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

In summary: Then x is in F.How do you know that every neighborhood of x contains points in F?Every neighborhood of x contains points in F. Use that a sequence of points in F converges to x. Then x is in F.How do you know that every neighborhood of x contains points in F?
  • #1
aaaa202
1,169
2

Homework Statement


Let (M,d) be a complete metric space and define a sequence of non empty sets F1[itex]\supseteq[/itex]F2[itex]\supseteq[/itex]F3[itex]\supseteq[/itex] such that diam(Fn)->0, where diam(Fn)=sup(d(x,y),x,y[itex]\in[/itex]Fn). Show that there [itex]\bigcap[/itex]n=1Fn 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[itex]\in[/itex]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[itex]\in[/itex]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.
 
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  • #2
aaaa202 said:

Homework Statement


Let (M,d) be a complete metric space and define a sequence of non empty sets F1[itex]\supseteq[/itex]F2[itex]\supseteq[/itex]F3[itex]\supseteq[/itex] such that diam(Fn)->0, where diam(Fn)=sup(d(x,y),x,y[itex]\in[/itex]Fn). Show that there [itex]\bigcap[/itex]n=1Fn 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[itex]\in[/itex]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[itex]\in[/itex]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.

You are almost there. Now use that the sets Fn are closed. What does closed mean in terms of sequences?
 
  • #3
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?
 
  • #4
aaaa202 said:
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.
 
  • #5
how do you prove that?
 
  • #6
@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
oops I forgot to say they were haha. But you are right I didn't say it and assumed it all along.
 
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  • #8
aaaa202 said:
how do you prove that?

How did you define closed and open?
 
  • #9
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.
 
  • #10
aaaa202 said:
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.
 
  • #11
Maybe? what makes you say that. I don't know
 
  • #12
aaaa202 said:
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.
 

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

What is a Cauchy sequence?

A Cauchy sequence is a sequence of real or complex numbers that gets arbitrarily close to each other as the sequence progresses. In other words, for any small distance, there exists a point in the sequence where all subsequent terms are within that distance of each other.

Why is the Cauchy sequence important?

The Cauchy sequence is important because it provides a rigorous definition of what it means for a sequence to converge. It is also a key concept in real analysis and is used in various mathematical proofs and applications.

How is the Cauchy sequence different from a convergent sequence?

A Cauchy sequence is a specific type of convergent sequence. While all Cauchy sequences are convergent, not all convergent sequences are Cauchy. A Cauchy sequence has the additional property that its terms get arbitrarily close to each other, not just to a single limit value.

What is the Cauchy sequence problem?

The Cauchy sequence problem is a famous mathematical problem that asks whether every Cauchy sequence is convergent. It was first posed by French mathematician Augustin-Louis Cauchy in the 19th century and remains an open problem in mathematics.

What are some applications of the Cauchy sequence?

The Cauchy sequence has numerous applications in mathematics, physics, engineering, and other fields. It is used in the study of real analysis, topology, and metric spaces. It also has applications in numerical analysis, where it is used to approximate irrational numbers, and in the study of differential equations and Fourier series.

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