# Real Analysis - The Cauchy Criterion

• Oct 22nd 2008, 06:08 PM
ajj86
Real Analysis - The Cauchy Criterion
Assume (an) and (bn) are Cauchy sequences. Use a triangle inequality argument to prove cn = |an - bn| is Cauchy.
• Oct 22nd 2008, 10:49 PM
Moo
Quote:

Originally Posted by ajj86
Assume (an) and (bn) are Cauchy sequences. Use a triangle inequality argument to prove cn = |an - bn| is Cauchy.

$\displaystyle (a_n) \text{ is a Cauchy sequence } \Leftrightarrow ~ \forall \varepsilon > 0,~\exists N_1 \in \mathbb{N},~ \forall m,n>N_1, ~ |a_m-a_n|<\varepsilon/2$
$\displaystyle (b_n) \text{ is a Cauchy sequence } \Leftrightarrow ~ \forall \varepsilon > 0,~\exists N_2 \in \mathbb{N},~ \forall m,n>N_2, ~ |b_m-b_n|<\varepsilon/2$

So :
$\displaystyle \forall \varepsilon,~ \exists N=\max(N_1,N_2),~ \forall m,n>N,~ |a_m-a_n|+|b_m-a_n|< \varepsilon/2+\varepsilon/2=\varepsilon$

$\displaystyle \varepsilon > |a_m-a_n|+|b_m-a_n|=|a_m-a_n|+|b_n-b_m|\geq |a_m-a_n+b_n-b_m|$ (by triangle inequality)

$\displaystyle \varepsilon > |(a_m-b_m) ~-~ (a_n-b_n)|=|c_m-c_n|$

Hence this is a Cauchy sequence.