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Bounded Metric

  1. Feb 16, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove that [itex] (\mathbb{R},d) [/itex], [itex] d = \frac{\mid x - y \mid}{1 + \mid x - y \mid} [/itex] is a complete metric space.


    2. Relevant equations



    3. The attempt at a solution
    If [itex] d_u = \mid x - y \mid [/itex], then I can prove this for the Cauchy sequences in [itex] (\mathbb{R},d) [/itex] that are also Cauchy in [itex] (\mathbb{R},d_u) [/itex]. But there may be Cauchy sequences in [itex] (\mathbb{R},d) [/itex] that are not Cauchy in [itex] (\mathbb{R},d_u) [/itex].
     
  2. jcsd
  3. Feb 16, 2013 #2

    Dick

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    Can you give me an example of one? I don't really think there are any.
     
  4. Feb 16, 2013 #3
    I couldn't find one either. How can I prove there are none i.e., there is one-to-one correspondence between the Cauchy sequences of each space?
     
  5. Feb 16, 2013 #4

    Dick

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    Suppose you could find constants m and M such that ##md(x,y) \le d_u(x,y) \le Md(x,y)##? You would have trouble with the M part of that, but remember when you are dealing with Cauchy sequences you only have to worry about small values of |x-y|.
     
  6. Feb 16, 2013 #5
    I don't quite understand what you are getting at, but I have that for n,m sufficiently large and [itex] x_n [/itex] Cauchy in (R,d), [itex] \frac{\mid x_n - x_m \mid}{1 + \mid x_n - x_m \mid} < \epsilon \rightarrow \mid x_n - x_m \mid < \frac{1}{1 - \epsilon} [/itex]. So [itex] x_n [/itex] is Cauchy in (R,d_u).
     
  7. Feb 16, 2013 #6
    Does this work?
     
  8. Feb 16, 2013 #7

    jbunniii

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    I don't see how that works, because ##1/(1-\epsilon)## does not approach 0 as ##\epsilon \rightarrow 0##.

    Suppose that
    $$\frac{|x-y|}{1 + |x-y|} \leq \frac{1}{2}$$
    Then we have
    $$\begin{align}
    |x-y| &= \frac{|x-y|}{1+|x-y|} \cdot (1 + |x-y|) \\
    &= \frac{|x-y|}{1+|x-y|} + \frac{|x-y|}{1+|x-y|} \cdot |x-y| \\
    &<= \frac{|x-y|}{1+|x-y|} + \frac{1}{2}|x-y| \\
    \end{align}$$
    So
    $$|x-y| \leq 2\frac{|x-y|}{1 + |x-y|}$$
    This should allow you to conclude that Cauchy in ##d## implies Cauchy in ##d_u##.

    By the way, did you already prove that ##d## is a metric?
     
  9. Feb 16, 2013 #8

    Dick

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    junniiii as already done a fine job of pointing out why it doesn't work. My point was that you only have to worry around small values of ε. If |x-y|<1 then d and d_u are almost the same. They differ by a factor of at most 2.
     
  10. Feb 17, 2013 #9
    Dick: Ok I understand now.

    jbunniii: Yes I did prove d is a metric.
     
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