Proving (i) for Cauchy Sequences in Completeness Theorem

In summary, the textbook suggests that to prove that the least upper bound principle holds, one first begins with any a1 in S and sets b1 as an upper bound for S. Then, one constructs a nested sequence In = [an,bn] where In = [an,bn] if c is the midpoint of I1 and I2 if c is not the midpoint of I1. Next, one shows that (an) and (bn) are Cauchy sequences with a common limit L. Finally, one proves that L = sup S.
  • #1
kingwinner
1,270
0

Homework Statement


Least Upper Bound (LUB) Principle: every nonempty subset S of R that is bounded above has a least upper bound.
Completeness Theorem: every Cauchy sequence of real numbers converges. So R is complete.

To prove that Completness Theorem implies the least upper bound property (i.e. take the copmleteness theorem as an axiom, and use it to prove the LUB principle), my textbook suggests the following:
"Begin with any a1 in S and let b1 be an upper bound of S.
Let I1 = [a1,b1] . Let c be the midpoint of I1 . If c is an upper bound of S, let I2 be the first half of I1, otherwise let I2 be the second half. Repeat the process to get a nested sequence In = [an,bn] , the interested reader can verify that:
(i) (an) and (bn) are Cauchy sequences with a common limit L
(ii) and that L = sup S. "

Homework Equations


N/A

The Attempt at a Solution


I'm struggling to prove (i).

1) By definition, an is Cauchy iff for all ε>0, there exists N s.t. if n,m≥N, then |an-am|<ε.
But I have no idea how to prove that (an) and (bn) in our problem are Cauchy. Can someone help me, please?

2) [solved] Now if we have proved that both an and bn converge, why does it follow that they must converge to a COMMON limit L? I can see that the legnth of the interval = bn-an ->0 as n->∞, so it's intuitvely believable that they both converge to a common limit, but how can we actually PROVE rigorously?

Thanks for any help!
 
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  • #2
Given [tex][a_{n-1}, b_{n-1}][/tex], there are only two possibilities for [tex]a_n[/tex]. What are they? In each case, what is [tex]|a_n - a_{n-1}|[/tex]? What does this tell you about [tex]|a_n - a_m|[/tex] for [tex]m > n[/tex]?

For the second part, if the sequences [tex](a_n)[/tex] and [tex](b_n)[/tex] converge to [tex]A[/tex] and [tex]B[/tex] respectively, and [tex]b_n - a_n \to 0[/tex] as you say, what is [tex]B - A[/tex]?
 
  • #3
ystael said:
Given [tex][a_{n-1}, b_{n-1}][/tex], there are only two possibilities for [tex]a_n[/tex]. What are they? In each case, what is [tex]|a_n - a_{n-1}|[/tex]? What does this tell you about [tex]|a_n - a_m|[/tex] for [tex]m > n[/tex]?

For the second part, if the sequences [tex](a_n)[/tex] and [tex](b_n)[/tex] converge to [tex]A[/tex] and [tex]B[/tex] respectively, and [tex]b_n - a_n \to 0[/tex] as you say, what is [tex]B - A[/tex]?
1) The two possibilities: either an=an-1 or an=(an-1+bn-1)/2.
Definition: an is contractive iff there exists c<1 s.t. |an-an+1| ≤ c |an-1-an| for all n≥2.

Are you trying to suggest that we should prove that an is contractive? (since any contractive sequence is Cauchy)

For example, if we can show that |an-an+1| ≤ (1/2) |an-1-an| for all n≥2, then this implies (an) is Cauchy and we're done.
But I'm facing a problem. In our setting, it's possible that a1=a2, but a3>a2 in which case it will not be contractive?



2) Is the following a correct way to prove that they converge to a common limit?
Suppose lim an=A, lim bn=B. Then lim(an-bn)=lim(an)-lim(bn)=A-B, but we know lim(an-bn)=0, so by uniqueness of limit, A-B=0, and hence A=B.


Thanks for any help!
 
  • #4
(2) is correct.

For (1), in the case that [tex]a_3 > a_2 = a_1[/tex], what is the size of [tex]|a_3 - a_2|[/tex]? How does it compare to the possible (not actual) size of [tex]|a_2 - a_1|[/tex]? Don't use the definition of "contractive" literally; try to come up with something related that will do what you want.
 
  • #5
ystael said:
(2) is correct.

For (1), in the case that [tex]a_3 > a_2 = a_1[/tex], what is the size of [tex]|a_3 - a_2|[/tex]? How does it compare to the possible (not actual) size of [tex]|a_2 - a_1|[/tex]? Don't use the definition of "contractive" literally; try to come up with something related that will do what you want.
1)
For the case [tex]a_3 > a_2 = a_1[/tex],

[tex]|a_3-a_2|=(1/4)(b_1-a_1)[/tex]

[tex]|a_2 - a_1|=0[/tex]
So in this case, (an) is NOT contractive, and I'm in trouble...
Then how can we prove that (an) is Cauchy?

Could someone kindly explain a little more, please? (I know the definition of Cauchy, but I am not too sure how to actually do proofs with it. I've always been struggling to prove that something is Cauchy, and it's one of my biggest weaknesses and frustrations so far...)
 
  • #6
OK, so now I see that |an+1-an|≤(1/2n-1)(b1-a1). With this, how can we prove an is Cauchy?

But in the definition of Cauchy, we have |an-am|. What should we do? (I'm always confused when I get to this step when proving something is Cauchy...I just don't know what to do...)

Thank you!
 
  • #7
Suppose [tex]m,n\ge N[/tex] with [tex]n > m[/tex]. We need to show that [tex]|a_n - a_m|[/tex] is small when N is large, can you find an upperbound for this using the bound you wrote in your last post (i.e. an upperbound on [tex]|a_{i+1} - a_i|[/tex]), together with the triangle inequality?
 
  • #8
Mandark said:
Suppose [tex]m,n\ge N[/tex] with [tex]n > m[/tex]. We need to show that [tex]|a_n - a_m|[/tex] is small when N is large, can you find an upperbound for this using the bound you wrote in your last post (i.e. an upperbound on [tex]|a_{i+1} - a_i|[/tex]), together with the triangle inequality?
WHY can we assume n>m in the first place?

If n>m,
|an - am|
≤|an-an-1|+|an-1-an-2|+...+|am+1-am|
≤(1/2n-2)(b1-a1)+(1/2n-3)(b1-a1)+...+(1/2m-1)(b1-a1)
But still I'm stuck...how can we prove that it is Cauchy??

Thanks!
 
  • #9
Well if m = n, then an upperbound is 0. If m > n, then you can switch m and n without loss of generality.

Now factor [tex](b_1 - a_1)[/tex] out of the last expression, and sum the geometric sum. Then try to bound the result by something that involves N, which will allow you to show that you can make the initial expression as small as you like by making N large enough.
 
  • #10
Mandark said:
Well if m = n, then an upperbound is 0. If m > n, then you can switch m and n without loss of generality.

Now factor [tex](b_1 - a_1)[/tex] out of the last expression, and sum the geometric sum. Then try to bound the result by something that involves N, which will allow you to show that you can make the initial expression as small as you like by making N large enough.
What do you mean by without loss of generality?


Should I sum the finite geoemtric series or infinite geometric series?
If it's the finite geoemtric series, then the expression will depend on BOTH n and m, what should I do? And how can I find N?

Thank you!
 
  • #11
Either should work. The entire infinite geometric series is still an upperbound. That deals with n, now to deal with m make use of the fact that [tex]m \ge N[/tex] to find an upperbound for the resulting expression in terms of N only.
 
  • #12
Mandark said:
Either should work. The entire infinite geometric series is still an upperbound. That deals with n, now to deal with m make use of the fact that [tex]m \ge N[/tex] to find an upperbound for the resulting expression in terms of N only.

If n>m,
|an - am|
≤|an-an-1|+|an-1-an-2|+...+|am+1-am|
≤(1/2n-2)(b1-a1)+(1/2n-3)(b1-a1)+...+(1/2m-1)(b1-a1)
=(b1-a1)(1/2n-2+1/2n-3+...+1/2m-1)
<(b1-a1)(1/2m-1+1/2m+...)
=(b1-a1)(1/2m-1)(1+ 1/2+...)
=(b1-a1)(1/2m-1)[1/(1-1/2)]
=(b1-a1)(1/2m-2)
But I don't see how to find N s.t. if n,m≥N, then |a_n-a_m|<ε.


thanks for any help.
 

Related to Proving (i) for Cauchy Sequences in Completeness Theorem

1. What is a Cauchy sequence?

A Cauchy sequence is a sequence of numbers in which the terms get closer and closer together as the sequence progresses. This means that for any positive number, there exists a point in the sequence where all subsequent terms are within that distance from each other.

2. How is a Cauchy sequence different from a convergent sequence?

A Cauchy sequence is a type of convergent sequence, but not all convergent sequences are Cauchy sequences. A convergent sequence has a limit, meaning that the terms in the sequence approach a specific number as the sequence progresses. A Cauchy sequence, on the other hand, is defined by the closeness of terms to each other, not a specific limit.

3. What is the importance of Cauchy sequences in mathematics?

Cauchy sequences are important in mathematics because they are used to define the concept of convergence, which is essential in many areas of mathematics, such as calculus, analysis, and topology. They also play a crucial role in the development of the real numbers and their properties.

4. How do you determine if a sequence is a Cauchy sequence?

To determine if a sequence is a Cauchy sequence, you can use the Cauchy criterion. This states that a sequence is Cauchy if and only if for any positive number, there exists a point in the sequence where all subsequent terms are within that distance from each other. In other words, if the terms in the sequence get closer and closer together without bound, the sequence is a Cauchy sequence.

5. Can a Cauchy sequence diverge?

Yes, a Cauchy sequence can diverge. A sequence can be a Cauchy sequence without converging, meaning that the terms get closer together but do not approach a specific limit. This can happen if the terms oscillate or if the sequence is not defined on the real numbers.

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