# Decreasing sequence of closed balls in COMPLETE metric space

1. Feb 7, 2010

### kingwinner

1. The problem statement, all variables and given/known data
Give an example of a decreasing sequence of closed balls in a complete metric space with empty intersection.
Hint 1: use a metric on N topologically equivalent to the discrete metric so that {n≥k} are closed balls. In={n,n+1,n+2,...}.

2. Relevant equations
N/A

3. The attempt at a solution
In the following post:
We showed that the metric d(m,n)=∑1/2k where the sum is from k=m to k=n-1, satisfies all the conditions required in the problem, except for completeness.
With that metric, we formed the closed balls by taking {n E N: d(k,n)≤1/2k-1} = {k-1,k,k+1,k+2,...} = Ik-1. And I1,I2,I3,... is a decreasing sequence of closed balls with empty intersection.

Now, we have to come up with another metric (possibly a modification of the above) that also satisfies completeness (i.e. every Cauchy sequence in N converges (in N)).

Does anyone have any idea?
Any help is greatly appreciated!

2. Feb 8, 2010

### kingwinner

If I define d to be

d(m,n)= 1/2 + ∑1/2k where the sum is from k=m to k=n-1, if m<n
d(m,n)=0, if m=n
d(m,n)=d(n,m), if m>n

Will this work or not?

Could someone kindly confirm this/correct me if I'm wrong.
Thank you!

3. Feb 8, 2010

### Dick

You know, I think that does work. {2,3,4,5,...} is no longer Cauchy, so the space is complete. Why don't you start filling in the other steps? Like showing 'd' is a metric and showing {n,n+1,...} is an interval?

4. Feb 8, 2010

### kingwinner

d(m,n)= 1/2 + ∑1/2k where the sum is from k=m to k=n-1, if m<n
d(m,n)=0, if m=n
d(m,n)=d(n,m), if m>n

d(1,2)=1, d(2,3)=0.75, d(3,4)=0.625
Define
...

Does this give the decreasing sequence of closed balls with B1={1,2,3,4,...}, B2={2,3,4,...}, B3={3,4,5,...}?

I am quite confident that B1={1,2,3,4,...}, B2={2,3,4,...}, B3={3,4,5,...} unless I've make some mistakes. But I would appreciate if someone can confirm that I'm producing the decreasing sequence of closed balls correctly.

thanks.

Last edited: Feb 8, 2010
5. Feb 8, 2010

### kingwinner

Using an upper bound based on the infinite geometric series, I think I've shown that my results in post #4 are correct.

But now how can we actually PROVE that the metric space (N,d) is COMPLETE? (i.e. every Cauchy sequence in N converges (in N))

For any n E N, the open ball of radius 1/2 about n = B(1/2,n) = {n}. But does this show that every Cauchy sequence in N converges (in N)) ??? I'm puzzled about this part, and I would appreciate if someone can help me out.

thanks.

6. Feb 8, 2010

### Dick

I'm liking your methodical approach here. Doing examples like this is the only way to get a feeling for what's going on. You've just got a typo in "B1=Closed ball of radius 0.1 about 2". You meant "B1=Closed ball of radius 1.0 about 2". Great work so far. Keep it up.

7. Feb 8, 2010

### Dick

Now it's getting easy again after the rude interruption from the last bogus hint. Pick epsilon=1/4. If {an} is Cauchy, then |an-am|<1/4 for n,m>N. That means an=am, right?