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Homework Help: Triangle inequality for distance to set

  1. Jan 15, 2009 #1


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    1. The problem statement, all variables and given/known data
    If X is a metric space and A is a subset of X, show that

    [tex]|d(x,A) - d(y,A)|\leq d(x,y)[/tex]

    for any x,y in X.

    2. Relevant equations
    Triangle inequality:
    [tex]|d(x,z) - d(y,z)|\leq d(x,y)[/tex],

    [tex]d(x,y)\leq d(x,z)+d(z,y)[/tex]

    3. The attempt at a solution
    Fidling around with the triangle inequality.

    Edit: This is the third problem in G. Bredon's Topology and Geometry book. The goals of the exercice is to establish the continuity of the map [itex]x\mapsto d(x,A)[/itex], but she gives as a hint, "Use the triangle inequality to show that [itex]|d(x,A) - d(y,A)|\leq d(x,y)[/itex])" from which the result clearly follows in view of the continuity of the map [itex]x\mapsto d(x,x_0)[/itex] for some fixed x_0...

    The point of this edit is to bring attention to the exact wording of her hint, from which one can arguably conclude that the result should follow essentially from the triangle inequality.
    Last edited: Jan 15, 2009
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  3. Jan 16, 2009 #2


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    As I understand it, [itex]
    |d(x,A) - d(y,A)|\leq d(x,y)[/itex] should follow from the triangle inequality for d, which is [itex]|d(x,z) - d(y,z)|\leq d(x,y)[/itex] for any x, y and z in X. We know the latter holds because d is a metric. What you want to prove is technically not a triangle inequality, because d(x, A) is not really a metric, in the sense that it doesn't map pairs of points to numbers.
  4. Jan 16, 2009 #3


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    Suppose A consists of two points. Label the points x and y so that x is farther from A than y. The only nontrivial case is that x is closer to a different point of A than y. So label the points in A={a,b} so that d(b,x)>d(a,x) and d(a,y)>d(b,y). Now assume the inequality isn't true. |d(x,A)-d(y,A)|>d(x,y) becomes d(x,a)-d(y,b)>d(x,y) -> d(x,a)>d(x,y)+d(y,b)>=d(x,b). That's a contradiction. For a general set A, you should be able to set up some epsilons and find points in A that represent the equivalent of a and b.
  5. Jan 25, 2011 #4
    You can't assume that there exists points in A such that d(x,A) = d(x,a) and similarly for y. The easiest example is if A is open. Then the inf is most likely not in A.
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