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Show the openness of a set

  1. Sep 25, 2007 #1
    I want to show that the set [tex]S=\{(x,y)\in \mathbb{R}^2 | xy\neq1\}[/tex] is open.

    I'm having trouble forming the open ball contained in S centered at each point in S. The idea I have is:

    Let [tex]q\in S[/tex]. Then select an open ball [tex]B_r(q), r\in\mathbb{R}[/tex]. Let [tex]P=\{(x,y)\in B_r(q)|xy=1\}[/tex]. If P is empty, we are done. If P is not empty, then create an open ball [tex]B_m(q),[/tex] where [tex] m=\min\{d(q-c)|c\in B_r(q)\}[/tex] (d is the distance function). Then we are done.

    However, I feel like this isn't sufficient because it has not been shown whether [tex]\{d(q-c)|c\in B_r(q)\}[/tex] actually does have a minimum, or if it contains elements tending to 0. Am I just missing a property of the reals somewhere? I hope so. Any help would be appreciated.
     
  2. jcsd
  3. Sep 25, 2007 #2
    Don't you think is easier to prove that [itex]S^c[/itex] is closed?
     
  4. Sep 25, 2007 #3
    In metric spaces, a set is defined to be closed if it's complement is open.
     
  5. Sep 25, 2007 #4

    Hurkyl

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    Is there any way you can leverage your knowledge of the topology of R to help?
     
  6. Sep 25, 2007 #5
    True, so why don't you prove that the complement of S is closed?
     
  7. Sep 25, 2007 #6
    Are you suggesting to prove by contradiction that it is closed? I'm not sure how to go about that because the only definition of closed I have is that it's complement is open.
     
  8. Sep 25, 2007 #7

    HallsofIvy

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    If x is in the set {(x,y)| xy is not equal to 1} then obviously either xy< 1 or xy>1. Why not look at the two cases separately?
     
  9. Sep 25, 2007 #8
    I could, but how would proving that the set is open be any easier when the cases are split up? I would still have to construct a ball such for each (x,y), no matter how close I make to the boundary.

    I'm looking at some theorems, and I have that any subset S of a metric space is bounded if it can be contained in a ball. I can easily contain the set of the d(s) in a ball, which then means it's bounded from above and below. So then I can create the open ball with radius from d(q, lower bound)
     
  10. Sep 25, 2007 #9

    Hurkyl

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    Just checking: have you yet learned what continuous means?
     
  11. Sep 25, 2007 #10
    I have not. (Not formally. I can't use it in the proof)
     
  12. Sep 25, 2007 #11

    Hurkyl

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    Ah. So you could not use the fact that multiplication is continuous, and the set of real numbers not equal to 1 is open. Oh well; that's the easy way to do this problem.
     
  13. Sep 25, 2007 #12

    morphism

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    Suppose it's not open. Then there exists a pair (x,y) in S such that for all r>0, Br [itex]\cap[/itex] Sc is nonempty. In particular, we can do this for all r=1/n (n=1,2,3,...), and thus we obtain for each n a pair (an, bn) satisfying two properties:
    (i) an * bn = 1, and
    (ii) d((an, bn), (x,y)) < 1/n.

    Can you show that this gives us a contradiction? (Hint: think sequences.)
     
    Last edited: Sep 25, 2007
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