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Equivalent Metrics

  1. May 19, 2008 #1
    1. The problem statement, all variables and given/known data
    Let [itex] (X,d) [/itex] be a metric space. Show that d is equivalent to the discrete metric if and only if every point in X is an isolated point

    2. Relevant equations

    Two metrics [itex] d, d' [/itex] are said to be equivalent if they have to same convergent sequences with the same limit points

    The discrete metric is [itex] d(x,y) = \begin{cases} 0 & \text{ if } x = y \\ 1 & \text{otherwise} \end{cases} [/itex]

    [itex] a \in X [/itex] is an isolated point if [itex] \exists r>0 [/itex] such that [itex] B(a;r) = \{a\} [/itex].

    3. The attempt at a solution

    I might just be getting hung up on technicalities but I can't really think of how to attack either end of the problem.

    In the [itex] \Rightarrow [/itex] direction, my problem is that I have to show it for any metric that is equivalent to the discrete metric - so I probably need to use something about convergent sequences under the discrete metric. But I can't really think of anything useful.

    In the [itex] \Leftarrow [/itex] direction, I wanted to use contradiction somehow. Since in a discrete metric space every subset is necessarily open, I want to say something about a non-discrete metric -equivalent having a closed subset, but I don't know if that's really true or not.

    I've seen in some literature a definition that states that a metric space can be broken up into a set of limit points and isolated points. In which case they say that a discrete metric space has no limit points. But I find this confusing, since isn't every element a limit point of a constant sequence?

    I would greatly appreciate any input.
     
  2. jcsd
  3. May 19, 2008 #2

    Dick

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    The only way you can have a convergent sequence x_n->a in the discrete metric is if there is an N such that for all n>N, x_n=a. Isn't that right? Start with that.
     
  4. May 19, 2008 #3
    I tried thinking about a way to use that earlier but I'm not terribly sure how it applies since I can't limit myself to only the discrete metric. I have to consider any metric that is equivalent to the discrete metric and I'm not entirely sure there's any way to relate the "constant subsequence" part to any other metric since we're not guaranteed an isometric mapping between the metrics exists.
     
  5. May 19, 2008 #4

    Dick

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    Nobody said they were isometric. But you said "Two metrics are said to be equivalent if they have to same convergent sequences with the same limit points." If the discrete metric has that property then so does the other one. Right?
     
  6. May 20, 2008 #5
    An excellent point. So I've been playing around with it a bit, and I've faced myself with two possible methods of attack.

    1) Let m be the minimal index such that [itex] x_m \neq a [/itex] and [itex] x_n = a \quad \forall n > m [/itex]. I then take [itex] r = d(x_m, x_{m+1}) [/itex] and create the open ball with that radius.

    However, I really don't like this method since there's nothing guaranteeing that there isn't a point that is "closer" to a.

    2) Let [itex] r = \min\{ d(x_n,a) | x_n \neq a \} [/itex] and construct the ball using this. I think the min is guaranteed to exist since the sequence must be constant after a finite number of indices.

    So is two the better method?

    Edit: Also, do you think my idea for the converse will work?
     
    Last edited: May 20, 2008
  7. May 20, 2008 #6

    Dick

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    Ok, so you know all convergent sequences of d are eventually constant. Suppose there is a point 'a' which is NOT isolated. Can you use that to construct a sequence converging to 'a' that is NOT eventually constant? That would show every point IS isolated. Now supposed that indeed every point IS isolated. What do you conclude about all convergent sequences?
     
  8. May 20, 2008 #7

    Dick

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    These both have the problem that you can't confine yourself to thinking about a single sequence converging to a. There may be other closer points in a DIFFERENT sequence.
     
  9. May 20, 2008 #8
    Ok, I think I should be able to work it from there. Thanks
     
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