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Let $ϵ>0$ be given. With loss of generality, we may assume $ϵ$ is rational. Suppose $a_n$ is a Cauchy sequence and $b_n, a_n$ are equivalent.

Choose N such that $|b_k−a_k|<\frac13\epsilon$ and $|a_n−a_m|<\frac13\epsilon$ whenever $k,m,n≥N.$ We can do this because an is Cauchy, and $a_n,b_n$ are equivalent.

Then if $n,m≥N$, we have

$|b_n−b_m|≤|b_n−a_n|+|a_n−a_m|+|a_m−b_m|≤\frac13\epsilon+\frac13\epsilon+\frac13\epsilon=ϵ.$ Hence $b_n$ is Cauchy.

Reversing the roles of $a_n,b_n$ finishes the proof.

What I cannot follow in the answer is why we can assume:


Am I missing something huge here or I am out of the loop on some standard inequality?

  • $\begingroup$ This is an easy inequality to prove. It's usually given as an exercise. What have you tried? $\endgroup$ – user122283 May 7 '14 at 3:43
  • $\begingroup$ What does "two sequences are equivalent" mean? That their difference converges to zero? $\endgroup$ – DonAntonio May 7 '14 at 3:47
  • $\begingroup$ Yes. That their distance from each other eventually gets less than any rational number. $\endgroup$ – atecce May 7 '14 at 3:50
  • $\begingroup$ Thanks @atecce I was just about to post my proof of this problem, but thanks to you i dont have to type it in. $\endgroup$ – Atif Farooq Jun 18 '18 at 22:20

This is a standard trick in analysis-y proofs: Add $0$ creatively, and use the triangle inequality. In particular, we have

$$|a - b| = |a - (c - c) - b| = |(a - c) + (c - b)| \le |a - c| + |c - b|$$

for any $a, b, c$. This can be generalized to $3$ (as in your case) or more terms.


It is the triangle inequality:


  • $\begingroup$ Thank you both! Sorry it was so easy for you guys. $\endgroup$ – atecce May 7 '14 at 3:52

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