can someone peek at a proof? (metric spaces, basic proof)

1. The problem statement, all variables and given/known data
Let A be a finite open subset of a metric space M. Prove that every point of A is an isolated point of M.

2. Relevant equations
A point x is isolated if there exists some r such that the open of radius r centered at x consists of x alone.

alternative definition of isolated:the set {x} is open

3. The attempt at a solution
Let A = {a1, a2, ... an}. Let r = min{D(ai, aj): for all i≠j, 1≤i<j≤n}. Then for all x in A, Sr(x)INTERSECT M = {x}, so every point is an isolated point of M.

So I think my teacher ok'd this, but now that I think about it, I have a question. What does it mean to be isolated in M? I get that every point in A is isolated, but why is it isolated in M?

Also, the first definition of isolated makes intuitive sense to me (the idea of isolated in english seems to fit this definition). The second definition does not make intuitive sense to me. Does anyone have an intuitive way to interpret the second definition?

This subject (metric spaces) has been a little difficult for me :(

 PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age
 Blog Entries: 1 Recognitions: Homework Help In a metric space, every open ball is an open set (just like in normal metric spaces you encounter, like Rn). So if there is a ball around x such that x is the only point in that ball, well that set, {x}, is open. Equivalently, if you're told that {x} is open, by definition for every point in {x} (of which there is only one) there is an open ball containing that point contained entirely in our set - {x}. So there is an open ball containing x, and which is a subset of {x} - the only possibility is that the open ball is {x} itself. I disagree with the proof you presented actually. This proves that every point of A is an isolated point of A, but does NOT prove that it is an isolated point in M. For example if M is the set {-1,2}U[0,1] and A={-1,2}, then every point of A is an isolated point of M, but your proof says that if we pick r=3, the ball of radius 3 around the point -1 contains only -1, which isn't true (since it will contain all of [0,1] as well). This set M is a pretty good example of a metric space to keep in mind when thinking about isolated points intuitively. To prove that every point in A is an isolated point of M, for x in A we need to find an open ball containing x, which contains no other point of M. The value of r that you calculated is important - certainly any such ball we use must have radius smaller than r. But you also need to use the fact that A is open - so there is a ball containing x which is a subset of A.
 Duh, I hate it when I forget to include all of my assumptions. But now I think I'm more confused. I don't know anything about M. I know that A is open, so yes, I know that for all x in A, there exists an open ball such that the ball is contained in A... But now I'm having a hard time even thinking of what an open finite set would look like. I usually think of open sets as intervals or open regions, but an interval wouldn't be finite.

Blog Entries: 1
Recognitions:
Homework Help

can someone peek at a proof? (metric spaces, basic proof)

An open finite set "looks like' the set {-1,2} as a subset of {=1,2}U[0,1] with the standard metric from the real numbers.

Then if we take the ball of radius 1/2 around the point -1, the only point in that ball is -1

 So I'm thinking that if A={0,1}, then A is finite , and it is open (A is discrete). But this does not hold with M=R... (0,1 are not isolated in M)... so I'm thinking I must be wrong, and A is not open...
 Let A be a finite open subset of a metric space M. Prove that every point is an isolated point of M. Assume there exists a point x in A that is not isolated in M. Then for all e>0, Se(x) contains points of M. But since A is open, there exists some r st Sr(x) is contained entirely in A. Since x is not isolated in M, there must exist an infinite amount of points in any e neighborhood, so this Sr(x) also contains an infinite amount of points in A. But A is finite. Therefore, every point in A is isolated in M.