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Math Help - Sequence related proof

  1. #1
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    Sequence related proof

    Suppose that \{ X_{n}\}_{n=1}^{\infty} is a sequence of real numbers that converges to x_{0} and that all x_{n} and x_{0} are non-zero.

    (i) Prove that there is a positive number B that |x_{n}|\geq B for all n.

    (ii)Using part (i), prove that \{\frac{1}{x_{n}}\} converges to \{\frac{1}{x_{0}}\}


    OK, so here is how I did part (i):

    Let |x_{s}| be the smallest absolute value in the given sequence, i.e. |x_{s}|=min(|x_{n}|) \forall n
    For \epsilon\geq 0,
    In N_{\epsilon}(|x_{s}|) there exists a number y such that y\leq x_{s}\to y\leq |x_{n}| for all n.

    (ii) [I have no idea how to do this one -- or even that the first part is right. So any help will be greatly appreciated. Thanks in advance]
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  2. #2
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    I'll give a couple of very nitpicking remarks.

    Quote Originally Posted by ashimb9 View Post
    Let |x_{s}| be the smallest absolute value in the given sequence
    When one says "let x be y", usually x is a fresh variable, a new name for the described object y. If x is not just a name, but an expression, then it is not clear that the right-hand side y can have the form x. For example, consider "let (x+2)(x-1) be -2". Does such x exist, and if it does, is it unique? So, wrapping x_s in the absolute value function, as in your case, is suspicious from the start.

    One should say, "Let s be such that |x_s|\le |x_n| for all n". Here, again, one has to describe why such s exists. And, in fact, it does not have to: consider x_n=1/n. In general, a set of real numbers bounded from below always has an infimum but does not always have a minimum (which by definition has to be an element of the set). This is an essential point of this problem.

    For \epsilon\geq 0,
    In N_{\epsilon}(|x_{s}|) there exists a number y such that y\leq x_{s}\to y\leq |x_{n}| for all n.
    It's not clear to me what N_{\epsilon}(|x_{s}|) is. Also, if such y exists, why is it positive, as required by the question?

    A correct way is to find N such that, say, |x_n|>1/2|x_0| for all n > N. Then the initial segment \{x_n\}_{n=0}^N is finite, and finite sets have not only infimum but minimum as well. Then you can find the positive lower bound for all |x_n|.

    For (ii), try to find an upper bound on |1/x_n-1/x_0| in terms of |x_n-x_0|, B, and |x_0|.
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