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Thread: Metric space

  1. #1
    Will power vvaann 's dreams are becoming reality. vvaann's Avatar
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    Metric space

    For every set S and every metric d on S, which of the following is a metric on S?

    A. 4 + d
    B. exp(d) - 1
    C. d - |d|
    D. d^2
    E. sqrt(d)

    Answer:
    SPOILER: E
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    Within my grasp! matroid radiates success. matroid's Avatar
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    Re: Metric space

    It's E by definition.

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    Re: Metric space

    What is the definition actually??
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    Within my grasp! matroid radiates success. matroid's Avatar
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    Re: Metric space

    Well, well, my favorite topology textbook says (without the important, but now not so very important details), d is a metric if it satisfies the following 2 axioms:

    i) d(a, b) = 0 if and only if a=b
    ii) d satisfies the triangle-inequality, ie. d(a, b) + d(a, c) >= d(b, c)

    Obviously, the second axiom is the tricky one. But using only the first axiom one can immediately eliminate the answers A and C.

    B and D will be wrong. Since these are convex functions, one can quickly find counterexamples. So E is just fine. (I hate selecting an answer by eliminating the other four... )

    Md.

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    Will power vvaann 's dreams are becoming reality. vvaann's Avatar
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    Re: Metric space

    Great again, matroid .

    BTW, we can show the second axiom satisfied by the answer E as follows:

    We have: d(a,b) + d(b,c) > d(a,c)

    <=> sqrt(d(a,b))^2 + sqrt(d(b,c))^2 > sqrt(d(a,c))^2

    => sqrt(d(a,b))^2 + sqrt(d(b,c))^2 + 2*sqrt(d(a,b))*sqrt(d(b,c))> sqrt(d(a,c))^2

    => [sqrt(d(a,b)) + sqrt(d(b,c))]^2 > sqrt(d(a,c))^2

    => sqrt(d(a,b)) + sqrt(d(b,c)) > sqrt(d(a,c))

    => sqrt(d) is a metric
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    Socratic realist Dragonfinity is a TestMagic guru. Show your respect! Dragonfinity is a TestMagic guru. Show your respect! Dragonfinity's Avatar
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    Re: Metric space

    There were 2 very important parts left our of the definition above, specifically that the function d is always greater than or equal to zero, and the function is not affected by the order of the points. The famous Munkres text on Topology gives the following definition:

    DEFINITION: A metric on a set Xis a function d:X x X -> R ("distance") such that
    (1) d(x,y) >= 0, for all x,y in X AND d(x,y) = 0 iff x = y.
    (2) d(x,y) = d(y,x)
    (3) d(x,y) + d(y,z) >= d(x,z)

    (a) is not a metric because d(x,x) = 4
    (b) is clearly not a metric because it violates the non-negativity of a metric
    (c) is not a metric because d(x,y) is nonnegative, so |d| = d, and d - |d| = 0, even if x is not equal to y
    (d) is not a metric because of the triangle inequality: Let S be the real number line, and let d be the standard metric, and denote d^2 by D. Then, D(1,2) + D(2,4) = 1 + 4 < 9 = D(1,4), in violation of the triangle inequality.

    It is unecessary to prove (e) as the question states that one of the choices is a metric and we have shown the other 4 are not metrics. Nevertheless, letting D denote sqrt(d):

    (1) d(a,b) >= 0 implies that D(a,b) >= 0, since we always assume that the square root function takes the positive square. Further, let D(a,b) = 0, then D^2 = 0, so d(a,b) = 0, and a = b. Similarly, if a = b, d(a,b) = 0, so D(a,b) = 0.
    (2) d(a,b) = d(b,a) implies sqrt(d(a,b)) = sqrt(d(b,a)) >= 0, again since the sqrt function takes the positive square; hence, D(a,b) = D(b,a).
    (3) vvann's proof looks correct, except he used > instead of >=; however, that substitution can be made in every line w/o altering the proof (including the addition of 2*sqrt(d(a,b))*sqrt(d(b,c))>, as we do not know if a = b or b = c).

    C

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    Within my grasp! matroid radiates success. matroid's Avatar
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    Re: Metric space

    Quote Originally Posted by Dragonfinity
    There were 2 very important parts left our of the definition above, specifically that the function d is always greater than or equal to zero, and the function is not affected by the order of the points. The famous Munkres text on Topology gives the following definition:

    DEFINITION: A metric on a set Xis a function d:X x X -> R ("distance") such that
    (1) d(x,y) >= 0, for all x,y in X AND d(x,y) = 0 iff x = y.
    (2) d(x,y) = d(y,x)
    (3) d(x,y) + d(y,z) >= d(x,z)
    I don't think I left out anything. What you write is correct, but if you write the triangle inequality the way I quoted, than the 2 "missing" properties follow from the other two axioms.

    For example, if we substitute b=c in ii) we get:
    d(a,b) + d(a,b) >= d(b,b) = 0, so 2d(a,b) >=0 which yields that d(a, b) >= 0

    Also if we substitute a=c we get:
    d(a,b)+d(a,a) >= d(b,a), so d(a,b)>=d(b,a). Since this holds for all a and b, it also holds when we swap the two variables. So d(a,b)>=d(b,a)>=d(a,b), and so we have proven that the function is not affected by the order of the points.

    Md.

  8. #8
    Socratic realist Dragonfinity is a TestMagic guru. Show your respect! Dragonfinity is a TestMagic guru. Show your respect! Dragonfinity's Avatar
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    Re: Metric space

    Interesting Matroid, thanks. Never really thought about it. Your reasoning is sound. Although defining the metric in this fashion makes more difficult the use of variants, such as quasi-metrics (which is obtained by dropping the property of symmetry).

    C

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