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Non-equivalent metrics on R^2

  1. Mar 29, 2010 #1
    Let's denote the ordinary metric with [itex]d[/itex], so that [itex]d(a,b) = \sqrt{(a_1-b_1)^2 + (a_2-b_2)^2}[/itex], and then let [itex]e[/itex] denote some other metric.

    Is it possible to define such [itex]e[/itex] that

    [tex]
    \sup_{x\in S_e(r)} d(x,0) = \infty
    [/tex]

    for all [itex]r>0[/itex], where

    [tex]
    S_e(r) := \{x\in\mathbb{R}^2\;|\; e(x,0)=r\}.
    [/tex]

    UPDATE: Mistake spotted. I didn't intend to define [itex]S_e(r)[/itex] so that it can be an empty set with some [itex]r[/itex]. Post #4 contains a new formulation for the original idea of the problem.
     
    Last edited: Mar 30, 2010
  2. jcsd
  3. Mar 30, 2010 #2
    Let e(x,y) = 0 if x=y ;
    = 1 otherwise.
    Clearly, the circles in e are not bounded with respect to the euclidean metric.
    I'm afraid I don't get the tenor of the question at all. If you mean to have a 'continuous' metric , it's not possible. See, however, the hyperbolic plane.
     
  4. Mar 30, 2010 #3
    I made a mistake in my original formulation. I should not have tried to define a boundary of a ball like I did for [itex]S_e(r)[/itex]. I should try to reformulate the problem in terms of open balls...
     
  5. Mar 30, 2010 #4
    I'll merely ask a new question now, which should aim for the same thing...

    If [itex]e[/itex] is a discrete metric, then it has a property that

    [tex]
    B_d(0,\epsilon) \nsubseteq B_e(0,\frac{1}{2})
    [/tex]

    for all [itex]\epsilon>0[/itex].

    Now, I'm interested to know that does there exist such metric [itex]e[/itex], and radius [itex]r>0[/itex], that

    [tex]
    B_e(0,\epsilon) \nsubseteq B_d(0,r)
    [/tex]

    for all [itex]\epsilon >0[/itex]?
     
  6. Mar 30, 2010 #5
    Let e(a,b) = |ay-by| if ax= bx ;
    = |ax -bx| otherwise.
    The 'circles' around the origin are a pair of parallel lines plus two points.
     
  7. Mar 30, 2010 #6
    Set a=(0,0), b=(1,1), c=(0,10).

    Don't you now get e(a,c) = 10, e(a,b) = 1, and e(b,c) = 1, so that the triangle inequality does not hold?
     
  8. Mar 30, 2010 #7
    Sorry for the slip.
    How about
    e(a,b) = 0 if a=b;
    = 1 + |ay-by|/(|ay-by|+1 ) if ax= bx & a not equal to b;
    = 2+|ax -bx| otherwise.
    This almost answers the question with r=1 : balls with radius less than one contain only the origin.
    Note that the origin itself lies in every ball centred at it (in any metric) & we can't get rid of the origin.
     
  9. Mar 31, 2010 #8
    I noticed similar problems earlier.

    My first idea to define a metric, such that some balls would become unbounded, was

    [tex]
    e(a,b) = |a_2-b_2| + |\tan^{-1}(a_1) - \tan^{-1}(b_1)|
    [/tex]

    Now [itex]B_e(0,\frac{\pi}{2})[/itex] stretches to infinity, but with radius [itex]r < \frac{\pi}{2}[/itex] the balls become bounded, and with [itex]r\to 0[/itex] the balls become arbitrarily small, and in the end the metric is equivalent with the Euclidean metric.

    The problem is not only to come up with such metric that some balls are unbounded. They must remain unbounded when the radius approaches zero.
     
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