Proving Cauchy Sequence Converges on Real Number Line

In summary: Then you have to prove that x is the supremum. You can do this by contradiction, and the argument will be along the lines of: If x were not the supremum, then you could find a member of the sequence that is greater than your x. But if you take any of those intervals that I mentioned, you can see that it must contain that member, and it must also contain all the members that come after it, since they're closer to x than that member is. So you end up with infinitely many members of the sequence inside a fixed interval, which is not possible. (You would have to use the fact that the sequence is Cauchy somewhere in this argument).It's been a long time since I've seen
  • #1
Bachelier
376
0
I know about the proof using lim inf and lim sup and the proof using a convergent subsequent, however I thought about this proof. Can you tell me if it is correct, and if not why?

Thank you



let Sn be Cauchy seq in R

Let S be its range. Then S is bounded.

Since R is complete, sup S exists. Let x= sup S

then for all ε > 0, ∃ N1 in N st
x - ε/2 <SN1 <= x​
⇒ x - ε/2 <SN1 < x + ε/2

so d(SN1, x) < ε/2​
now for all ε > 0, ∃ N2 in N st for all n=>N2 implies d(SN2, Sn)<ε/2

let N = max (N1, N2)

then d(Sn, x) <= d(SN, Sn) + d(SN, x)< ε

Hence Sn converges to x.

and we're done.



 
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  • #2
any ideas?
 
  • #3
You should be very suspicious of the result since you found that an arbitrary Cauchy sequence converges to its least upper bound. Think about 1/n for example.
 
  • #4
Fredrik said:
You should be very suspicious of the result since you found that an arbitrary Cauchy sequence converges to its least upper bound. Think about 1/n for example.

But 1/n is not Cauchy.
 
  • #5
Yes, it is. 1/n is convergent, and thus Cauchy...
 
  • #6
You assume that the real numbers are complete. By definition a complete metric space is one in which every Cauchy sequence converges. So it seems that you assumed what you wanted to prove.

The real numbers may be defined as the completion of the rationals under the absolute value metric. If you take this as the definition, then maybe you still need to prove that a Cauchy sequence of real numbers is equivalent to a Cauchy sequence of rationals.
 
  • #7
then for all ε > 0, ∃ N1 in N st
x - ε/2 <SN1 <= x

micromass's example shows this is not true. Perhaps your are thinking of "lim sup" instead of "sup".
 
  • #8
micromass said:
Yes, it is. 1/n is convergent, and thus Cauchy...

Here's why I don't think it's Cauchy.

Assume Sn = 1/n seq in R

assume n>m , and n, m are large enough natural numbers

now |Sn - Sm| = [1/(m+1) + 1/(m+2) + ...+ 1/n] > (n-m)/n

now let n = 2m. then |Sn - Sm| > 1/2

Hence for large enough values (bigger than some pos. integer N) , I found 2 sequences such that d(Sn , Sm) is always bigger than 1/2.

hence 1/n is not Cauchy.
 
  • #9
Bachelier said:
now |Sn - Sm| = [1/(m+1) + 1/(m+2) + ...+ 1/n]

Why is this true? To my knowledge, we have [tex]|S_n-S_m|=|1/n-1/m|[/tex].

I think you're being confused with the harmonic series, which is something completely different. And this series indeed is not Cauchy. But here we're working with the sequence 1/n...
 
  • #10
Stephen Tashi said:
micromass's example shows this is not true. Perhaps your are thinking of "lim sup" instead of "sup".

How so?

for ε= π

then x - π/2 < 1/n <= x
 
  • #11
micromass said:
Why is this true? To my knowledge, we have [tex]|S_n-S_m|=|1/n-1/m|[/tex].

I think you're being confused with the harmonic series, which is something completely different. And this series indeed is not Cauchy. But here we're working with the sequence 1/n...

yeah, while you were typing your answer, I thought about it.

I was mixing up sequences. The one I used in my proof is the following:

1 + 1/2 + 1/ 4 + ... + 1/n (harmonic series indeed : Σ 1/k)

You are right. 1/n is cauchy. :smile:
 
  • #12
micromass said:
Why is this true? To my knowledge, we have [tex]|S_n-S_m|=|1/n-1/m|[/tex].

I think you're being confused with the harmonic series, which is something completely different. And this series indeed is not Cauchy. But here we're working with the sequence 1/n...

so micromass, what do you think about the proof?
 
  • #13
Bachelier said:
so micromass, what do you think about the proof?

Like Frederik and others point out, there has to be something wrong with the proof since the result isn't correct. 1/n is cauchy and doesn't converge to it's supremum.

Try to work out thesame proof with 1/n instead of Sn to see where it goes wrong...
 
  • #14
lavinia said:
You assume that the real numbers are complete.

I did, but I used the fact that in a complete ordered field such as R, a bounded sequence (such as cauchy, proof is easy and can be added to the whole proof) has a supremum.
 
  • #15
micromass said:
Like Frederik and others point out, there has to be something wrong with the proof since the result isn't correct. 1/n is cauchy and doesn't converge to it's supremum.

Try to work out thesame proof with 1/n instead of Sn to see where it goes wrong...

good point. Thanks
 
  • #16
Bachelier said:
How so?

for ε= π

then x - π/2 < 1/n <= x

For the sequence {1/n} the supremum of the range (as a set of numbers) is 1, not zero.
 
  • #17
Thanks Stephen. I found the fallacy in my argument.

I think this argument will only work for bounded non-decreasing (increasing) sequences.

The fallacy is in assuming that d(SN, Sn) is less than ε/2 (for N = max (N1, N2)).
 
  • #18
Bachelier said:
The fallacy is in assuming that d(SN, Sn) is less than ε/2 (for N = max (N1, N2)).
Right, because [itex]d(S_{N_1},x)<\varepsilon/2[/itex] doesn't imply [itex]d(S_N,x)<\varepsilon/2[/itex].

I think the first thing you should do is another exercise, which I believe you will find quite easy: Prove that every convergent sequence is Cauchy. (Just use the definitions and the triangle inequality).

Now, regarding what you're trying to prove, I think that you should be looking for a sequence of closed intervals such that each contain all but a finite number of members of the Cauchy sequence, and such that the lengths of those intervals go to zero. Then you define your x as the intersection of all those intervals. (You either have to prove that such an intersection is non-empty, or refer to a theorem that tells you that it is).
 

1. What is a Cauchy sequence?

A Cauchy sequence is a sequence of numbers where the difference between any two terms can be made arbitrarily small by choosing terms far enough along the sequence. It is named after the French mathematician Augustin-Louis Cauchy.

2. How is the convergence of a Cauchy sequence proven?

To prove that a Cauchy sequence converges, we need to show that the terms of the sequence approach a single limit, or a specific value. This can be done by showing that the terms get closer and closer to each other, or by using the Cauchy convergence criterion, which states that a sequence is convergent if it is a Cauchy sequence.

3. What is the importance of proving the convergence of a Cauchy sequence?

Proving the convergence of a Cauchy sequence is important because it allows us to understand the behavior of a sequence of numbers and make predictions about its limit. This is crucial in many areas of mathematics, including calculus, analysis, and number theory.

4. Can a Cauchy sequence converge to a non-real number?

No, a Cauchy sequence can only converge to a real number. This is because the Cauchy convergence criterion only applies to sequences of real numbers, and the concept of convergence is defined in terms of real numbers.

5. Are all convergent sequences Cauchy sequences?

No, not all convergent sequences are Cauchy sequences. While all Cauchy sequences are convergent, there are sequences that are convergent but not Cauchy. For example, the sequence {1, 1/2, 1/3, 1/4, ...} converges to 0, but it is not a Cauchy sequence because the terms do not get arbitrarily close to each other.

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