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Thread: Cauchy sequences proof (from Intro to Real Analysis class)

  1. #1
    Member ilikedmath's Avatar
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    Question Cauchy sequences proof (from Intro to Real Analysis class)

    Problem:
    If {$\displaystyle a_n$} and {$\displaystyle b_n$} are Cauchy sequences in $\displaystyle \mathbb{R}$, prove that {$\displaystyle a_n + b_n$} and {$\displaystyle a_nb_n$} are also Cauchy sequences.
    You cannot use the Theorem 2.6.4 which states, "Every Cauchy sequence of [tex]\mathbb{R}/MATH] converges.

    My work:
    We can't use Theorem 2.6.4, but Theorem 2.6.2 says, "Every convergent sequence of $\displaystyle \mathbb{R}$ is a Cauchy sequence." So in order for me to prove {$\displaystyle a_n + b_n$} and {$\displaystyle a_nb_n$} are also Cauchy sequences, I need to prove that {$\displaystyle a_n + b_n$} and {$\displaystyle a_nb_n$} converge.

    I found this:


    But our professor told us to prove this claim directly by using $\displaystyle \epsilon$, and think that proof above uses the theorem 2.6.4 which we are not allowed to use.

    My work so far:
    Assume {$\displaystyle a_n$} and {$\displaystyle b_n$} are Cauchy sequences in $\displaystyle \mathbb{R}$. Thus $\displaystyle \forall$ $\displaystyle \epsilon$ > 0, $\displaystyle \exists$ $\displaystyle n_o$ $\displaystyle \in$ $\displaystyle \mathbb{N}$ such that $\displaystyle |a_n - a_m|$ < $\displaystyle \epsilon$ $\displaystyle \forall$ $\displaystyle n,m \geq$ $\displaystyle n_o$. And also, $\displaystyle \forall$ $\displaystyle \epsilon$ > 0, $\displaystyle \exists$ $\displaystyle k_o$ $\displaystyle \in$ $\displaystyle \mathbb{N}$ such that $\displaystyle |b_v - b_w|$ < $\displaystyle \epsilon$ $\displaystyle \forall$ $\displaystyle v,w \geq$ $\displaystyle k_o$.

    Let $\displaystyle \epsilon$ > 0. Since $\displaystyle |a_n - a_m|$ < $\displaystyle \frac{\epsilon}{2}$, also $\displaystyle |b_v - b_w|$ < $\displaystyle \frac{\epsilon}{2}$. Thus we have $\displaystyle |a_n - a_m + b_v - b_w|$ $\displaystyle \leq$ $\displaystyle |a_n - a_m|$ + $\displaystyle |b_v - b_w|$ < $\displaystyle \frac{\epsilon}{2}$ + $\displaystyle \frac{\epsilon}{2}$ = $\displaystyle \epsilon$.

    This is where I got stuck/confused: How to prove {$\displaystyle a_nb_n$} converges using epsilon because when I multiply $\displaystyle |a_n - a_m|$ $\displaystyle |b_v - b_w|$, I get lost because I'm seeing too many different subscripts . How do I go about finishing the proof using epsilons to proof convergence so I can say the sequences are Cauchy sequences?

    Any help, suggestions, corrections, and tips are greatly appreciated.
    Thank you for your time.
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  2. #2
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    Because $\displaystyle a_n$ is a Cauchy sequence we have
    $\displaystyle \begin{gathered} \left( {\exists N_a } \right)\left[ {m,n \geqslant N \Rightarrow \left| {a_m - a_n } \right| < 1} \right] \hfill \\ \left| {a_n } \right| < 1 + \left| {a_{N_a } } \right| \hfill \\
    \end{gathered}$.
    In other words, Cauchy sequences are bounded.
    So your sequences are bounded. $\displaystyle \left( {\forall n} \right)\left[ {\left| {a_n } \right| \leqslant A\,\& \left| {b_n } \right| \leqslant B\,} \right]$.

    Now,
    $\displaystyle \begin{gathered} \varepsilon > 0\,\& \,C = A + B \hfill \\
    \left( {\exists N_a } \right)\left[ {m,n \geqslant N_a \Rightarrow \left| {a_m - a_n } \right| < \frac{\varepsilon }{{3C}}} \right] \hfill \\ \left( {\exists N_b } \right)\left[ {m,n \geqslant N_b \Rightarrow \left| {b_m - b_n } \right| < \frac{\varepsilon }{{3C}}} \right] \hfill \\ N = N_a + N_b \hfill \\ \end{gathered}
    $.

    $\displaystyle \begin{gathered} n,m \geqslant N \hfill \\ \left| {a_n b_n - a_m b_m } \right| \leqslant \left| {a_n b_n - a_m b_n } \right| + \left| {a_m b_n - a_m b_m } \right| < \left| {a_m - a_n } \right|\left| {b_n } \right| + \left| {b_m - b_n } \right|\left| {a_m } \right| < \varepsilon \hfill \\
    \end{gathered} $
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  3. #3
    Member ilikedmath's Avatar
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    So what I did was not a legitimate approach to proving the problem?
    I didn't look at the second part of Theorem 2.6.2 which states "Every Cauchy sequence is bounded." And that's what you used.

    Thanks!
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  4. #4
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    Quote Originally Posted by ilikedmath View Post
    So what I did was not a legitimate approach to proving the problem?
    ??
    No, you did the sum.
    I tried to help you with the product.
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  5. #5
    Member ilikedmath's Avatar
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    Quote Originally Posted by Plato View Post
    ??
    No, you did the sum.
    I tried to help you with the product.
    Oh, okay. Got it. Thanks again.
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