# Cauchy sequence problem

1. Jun 5, 2013

### aaaa202

1. The problem statement, all variables and given/known data
Let (M,d) be a complete metric space and define a sequence of non empty sets F1$\supseteq$F2$\supseteq$F3$\supseteq$ such that diam(Fn)->0, where diam(Fn)=sup(d(x,y),x,y$\in$Fn). Show that there $\bigcap$n=1Fn is nonempty (contains one element).

2. Relevant equations

3. The attempt at a solution
We wonna use the completeness of M somehow. Let (xn) be a sequence of elements such that xn$\in$Fn. Then as diam(Fn)->0 we must have for a specific N that lxn - xml < ε for all m,n>N. Thus the sequence of (xn) is a Cauchy sequence and must be convergent in M due to the assumed completeness. Denote the limit by x. We must show that x$\in$Fn for all n. But I am unsure how to this. And is this even the right approach? I don't have a lot of experience with proofs.

2. Jun 5, 2013

### Dick

You are almost there. Now use that the sets Fn are closed. What does closed mean in terms of sequences?

3. Jun 5, 2013

### aaaa202

I dont know what you are referring to sorry.
The only thing I can come up with is something like: Assume x is not a member of all Fn. Then we can pick diam(Fn)<ε and lx-xnl ≥ ε. But that is a contradiction. But I don't know if that is the right way to do it.
What did you mean by closed in terms of sequences?

4. Jun 5, 2013

### Dick

If a set F is closed and {xn} is a sequence in F that converges to x, then x is also in F. That's what I mean.

5. Jun 5, 2013

### aaaa202

how do you prove that?

6. Jun 5, 2013

### LCKurtz

@aaa202 Your theorem is false as stated. Consider $F_n = (0,\frac 1 n)$. Given the discussion so far, haven't you noticed that you haven't assumed $F_n$ is closed?

7. Jun 5, 2013

### aaaa202

oops I forgot to say they were haha. But you are right I didn't say it and assumed it all along.

Last edited: Jun 5, 2013
8. Jun 5, 2013

### micromass

Staff Emeritus
How did you define closed and open?

9. Jun 5, 2013

### aaaa202

Something like: An element is said to be on the boarder (I dont know the appropriate term in english) in a set if each sphere around it contains at least one element of the set. A set is closed if it contains all its boarder elements. Im sorry if this is not well translated - i hope you can understand.

10. Jun 5, 2013

### Dick

Approximately right word, but it's spelled 'border'. 'boarder' is something else. In the sequence argument, isn't x on the border of F? And the F's being closed is so important, I guess I didn't even notice the problem statement didn't say that.

11. Jun 5, 2013

### aaaa202

Maybe? what makes you say that. I don't know

12. Jun 5, 2013

### Dick

Every neighborhood of x contains points in F. Use that a sequence of points in F converges to x.