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Math Help - Real Anlysis: Cauchy 1

  1. #1
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    Real Anlysis: Cauchy 1

    Here is the problem:


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  2. #2
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    you now that both (a_n) and (b_n) are Cauchy, so given that m,n>N\in\mathbb{N} and \epsilon >0 we have that

    \|a_n-a_m\|<\frac{\epsilon}{2} and
    \|b_n-b_m\|<\frac{\epsilon}{2} whenever m,n>N

    now you need to look at

    \|c_n-c_m\|=\| |a_n-b_n|-|a_m-b_m|\|

    and you need to show that (using the triangle inequality)

    \|c_n-c_m\|<\|a_n-a_m\|+\|b_n-b_m\|<\epsilon

    to show that (c_m) is Cauchy
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  3. #3
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    thank you!! will give that a try
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  4. #4
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    Quote Originally Posted by Phyxius117 View Post
    Here is the problem:


    |c_{n}-c_{m}| =||a_{n}-b_{n}|-|a_{m}-b_{m}||\leq |(a_{n}-b_{n})-(a_{m}-b_{m})|= |(a_{n}-a_{m})+(b_{m}-b_{n}|\leq |a_{n}-a_{m}|+|b_{n}-b_{m}|

    And since { a_{n}} and { b_{n}} are Caychy sequences ,given ε>0 there exist natural Nos k_{1} k_{2} such that:

    for n,m \geq k_{1} ,then |a_{n}-a_{m}|<\frac{\epsilon}{2} ,and

    for n,m \geq k_{2} ,then |b_{n}-b{m}|<\frac{\epsilon}{2}.

    Now choose k = max( k_{1}, k_{2})

    e.t.c e.t.c
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