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Prove that a set is an open set

  1. Apr 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that {(x,y)ℝ^2/1<x^2+y^2<7} is an open set.


    2. Relevant equations



    3. The attempt at a solution

    =(
     
  2. jcsd
  3. Apr 19, 2012 #2
    Sometimes a picture helps me a lot with these kinds of problems, so let's try and see what this set looks like.

    Well, we know that [itex]x^2 + y^2[/itex] is the points distance from the origin squared, so it should look something like a donut with outer radius [itex]\sqrt{7}[/itex] and inner radius [itex]1[/itex]. Also, since those are not equal, the boundaries there are not contained in this set.

    There are a couple different definitions for open, but when talking about metric spaces, to say a set is open typically means that for any point in the set, you can find an open ball around that point that is contained within the set. So, if I were to give you a point [itex]\left( x, y \right)[/itex] in this donut, could you find a radius small enough such that a ball of that radius centered at that point was entirely contained in the donut?
     
  4. Apr 19, 2012 #3
    I think r < min {√7-|(x,y)| ; |(x,y)| - 1} works.

    I tried to separate the problem in two sets, because the intersection of two open sets is an open set, and proved that x^2+y^2<7 is open, but the method I used with that proof don't work with the set 1 < x^2 + y^2
     
  5. Apr 19, 2012 #4
    That radius r is good. You may need to also prove that such an r > 0 exists for each (x, y) in the set, but that shouldn't be too bad.

    Also, what method did you use to show that x^2 + y^2 < 7 is open?
     
  6. Apr 19, 2012 #5
    if B(x,y) is an open ball with center (x,y) and radius r, i show that if (a,b) belongs to the ball B, then (a,b) belongs to the set, so a^2 + b^2 < 7

    |(a,b)| = |(a,b) - (x,y) + (x,y)| = |(a-x, b-y) + (x,y)| ≤ |(a-x, b-y)| + |(x,y)|

    |(a-x, b-y)| < r, so

    |(a-x, b-y)| + |(x,y)| < r + |(x,y)|

    r = √7 - |(x,y)|, so

    r + |(x,y)| = √7 - |(x,y)| + |(x,y)| = √7

    Finally, I get

    |(a,b)| < √7

    √(a^2 + b^2) < √7

    a^2 + b^2 < 7

    and that proves that if (a,b) is in the ball then (a,b) is in the set x^2 + y^2 < 7
     
  7. Apr 19, 2012 #6
    Okay, well the proof that [itex]\left\{ \left( x, y \right) \in \mathbb{R}^{2} : x^{2} + y^{2} > 1 \right\}[/itex] should be similar.

    Perhaps if you consider that [itex]\left| \left( x, y \right) \right| < \left| \left( x, y \right) - \left( a, b \right) + \left( a, b \right) \right|[/itex]. It's similar to what you were using before, but you're using the triangle inequality in a different order.
     
  8. Apr 19, 2012 #7
    I solved it that way!! Thanks!!!
     
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