Is the Intersection of Nested Sets in a Complete Metric Space Nonempty?

[aaaa202]

Homework Statement

Let (M,d) be a complete metric space and define a sequence of non empty sets F1\supseteqF2\supseteqF3\supseteq such that diam(Fn)->0, where diam(Fn)=sup(d(x,y),x,y\inFn). Show that there \bigcap_n=1^∞Fn is nonempty (contains one element).

Homework Equations

The Attempt at a Solution

We wonna use the completeness of M somehow. Let (xn) be a sequence of elements such that xn\inFn. 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\inFn 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.

 

[Dick]

Homework Statement

Let (M,d) be a complete metric space and define a sequence of non empty sets F1\supseteqF2\supseteqF3\supseteq such that diam(Fn)->0, where diam(Fn)=sup(d(x,y),x,y\inFn). Show that there \bigcap_n=1^∞Fn is nonempty (contains one element).

Homework Equations

The Attempt at a Solution

We wonna use the completeness of M somehow. Let (xn) be a sequence of elements such that xn\inFn. 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\inFn 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.

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

 

[aaaa202]

I don't 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?

 

[Dick]

I don't 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?

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.

 

[aaaa202]

how do you prove that?

 

[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?

 

[aaaa202]

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

 

[micromass]

how do you prove that?

How did you define closed and open?

 

[aaaa202]

Something like: An element is said to be on the boarder (I don't 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. I am sorry if this is not well translated - i hope you can understand.

 

[Dick]

Something like: An element is said to be on the boarder (I don't 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. I am sorry if this is not well translated - i hope you can understand.

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.

 

[aaaa202]

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

 

[Dick]

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

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