[Math] Two sequences are equivalent. Prove that one is Cauchy iff the other is Cauchy.

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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:

$|b_m-b_n|\le|b_n-a_n|+|a_n-a_m|+|a_m-b_m|$

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

Best Answer

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.

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