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  • 1 Post By hollywood

Math Help - Check my proof. Please :)

  1. #1
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    Check my proof. Please :)

    So I'm studying analysis on my own which has the disadvantage that there aren't many people capable of checking proofs in my neighborhood. So if you would be so kind, I would truly appreciate it!

    Problem:
    Suppose that \{a_n\} is a Cauchy sequence. Prove that \{a_{n}^{2}\} is a Cauchy sequence.

    I came up with two proofs.

    Proof 1:
    Since \{a_n\} is a Cauchy sequence, and the real numbers are complete, \lim_{n\rightarrow \infty}a_n=a for some a\in\Re. Since the limit of a product of convergent sequences is the product of the limits of the convergent sequences, we have

    \lim_{n\rightarrow \infty}\(a_{n}^{2}\)=\({\lim_{n\rightarrow \infty}a_{n}}\)^2=a^2.

    Since \{a_{n}^{2}\} converges to a finite limit, it is a Cauchy sequence. QED.

    Proof 2:
    Since \{a_n\} is a Cauchy sequence, it converges to a finite limit. Since it converges to a finite limit, it is bounded. That is, there exists a number M so that \mid a_n\mid \leq M for all n. Using the triangle inequality, we see that

    \mid a_n+a_m \mid \quad \leq \quad \mid a_n\mid+\mid a_m \mid \quad \leq \quad 2M for all m and n.

    Let \varepsilon > 0 be given. Since \{a_n\} is a Cauchy sequence, there exists an N so that n\geq N and m\geq N implies \mid a_n-a_m \mid \leq \frac{\varepsilon}{2M}. But this means that
    \mid a_n^2-a_m^2\mid \quad = \quad \mid a_n+a_m \mid \mid a_n-a_m\mid \quad \leq \quad 2M\mid a_n-a_m \mid \quad \leq \quad \varepsilon.

    Therefore, \{a_n^2\} is a Cauchy sequence. QED.
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  2. #2
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    Re: Check my proof. Please :)

    Both of your proofs assume that you have a sequence of real numbers, but that isn't specified in the problem.

    In proof 2, you only use the completeness of the real numbers to show that the sequence is bounded. I think you can show that without assuming completeness.

    - Hollywood
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  3. #3
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    Re: Check my proof. Please :)

    Both of your proofs assume that you have a sequence of real numbers, but that isn't specified in the problem.

    In proof 2, you only use the completeness of the real numbers to show that the sequence is bounded. I think you can show that without assuming completeness.

    - Hollywood
    Hi Hollywood,

    Thank you for taking time to look at my proofs. I really appreciate it.

    I guess I should have prefaced the post with the fact that I'm studying real analysis, so it is implied that I'm dealing with a sequence of real numbers.

    In proof 2, I used completeness to show the sequence converges. Which in turn I used to show the sequence is bounded.
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