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Metric Spaces and Topology

  1. Oct 5, 2005 #1
    Hi.

    I was trying to figure out if the following is a metric space in R x R (Cartesian product).

    D[(x1,y1),(x2,y2)] = min( abs(x1-x2), abs(y1-y2) )

    I know there are four properties to confirm that the following is a metric space but I'm having trouble with the "triangle inequality" for the distance function.

    So, I have D[(x1,z1),(x2,z2)] should be less than or equal to D[(x1,y1),(x2,y2)] + D[(y1,z1),(y2,z2)].

    D[(x1,z1),(x2,z2)] = min( abs(x1-x2), abs(z1-z2) ) and basically I'm stuck on the next step. How do I configure it such that I could put it in the form of D[(x1,y1),(x2,y2)] + D[(y1,z1),(y2,z2)]?
     
  2. jcsd
  3. Oct 5, 2005 #2
    Anybody could help...?
     
  4. Oct 5, 2005 #3

    Tom Mattson

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    This isn't my area of expertise, but here's a suggestion: Try to reduce your problem to triangle inequalities involving distances in [itex]\mathbb{R}[/itex]. First let [itex]|x_1-x_2| \leq |y_1-y_2| \leq |z_1-z_2|[/itex]. In that case we have:

    [itex]d[(x_1,y_1),(x_2,y_2)]=|x_1-x_2|[/itex]
    [itex]d[(x_1,z_1),(x_2,z_2)]=|x_1-x_2|[/itex]
    [itex]d[(y_1,z_1),(y_2,z_2)]=|y_1-y_2|[/itex]

    At this point you can set up the triangle inequalities for the absolute values from the right hand sides, and then appropriately substitute the metric functions from the left hand sides.
     
  5. Oct 5, 2005 #4

    Hurkyl

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    If you're having trouble proving something, try looking for a counterexample! Even if you can't find one, the search might help you figure out the missing step of your proof.
     
  6. Oct 5, 2005 #5

    AKG

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    One thing the triangle inequality says is that if you have a right-angled triangle, then the "length of the hypoteneuse" (by length I mean D(x,y), where x and y are the endpoints of the hypoteneuse) is less than or equal to the sum of the lengths of the other two sides of the triangle. Is this the case for your D?
     
  7. Oct 5, 2005 #6

    benorin

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    I have an analysis book in which a distance function like the one you gave, only with max instead of min, is a metric, if that helps
    .
     
  8. Oct 5, 2005 #7

    benorin

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    It's NOT a metric, it's not positive definite. For if x1=x2 and y1 > y2, then D[(x1,y1),(x2,y2)] = 0 but (x1,y1) does not equal (x2,y2).
     
    Last edited: Oct 5, 2005
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