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Inverse of a Metric

  1. Jan 29, 2013 #1
    In Topology:

    is the multiplicative inverse of a metric, a metric?

    How do we define the Schwarz inequality then?

    if ##d(x,z) ≤ d(x,y) + d(y,z)## the inverse ##1/d(x,z)## would give the opposite?
     
  2. jcsd
  3. Jan 29, 2013 #2

    chiro

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    Hey Bachelier.

    Try taking the reciprocal (also flip the inequality) and collect terms.
     
  4. Jan 31, 2013 #3
    Hi Chiro:

    I don't know if what you mean is taking the reciprocal of ##d(x,y)##

    indeed if I start with ##d(x,z) ≤ d(x,y) + d(y,z)##

    then ##\frac{1}{d(x,z)} ≥ \frac{1}{d(x,y) + d(y,z)}##

    would lead to ##d(x,z) ≤ d(x,y) + d(y,z)## again.

    But what I feel like is the need to show that: my new metric defined as ##D(x,y) = \frac{1}{d(x,y)}## respects the triangle inequality.

    ##i.e. D(x,z) ≤ D(x,y) + D(y,z)## (result A)

    but the way the metric ##D## is defined would give me

    ##D(x,z) ≥ D(x,y) + D(y,z)## not result A

    Failing this will not make ##D## a metric.
     
  5. Jan 31, 2013 #4

    chiro

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    I don't think your D metric will be a metric at all.
     
  6. Feb 1, 2013 #5
    I thought so. I heard someone talking about the inverse of a metric as being a metric on the same space.
     
  7. Feb 2, 2013 #6

    jbunniii

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    Consider the standard euclidean metric on the real numbers d(x,y) = |x-y|, and let D(x,y) = 1/d(x,y). Even overlooking the problem that this is not defined when x=y, we can find a simple counterexample. Let x = 0, y = 3, z = 1. Then D(x,y) = 1/3, D(y,z) = 1/2, and D(x,z) = 1, so the triangle inequality D(x,z) <= D(x,y) + D(y,z) is not satisfied.
     
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