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Math Help - Equivalence relations!

  1. #1
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    Equivalence relations!

    Which of the following relation on the given sets (S) define equivalence relations? Describe the equivalence classes where appropriate:

    a) S= R3
    x~y if there is a 1 dimensional subspace of R3 which contains x and y

    b) S= R3 - {0}
    x~y if there is a 1 dimensional subspace of R3 which contains x and y

    c) S = R3
    x~y if there is a 2D subspace of R3 which contains x and y

    My Answers:

    I always get confused with equivalence relations so I just wanted to give my thoughts to see if I am on the right lines!

    a) I think this is an equivalence relation as it is reflexive, symmetric and transitive.
    I think the equivalence classes would be the individual lines that the two points lie on. However this would mean that there would be infinitely many equivalence classes so seems a bit suspect?

    b) I dont think this relation is an equivalence relation as it is not transitive. I looked at the case of a triangle with two of the lines going through the origin, but the other therefore wouldnt.

    c) I assumed this question would be similar to a. Reflexive, Symmetric and Transitive, with the equivalence classes being the individual planes that the lines lie within.

    [/b]
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  2. #2
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    Be careful with the equivalence classes.

    .
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  3. #3
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    I needed to edit this post, because I was in a hurry last time, and overlooked the zero vector. Here is the corrected solution:

    First of all, I will be assuming that your "if"s are "iff"s, and that "subspace" refers to a vector subspace. That said...

    (a) Since we can choose x,y\in\mathbb{R} with x\not\sim y (for example (1,0,0),(0,1,0)), then x,y\sim\textbf{0}, and so transitivity does not hold. So \sim is not an equivalence relation.

    (b) Reflexivity and symmetry are obvious here. For transitivity, recall that (x_1,x_2,x_3)\sim(y_1,y_2,y_3) iff (y_1,y_2,y_3)=a(x_1,x_2,x_3) for some a\in\mathbb{R} (the underlying field). Suppose (x_1,x_2,x_3)\sim(z_1,z_2,z_3). Then there are a,b\in\mathbb{R} with a(y_1,y_2,y_3)=b(z_1,z_2,z_3), which means (z_1,z_2,z_3)\sim(y_1,y_2,y_3). So transitivity holds, and \sim is an equivalence relation.

    (c) Transitivity is again the issue, here, but this time it might work even with the zero vector. Indeed, for any x,y\in\mathbb{R}^3, we have x,y\in\text{span }\{x,y\}, where \text{span }\{x,y\} is a two-dimensional vector space. So x\sim y, and transitivity holds such that \sim is an equivalence relation (and \mathbb{R}^3/\sim=\{\mathbb{R}^3\}).
    Last edited by hatsoff; January 16th 2010 at 04:20 PM.
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  4. #4
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    a) S= R3
    x~y if there is a 1 dimensional subspace of R3 which contains x and y
    First, who would say this instead of something like "the line segment (x, y) is a subset of S"?

    Anyway, I agree that (a) and (c) are equivalence relations and (b) is not.

    (a) Reflexivity and symmetry are obvious here. For transitivity, recall that iff for some .
    I am not sure about this. E.g., (0,0,1) ~ (0,1,0) because they lie on the same line, but there is no such a that (0,0,1) = a * (0,1,0).

    (b) Everything is the same from part (a). Please note that wording is important here, since there are plenty of subspaces of , but no subspaces of . In that case, is still an equivalence relation, and indeed .
    The relation ~ in (b) is not empty; it is already defined and, for example, (0,0,1) ~ (0,0,2).

    Concerning equivalence classes, in (a) it's not lines because any three points are equivalent, and similarly for (c).
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  5. #5
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    Quote Originally Posted by emakarov View Post
    First, who would say this instead of something like "the line segment (x, y) is a subset of S"?
    I think he means vector spaces. So, it is not enough that x,y be in the same line, but they have to be scalar multiples of each other. If I am mistaken in this assumption, then of course my results will be invalid.
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  6. #6
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    Quote Originally Posted by hatsoff View Post
    I think he means vector spaces. So, it is not enough that x,y be in the same line, but they have to be scalar multiples of each other. If I am mistaken in this assumption, then of course my results will be invalid.
    I think that your reading of this question is correct.
    I have been reluctant to edit this thread.

    The question is very poorly worded.
    But now I think that it means that two points are “equivalent” is they belong to the same ‘span’.

    What can we do with this?
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