Let
d(m,n)= 1/2 + ∑1/(2^k) 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
B1=Closed ball of radius 1 about 2={1,2,3,4,...}
B2=Closed ball of radius 0.75 about 3={2,3,4,...}
B3=Closed ball of radius 0.625 about 4={3,4,5,...} <---Is this correct? Could someone confirm this?
...
If so, then it's a decreasing sequence of closed balls with empty intersection.
Now the problem is whether this metric space (N,d) is complete or not. (complete means 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 with this part.
Thanks for any help!
The following satisfies every single requirement in the original question, so it is a valid counterexample.
Define the metric
d(m,n)= 1/2 + ∑1/(2^k) 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
B1=Closed ball of radius 1 about 2={1,2,3,4,...}
B2=Closed ball of radius 0.75 about 3={2,3,4,...}
B3=Closed ball of radius 0.625 about 4={3,4,5,...}
...etc
This is a decreasing sequence of closed balls with empty intersection.
Claim: the metric space (N,d) is complete. (complete means every Cauchy sequence in N converges (in N))
Proof:
For any n E N, the open ball of radius 1/2 about n = B(1/2,n) = {n}.
Let {a_n} be any Cauchy sequence in N, then there exists K s.t. d(a_n,a_m)<1/4 for all n,m>K.
=> for all n,m>K, a_n=a_m
=> every Cauchy sequence in N is eventually constant, and hence converges. Thus (N,d) is complete.
I think every step is properly justified, so this is indeed a counterexample. Please feel free to point out any mistakes if there is one.
I think your example is indeed correct. Sorry for misleading you with my other reply. I said "in a compact space (in this case, the first ball), any intersection of non-empty nested closed sets is non-empty.". The statement is true, but... a closed ball need not be compact xD Sorry for that And good job ^^
Thanks for confirming!
There is a theorem that says: a metric space is complete iff every decreasing sequence of closed ball with radii going to zero has nonempty intersection.
And as Drexel28 pointed out, if it's diameter does not approach zero then it must contain MORE elements then if it's diameter approaches zero. This makes perfect sense to me as well.
But I indeed have constructed a concrete counterexample.
What explains this inconsistency?