Proving Cauchy Sequence Converges on Real Number Line

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  • #1
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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.



 

Answers and Replies

  • #2
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any ideas?
 
  • #3
Fredrik
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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
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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
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Yes, it is. 1/n is convergent, and thus Cauchy...
 
  • #6
lavinia
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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
Stephen Tashi
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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
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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
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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
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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
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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
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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
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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
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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
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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
Stephen Tashi
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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
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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
Fredrik
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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).
 

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