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Thread: 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 $\displaystyle (a_n)$ and $\displaystyle (b_n)$ are Cauchy, so given that $\displaystyle m,n>N\in\mathbb{N}$ and $\displaystyle \epsilon >0$ we have that

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

    now you need to look at

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

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

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

    to show that $\displaystyle (c_m)$ 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:


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

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

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

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

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

    e.t.c e.t.c
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