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Homework Help: Find all the metrics for a set of two points

  1. Dec 17, 2011 #1
    Hi there. I'm working with some notions on functional analysis. I have this exercise, which says: Find all the metrics on a set X consisting of two points.

    So I have this set
    (I don't know if this is the right way on defining a set, so I'm just trying).

    The metric's d which I must find, should accomplish:
    M1) d is real valued, finite and nonnegative.
    M2)d(x,y)=0 if and only if x=y.
    M4)[tex]d(x,y)\leq d(x,z)+d(z,y)[/tex]

    So here is the deal, as I have only two points, (x,y) on my set, I think that M4) could never be accomplished, so there are no possible metrics for this set.

    Is this right?
  2. jcsd
  3. Dec 17, 2011 #2
    No it is not correct. The triangle inequality says something that must hold for all x,y,z.

    Firstly, it does not claim that there must exists three different points x,y and z. Even the empty set can be a metric space!!

    Secondly, we can have x=y or x=y=z. This is allowed.
  4. Dec 17, 2011 #3
    Alright. Thank you. So, how can I state that there are a finite set of possible metrics? how should I proceed? should I define an arbitrary metric such as d(x,y) and work with it to get the conditions it must accomplish?

    Thanks for your answer.
  5. Dec 17, 2011 #4
    There are infinitely many possible metrics.

    What are the possible functions [itex]X\times X\rightarrow X[/itex].
  6. Dec 17, 2011 #5
    d(x,y)=|x-y| would be one I think, right? but it asks me for all the possible metrics.

    I see that what I said is just wrong. Give me a second. I didn't realize the implications of the definition you gave, and I didn't have it in mind before.

    So I must find d(x,y) such that d(x,y)=x or d(x,y)=y, right?
  7. Dec 17, 2011 #6
    No, | | is only defined on [itex]\mathbb{R}[/itex]. You aren't working in this space here.

    In order to specify a metric uniquely, you need to specify 4 things:


    But some of these things are known.
  8. Dec 17, 2011 #7
    Alright, thanks, I'm getting some insight on your guide.

    d(x,y)=x, &/or d(x,y)=y
    d(y,x)=d(x,y) so, the condition before must be d(x,y)=x=y?
  9. Dec 17, 2011 #8
    This makes no sense at all. d(x,y) is a number, x is a point in your space. How can possible d(x,y)=x?????
  10. Dec 17, 2011 #9
    Sorry for that. I'm just starting with this topic. You're right. A point would be like a collection of numbers, I know that functions and successions can be defined as points in a set too to define a space. The thing is, as you defined it, the distance function must be included in the set. So it should belong to one of the numbers for those points? I'm sorry, I don't get it. I mean, okey, the distance must be a positive real number, and that's all.

    I see now that functions and successions seems to include the entire reals. So there will be no problem with that.
  11. Dec 17, 2011 #10
    Lets start again.
  12. Dec 17, 2011 #11
    Indeed, so for every choice of d(x,y), you got a metric right??

    What can d(x,y) be??
  13. Dec 17, 2011 #12
    I think it can be any real valued and bounded function between those points, but I think my answer doesn't make much sense. It must accomplish the triangle inequality too, is that trivial? I know this must be quiet easy, and I'm starting to desperate you, sorry for that.

    Ok, I'm stupid, it can't be any real valued function, I tried one before and it was wrong. Damn.
  14. Dec 17, 2011 #13


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    It's not clear to me that you understand that x and y are specific points in X. For example, if X={1,2}, the four lines you wrote in post 10 would be

    d(1,1) = 0
    d(1,2) = d(2,1) = …
    d(2,2) = 0

    d(1,2) is a single real number. What are the allowed values for that number?
  15. Dec 17, 2011 #14
    I'm tempted to say "1", but I think that it depends on the metric. If I say one I'd be using the euclidean metric I think.
  16. Dec 17, 2011 #15
    I think that definition is wrong, I think it would be:

    [itex]X\times X\rightarrow ℝ[/itex]
    Isn't it? I mean, the positive not infinite reals.
  17. Dec 17, 2011 #16


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    Don't be misled by the specific example. My choice of X={1,2} was completely arbitrary. X just has to have two elements, and the elements aren't necessarily numbers.

    Look at the conditions required for d to be a metric.
  18. Dec 17, 2011 #17
    Yes, that is indeed correct. I made a mistake :blushing:
  19. Dec 17, 2011 #18
    Alright, it just must be a real number, right? any metric satisfying

    d(x,x)=0, d(x,y)=d(y,x)=k (being k certain positive real constant), and d(y,y)=0, that would suffice to get a metric, right?
  20. Dec 17, 2011 #19


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    Of course it would.
    Last edited: Dec 17, 2011
  21. Dec 19, 2011 #20
    Thank you all guys :)
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