# (dis)prove an if and only if statement of a Cauchy sequence and and interval.

1. Apr 6, 2012

### Hodgey8806

1. The problem statement, all variables and given/known data
A sequence (Xn) is Cauchy if and only if, for every ε>0, there exists an open interval length ε that contains all except for finitely many terms of (Xn).

2. Relevant equations
The Cauchy Definition is:
A sequence X = (xn) of real numbers is said to be a Cauchy sequence if for every ε>0 there exists a natural number H(ε) such that for all natural numbers n,m≥H(ε) satisfy |xn - xm| < ε.

I want to check my contrapositive for the left-to-right implication:
There exists ε>0, such ALL open intervals ε contain all except for finitely many terms of (xn) => (xn) is not Cauchy.

3. The attempt at a solution
[STRIKE]I want to say this is false. As the right side holds the criterion that there is an interval around some point p such that the interval bounds are (p-ε/2 , p+ε/2) while the first only requires that the terms of xn only be between (xm - 1, xm +1). [/STRIKE] I'm certain now that this is true

[STRIKE]However, I cannot come up with a counter example as when I try it seems that both sides imply each other.[/STRIKE] So then I assumed it was true and to prove it. Now, I believe that the right side does indeed imply Cauchy. So for the left to right implication, I decided to prove it with the contrapositive:

If the contrapositive is the correct form, then it is easy to see that the hypothesis is false as now finite interval that can be any where on the number line can contain all the terms of a sequence.

Does this make sense? If not, please let me know what I could clarify. Thank you!

Last edited: Apr 6, 2012
2. Apr 6, 2012

### HallsofIvy

Staff Emeritus
A sequence of real numbers is Cauchy if and only if it converges.

3. Apr 6, 2012

### Hodgey8806

Thank you very much for this. This does help me see that the statement is true.
Unfortunately, and I apologize for not mentioning this, we are not allowed to refer to convergence. So that theorem isn't allowed for this problem. Thank You!

4. Apr 6, 2012

### Dick

There exists ε>0, such that for ALL open intervals of size ε there are an infinite number of terms of (xn) outside of the interval => (xn) is not Cauchy.

But I'm not sure using the contrapositive is the best way to prove this. Try it directly.

Last edited: Apr 6, 2012
5. Apr 6, 2012

### Hodgey8806

Oh wow, I think it was much easier than I was thinking. Using my beginning definition of Cauchy, I will prove it left to right first:

Spse (xn) is Cauchy.
Let ε/2 > 0, there exists a natural number H s.t. for all n,m≥H
|xn - xm| < ε/2,
Taking m = H
|xn - xH| < ε/2 if and only if -ε/2 + xH<xn< ε/2 + xH for all n≥H
Thus, the number of terms that can be outside of this interval is H-1.
Since ε/2>0 is arbitrary:
For all ε>0, there exists an open interval of length ε that contains all except for finitely many terms of (xn).
Q.E.D (1)

Boom! How is that?

Proof in the right-to-left direction:
Spse for all ε>0, there exists an open interval of length ε containing all except for finitely many terms of (xn).
This implies there exists a natural number H s.t. for all n,m≥H, -ε/2 <xn - xm< ε/2 if and only if |xn - xm| < ε/2 for all n,m≥H.
Since ε/2 > 0 is arbitrary, we conclude that (xn) is Cauchy.
Q.E.D (2)

Last edited: Apr 6, 2012
6. Apr 6, 2012

### Dick

Pretty good. You might want to explicitly say what H is in the second part.

7. Apr 6, 2012

### Hodgey8806

Thank you :) I can't believe that I over-thought that so much. However, I'm a bit confused as to what you mean? I don't quite know what to state that H is apart from a natural number that serves as the "beginning natural index" for n and m.

8. Apr 6, 2012

### Dick

I mean that you know there are only a finite number of (xn) that are outside of your interval. Call them {x_n1,x_n2,...x_nk}. Pick H to be the max of {n1,n2,...,nk}. That works, yes?

9. Apr 6, 2012

### Hodgey8806

Oh ok, I understand that. But can we let H be that max since n or m could actually equal H? I would think the max of that set is not necessarily within that interval so n,m cant necessarily be ≥ H.

How about something like this: "Thus at most there are H-1 terms outside of the given interval. Let xH be the first term in the interval such that for all n,m≥H", (continue the proof set up). That does make more sense though, thank you for pointing that out to me.

10. Apr 6, 2012

### Dick

Well, ok. Pick H=max{n1,n2,...,nk}+1, since you have a >= sign. All of this is just fussing around. You certainly have the right idea.

11. Apr 6, 2012

### Hodgey8806

Oh ok, great! That makes sense! My teacher, will knock huge points quick for mislabeling and I probably would haha. I really appreciate the help, again! Have a great weekend!