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Show that this is a metric.

  1. Feb 1, 2014 #1
    1. The problem statement, all variables and given/known data
    A sequence [itex]\{x_{n}\}[/itex] of real numbers is called bounded if there is a number M such that |[itex]x_{n}[/itex]| ≤M for all n. Let X be the set of all bounded sequences, show that
    [itex]d(\{x_{n}\},\{y_{n}\})=sup \{|x_{n}-y_{n}| :n \in N \}[/itex] is a metric on X.

    The only part I am struggling with is the transivity part. As I see it, I have to show that:

    [itex]sup \{|x_{n}-y_{n}| :n \in N \}≤sup \{|x_{n}-z_{n}| :n \in N \}+sup \{|z_{n}-y_{n}| :n \in N \}[/itex]

    Do you guys have any tips on how to show this? The problem is that I can not be sure that I have an n where I get the sup value, and even if I did, this n might be different for the three parts.
    Last edited: Feb 1, 2014
  2. jcsd
  3. Feb 1, 2014 #2


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    Start with ##|x_n-y_n|\le |x_n-z_n|+|z_n-y_n|## and take the sup of both sides. Then compare what that gives with the right side of your expression.
  4. Feb 2, 2014 #3
    It became a little messy, can I simplify?

    |x_{n}-y_{n}|≤ |x_{n}-z_{n}|+|z_{n}-y_{n}| \forall n ## (1)
    |x_{n}-z_{n}| ≤ sup\{|x_{n}-z_{n}| : n \in N \} \forall n ## (2)
    ##|z_{n}-y_{n}| ≤ sup\{|z_{n}-y_{n}| : n \in N \} \forall n ## (3)

    So if I put 2 and 3 in 1 I get:
    ##|x_{n}-y_{n}|≤ sup\{|x_{n}-z_{n}| : n \in N \} + sup\{|z_{n}-y_{n}| : n \in N \} \forall n ## (4)

    Hence ##sup\{|x_{n}-z_{n}| : n \in N \} + sup\{|z_{n}-y_{n}| : n \in N \} ## is an upper bound for ##\{|x_{n}-y_{n}| :n\in N\}##. So it must also be bigger than the least upper bound for this set. Hence:
    ##sup\{|x_{n}-y_{n}| : n \in N \} ≤ sup\{|x_{n}-z_{n}| : n \in N \} +sup\{|z_{n}-y_{n}| : n \in N \} ##

    Is this the simplest way do it, or is it a simpler way?
    Last edited: Feb 2, 2014
  5. Feb 2, 2014 #4


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    There may be slightly nicer ways to write it up but I see nothing wrong with your argument.
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