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Symmetry and Transitivity

  1. Nov 17, 2015 #1


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    The source I'm using is:
    http://inperc.com/wiki/index.php?title=Homology_classes [Broken]
    And they say
    Symmetry: A∼B⇒B∼A . If path q connects A to B then p connects B to A ; just pick p(t)=q(1−t),∀t .
    Transitivity: A∼B , B∼C⇒A∼C . If path q connects A to B and path p connects B to C then there is a path r that connects A to C ; just pick:
    ##r(t)= \left [ \begin{array}{c}
    q(2t) & \text{for} & t∈[0,1/2],\\
    p(2t−1) & \text{for} & t∈[1/2,1] \\
    \end{array} \right ] ##

    and the problem I'm having is with the symmetry part.
    Using their function definition, p(t) = q(1-t) for all t:
    Let the two points A, and B, in question be (0,0) and (1,1), respectively, connected by the path q(t) = t
    Then the function p(t)=q(1-t) should give me a line from (1,1) to (0,0) (B->A), but plugging it in gives me p(t) = q(1-t) = 1-t.
    Neither point is on that line. I'm assuming I'm missing something, here, but I don't know what.


    *I should add, that it makes sense to me, conceptually, but this example just doesn't seem to work for me, and I want to make sure I have a solid foundation before continuing.
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Nov 17, 2015 #2


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    Since you take two points in ##\mathbb R^2##, your path should also be a function into ##\mathbb R^2##.
    So you have ##q(t)=(t,t)## and then ##p(t)=(1-t,1-t)##
  4. Nov 17, 2015 #3


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    Oh, wait, I think I get it now.
    So the function needs to be on t = [0,1] and [1,0] respectively, so when you plug in 0 and 1 in q(t)=t you get 0 and 1 out, so (0,0) and (1,1) and when you plug in [1,0] into p(t)=1-t you get 1-1 = 0 and 1-0 = 1.

    Is this correct?
  5. Nov 17, 2015 #4


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    Ok, yea, for some reason I wasn't thinking parametrically D=
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