# More metric spaces

• Oct 14th 2008, 03:08 PM
EricaMae
More metric spaces
So this has 3 parts. I've gotten somewhere on it, but I'm stuck.

Let (X,d) be a metric space. We say that X is d bounded if there is a point a in X and r>0 such that the ball of radius r centered at a, B(a,r)={x in X|d(x,a)<r} contains X. i.e. the ball contains X.

1)Show that if X is d bounded then for every b in X there is an r_b>0 such that X=B(b,r_b)

- On this part, I understand that the Ball B(a,r) contains X implies that B(a,r) is a neighborhood of X. Does that mean that X is a neighborhood of the ball inside of it B(b,r_b) would imply that X contains this ball, but I can't get how X=B(b,r_b).

2) Show that X is NOT d bounded if and only if for every r>0, there exists a sequence {a_n} n=1,2,... such that the collection of sets {B(a_i,r)} is pairwise disjoint.

- I know that pairwise disjoint means that for every i,j i not equal to j, B(a_i, r) intersected with B(a_j, r) is the empty set. This all means that X is made up of the sets of balls that are not joined in any way. Does this mean that all of these balls make up one big ball that is the neighborhood of the set of neighborhoods? If so, how will I know that X contains this ball, which by definition given X must contain the ball B(a,r) to not be d bounded.

3) Give an example of a d bounded metric space (X,d) for which there is an r>0 and a sequence {a_i}, n=1,2,... such that the collection of sets {B(a_n,r)} is pairwise disjoint.

-Basically just give an example of part B. I'm not really sure what exactly is wanted here?

Again, anything will help here! thanks!
• Oct 15th 2008, 11:44 AM
Opalg
Quote:

Originally Posted by EricaMae
Let (X,d) be a metric space. We say that X is d bounded if there is a point a in X and r>0 such that the ball of radius r centered at a, B(a,r)={x in X|d(x,a)<r} contains X. i.e. the ball contains X.

1)Show that if X is d bounded then for every b in X there is an r_b>0 such that X=B(b,r_b)

The condition of being d bounded means that every point in the space X is within distance r of a. It follows (from the triangle inequality) that any two points in the space are within distance 2r of each other. In particular, every point is within distance 2r of b. So B(b,2r) contains the whole space.

Quote:

Originally Posted by EricaMae
3) Give an example of a d bounded metric space (X,d) for which there is an r>0 and a sequence {a_i}, n=1,2,... such that the collection of sets {B(a_n,r)} is pairwise disjoint.

Take any infinite set with the discrete metric (where d(x,y)=1 whenever x≠y). Then the balls B(x,r) are all disjoint provided that r<1/2. But the space is d bounded because any ball of radius greater than 1 contains the whole space.