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Indexed Collection of Sets /

  1. Sep 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Let I denote the interval [0,[itex]\infty[/itex]). For each r [itex]\in[/itex] I, define
    Ar = {(x,y) [itex]\in[/itex] RxR : x2+y2 = r2},
    Br = {(x,y) [itex]\in[/itex] RxR : x2+y2 [itex]\leq[/itex] r2},
    Cr = {(x,y) [itex]\in[/itex] RxR : x2+y2 > r2}

    a) Determine [itex]\bigcup[/itex]r[itex]\in[/itex]IAr and [itex]\bigcap[/itex]r[itex]\in[/itex]IAr
    b) Determine [itex]\bigcup[/itex]r[itex]\in[/itex]IBr and [itex]\bigcap[/itex]r[itex]\in[/itex]IBr
    c) Determine [itex]\bigcup[/itex]r[itex]\in[/itex]ICr and [itex]\bigcap[/itex]r[itex]\in[/itex]ICr


    2. Relevant equations
    Well, I'm not sure on relevant equations in this situation, other than each one of the equations in a,b,c are of circles with radius r.

    3. The attempt at a solution
    Well, I know for A the union is all points (x,y) in the x,y plane. This is because infinitely many circles can be drawn from the origin, and their union would include all points. Now, I also know that the intersection for A is nothing, so the empty set.
    b) I'm a bit confused as to how to write my answer here. I know B contains all the points on the edge of and within the circle of radius r, drawn from the origin. So, the union, again, is every point (keep drawing circles with an increasing radius starting from the origin, and the union would include every point). Now, for the intersection, I'm confused. I know the only thing they would have in common would be the smallest circle drawn, as it would be contained within each of the larger ones. I just dont know how to write this answer / would it be the circle of radius 0 (i.e. they just have the origin in common)?
    c) The union is all points greater than the origin, correct? As, starting from 0, it would be all points greater than (0,0). The intersection, would it be the empty set, or would it be infinity? As the points included in the set are those outside the radius of the circle, so as the circles get larger and infinitely larger, the only area in common is infinity / the area not enclosed by the largest circle's circumference.

    Thanks!
     
  2. jcsd
  3. Sep 1, 2011 #2

    micromass

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    Correct.

    Indeed, the union is everything. The intersection is the origin.

    There is no point called infinity. Infinity is not a member of [itex]\mathbb{R}^2[/itex]. So the circle cannot have infinity in common.
     
  4. Sep 1, 2011 #3

    LCKurtz

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    What does "greater than (0,0)" mean. Possibly you mean "except (0,0)".
     
  5. Sep 1, 2011 #4
    LCKurtz- that is what I meant. My bad there. I guess this is why UGA added "Intro to Advanced Mathematics and Proof Writing" as a course. I know in my head what my answer is, but conveying it in proper mathematical terms is where I mess up. Thanks!

    Micromass, I see what you're saying. I guess, in this case, the answer would be the empty set, like in part A? I guess I want to check my reasoning, too. Was I correct in saying that, if circles were to be drawn with a constantly increasing radius from the origin, the only points in common would be those beyond the circumference of the largest circle?
     
  6. Sep 1, 2011 #5

    micromass

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    You don't mean circles here, do you? You mean everything "beyond" the circle?

    In that case, the answer is that there is no largest circle. So the answer is indeed the empty set.

    In the case where r is in [0,5], for example. Then you're correct in saying that the answer is everything beyond the circle with radius 5...
     
  7. Sep 2, 2011 #6
    Awesome. All clear now. Mark this one solved. Thanks for all the help!
     
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