r) that contains a point, a, not equal to c?
I don't understand what d(b,A) means? How can I measure the distance from a point to a whole set? Would that mean the distance of b to the boundry of A? If so, I don't get why d(b,A)=d(b,a)
The distance d(c,A)=0 would mean that c is on the boundry of A, does that mean that it is an accumulation point? If so, then d(c,A)=0, d(c,a)= r, d(b,a)>r, d(b,c)=d(b,A)?
I'm not sure where this takes me?
The way I have the picture drawn is that there is a metric space X, that contains a space A. I have b in X and not in A, and a in A.
Assuming that I say that a is a limit point because it is in A, and b is not a limit point because it is not in A. How would I show that b is not a limit point?
Okay,
So could I say that b is in X and not A. there is a ball B(b,r) that is completely contained in X such that B(b,r) intersect A= the empty set. This directly implies that b is not a limit point.
Assuming that c is a limit point in A, there is a ball B(a,E) and a not equal to c. So d(a,c)<E implies D(A,c)<E so D(A,c)=0?
My problem is that b is not in A. So I'm not sure how to relate d(a,c) and D(A,c) with d(a,b) and D(A,b) because c is in A, and b is not.
I am sorry to say but I don’t think that you have any idea what this material is all about.
Now, don’t be offended by that remark! It is meant as a very constructive remark.
What you need to do is to go to the instructor and have a one-to-one tutorial with him/her. Any competent lecturer would be glad to help you. I say this having taught this material for more that thirty years.
The crux of this problem is the following theorem.
In any metric space, the closure of any set is (A,x) = 0} \right\}" alt="\overline A = \left\{ {x(A,x) = 0} \right\}" />.
A point is in the closure of a set iff its distance to the set is zero.