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Analysis: is a particular function a metric

  1. Feb 10, 2010 #1
    1. The problem
    (this is from Rudin's Principles of Mathematical Analysis, p. 43 #11)
    Determine if the following is a metric

    d(x,y) = abs(x-y)/(1 + abs(x-y))
    where the set in question is the real number line (R1)

    2. Relevant equations
    the definition of a metric on a set requires that it be:
    (1) positive definite
    (2) symmetric
    (3) satisfies the triangle inequality, that is d(x,y) ≤ d(x,r) + d(r,y) for all x, y, r

    3. The attempt at a solution

    the positive definiteness of d(•,•) is apparent since both the numerator and denominator are positive for all x, y such that x≠y. and 0 when x=y.

    The symmetry likewise follows quickly from the symmetry of the absolute value function.

    The triangle inequality is more difficult:
    the problem is given x,y, z in R1 is the following satisfied:

    abs(x-y)/(1+ abs(x-y)) ≤ abs(x-r)/(1+ abs(x-r)) + abs(r-y)/(1+abs(r-y).

    If we look only at the numerator we see that this is the ordinary triangle inequality for the standard metric on R1. I started with the simple case with y=-x and r=0
    which gives:

    2|x|/(1+|2x|) for the left hand side and:
    |x|/(1+|x|) +|x|/(1+|x|) for the right hand side which reduces to:
    2|x|/(1+|x|) which is obviously greater than the left (since the denominator for the LHS is strictly greater than that for the RHS)

    when I branch out from this very simple case I can't either compare the quantities or they conform to the triangle inequality (I can post the other values I used if they help at all).
  2. jcsd
  3. Feb 11, 2010 #2
    Hey guys im having trouble with this proof for my real analysis homework. The problem is

    Translate and Prove:
    Say (Epsilon)>0. For some (Delta)>0, if abs.x<(Delta), then [(abs.x)^1/2.(abs.cos(x))]<(Epsilon).
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