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Math Help - 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 { a_n} and { b_n} are Cauchy sequences in \mathbb{R}, prove that { a_n + b_n} and { 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 \mathbb{R} is a Cauchy sequence." So in order for me to prove { a_n + b_n} and { a_nb_n} are also Cauchy sequences, I need to prove that { a_n + b_n} and { a_nb_n} converge.

    I found this:


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

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

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

    This is where I got stuck/confused: How to prove { a_nb_n} converges using epsilon because when I multiply |a_n - a_m| |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 a_n is a Cauchy sequence we have
    \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 \\ <br />
\end{gathered}.
    In other words, Cauchy sequences are bounded.
    So your sequences are bounded. \left( {\forall n} \right)\left[ {\left| {a_n } \right| \leqslant A\,\& \left| {b_n } \right| \leqslant B\,} \right].

    Now,
    \begin{gathered}  \varepsilon  > 0\,\& \,C = A + B \hfill \\<br />
  \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} <br />
.

    \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 \\ <br />
\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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