Proving Cauchy Sequences: Infinite Subsequences

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SUMMARY

The discussion focuses on proving that any infinite subsequence {xnk} of a Cauchy sequence {xn} is also a Cauchy sequence and is equivalent to {xn}. The definition of a Cauchy sequence states that |an - am| approaches 0 as m and n approach infinity. The user demonstrates that if {an} converges to a limit L, then the subsequence must also converge to L, confirming the equivalence. The user expresses concern over the simplicity of the proof, but the logic is sound, as any convergent sequence is inherently a Cauchy sequence.

PREREQUISITES
  • Understanding of Cauchy sequences and their properties
  • Familiarity with limits and convergence in real analysis
  • Knowledge of epsilon-delta definitions in mathematical proofs
  • Basic concepts of subsequences in sequences
NEXT STEPS
  • Study the formal definition of Cauchy sequences in detail
  • Explore the relationship between convergence and Cauchy sequences
  • Learn about null sequences and their implications in analysis
  • Review proofs involving subsequences and their convergence properties
USEFUL FOR

Mathematics students, particularly those studying real analysis, educators teaching sequence convergence, and anyone interested in the foundational concepts of Cauchy sequences and their properties.

clg211
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Hi,

I need to prove that any infinite subsequence {xnk}of a Cauchy sequence {xn}is a Cauchy sequence equivalent to {xn}.

My problem is that it seemed way too easy, so I'm concerned that I missed something. Please see the attachment for my solution, and let me know what you think.

Thanks.
 

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The definition of "Cauchy" sequence is that |an- am| goes to 0 as m and n go to infinity, independently. That can be stated as "Given any \epsilon> 0, there exist N such that if m> N and n> N, then |a_n- a_m|< \epsilon. If an and am are from a subsequence, then they are also in the original sequence so that must be true.

Showing that the two sequences are "equivalent" means showing that the converge to the same limit. Again, if {an} converges to L, then, for any \epsilon> 0, there exist N such that if n> N then |a_n- L|< \epsilon. If an is from a subsequence, it is from the sequence and so that is still true.

Strictly speaking, you only need to prove the second because any convergent sequence is a Cauchy sequence.
 
For the the equivalence part, I want to show that the difference of the Cauchy sequence and its subsequence is a null sequence. Let xn and xnk exist in the Cauchy sequence where xnk is also an element in the subsequence. Therefore, |xn - xnk| < epsilon for n, nk < N, so Cauchy sequence - subsequence is null. Sound legitimate? Just seems too simple.

I hear where you're coming from with the limits being the same for equivalence, but we're supposed to use the "difference is null" definition.
 

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