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Abstract math prove involwing sets

  1. Feb 27, 2012 #1
    1. The problem statement, all variables and given/known data
    Let Ts denote the set of points in the x; y plane lying on the square whose
    vertices are (-s; s), (s; s), (s;-s), (-s;-s), but not interior to the square. For
    example, T1 consists of the vertices (-1; 1), (1; 1), (1;-1), (-1;-1) and the
    four line segments joining them. Let
    S = union of Ts, where s is an element of positive real numbers
    Determine a set J, that is not de ned in terms of unions, that equals S. Prove
    that S and J are equal.
    Please help. I have no idea how to start that problem. What I figure out is that J=AuBuCuD, where set A={ (x,y)| (x,y)=(s,y), for -s≤y≤s} B={(x,y)| (x,y)=(-s,y), for -s≤y≤s}, C={(x,y)| (x,y)=(x,s), for -s≤x≤s}, D={(x,y)| (x,y)=(x,-s), for -s≤x≤s}
    I do not know how to write down set S in different form and how to prove that J=S

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 27, 2012 #2

    jbunniii

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    Try explaining in words what points of the plane are contained in [itex]\cup T_s[/itex]. Alternatively, it might be easier to identify what points are NOT contained in the union.
     
  4. Feb 28, 2012 #3

    Deveno

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    here are a couple of hints:

    suppose (x,y) lies on some square Ts. if (x,y) is in the 4th quadrant, then (x,-y) is in the 1st quadrant, on the same square Ts. if (x,y) is in the 3rd quadrant, then
    (-x,-y) is on the same square in the first quadrant, if (x,y) is in the 2nd quadrant, then (-x,y) is in the 1st quadrant.

    so you may as well just consider (x,y) in the first quadrant, from symmetry considerations.

    now show that if x = y, (x,y) is the corner of some square (which one?), if x > y, (x,y) is on the right vertical side of some square (again: which one, that is...what is s?), and if y < x, then (x,y) is on the top edge of some square (and s is...?).

    of course, not every point of the plane is in a quadrant. some points lie on the x-axis, and some points lie on the y-axis. consider these points seperately.

    last hint: is a square with sides of length 0 allowed?
     
  5. Feb 28, 2012 #4
    So in my set s cannot equal -s. Would that take care of the problem with 0 side length? But in that problem it does not say that it can't but it says that s is a positive real number, so 0 it is not included because 0 is not negative nor positive.
     
    Last edited: Feb 28, 2012
  6. Feb 29, 2012 #5

    Deveno

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    you are correct that 0 is not a positive real number.
     
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