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Math Help - Real Analysis - The Cauchy Criterion

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    Real Analysis - The Cauchy Criterion

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

    \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)

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

    Hence this is a Cauchy sequence.
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