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An Open Set

  1. Sep 21, 2008 #1
    The problem statement, all variables and given/known data
    Prove that the set A = {(x,y) in R2 : xy ≠ 1} is open in the metric space (R2, d), where d is the Euclidean metric.

    The attempt at a solution
    A is open if for any p in A, I can find an open ball centered at p that is contained in A. This would be easy if I could find the closest point q = (x,y) to p that is not in A. Then I would just choose the open ball whose radius is d(p,q). Heck, I don't even need to find q: It suffices to know that such a q exists. This is where I'm stumped. Any tips?
     
  2. jcsd
  3. Sep 21, 2008 #2

    morphism

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    How about showing that the complement of A is closed? If you use sequences this should be pretty straightforward.
     
  4. Sep 21, 2008 #3
    True, but this problem is given before the problems on sequences, so I think that would be unfair.
     
  5. Sep 21, 2008 #4

    morphism

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    Fine. Let's take an element p=(a,b) in A, and suppose that every neighborhood of p contains elements of the form (x,1/x) for some (nonzero) x in R. This means that we can make the quantity |(a,b) - (x,1/x)|^2 arbitrarily small by choosing an appropriate x. But |(a,b) - (x,1/x)|^2 = (a-x)^2 + (b-1/x)^2, and both (a-x)^2 and (b-1/x)^2 are nonnegative, so ...
     
  6. Sep 21, 2008 #5
    It seems to me that I must find the minimum of (a-x)2 + (b-1/x)2, but that requires calculus, which I'm not allowed to use yet. I can't think of anything else I can do here.
     
  7. Sep 21, 2008 #6

    Dick

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    I don't know where you are getting these restrictions on what you can and cannot use. You HAVE to use something. There are continuous curves in R^2 that are dense. What can you use? Why can't you use calculus?
     
  8. Sep 21, 2008 #7

    Hurkyl

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    Even if you can't use calculus in your proof, you can use it in your scratchwork. Once you know what the answer is supposed to be, it shouldn't be difficult to prove it with ordinary algebra of inequations.
     
  9. Sep 21, 2008 #8
    Let's see, I can use the fact that the reals are a complete ordered field, the definition of metric space, the Schwarz inequality, the inequality

    [tex]\sqrt{(a_1+b_1)^2 + \cdots + (a_n+b_n)^2} \le \sqrt{a_1^2 + \cdots + a_n^2} \, + \, \sqrt{b_1^2 + \cdots + b_n^2}[/tex],

    the inequalities

    [tex]d(p_1,p_n) \le d(p_1,p_2) + \cdots + d(p_{n-1},p_n)[/tex] and

    [tex]|d(p,r)-d(r,q)| \le d(p,q)[/tex]

    the definitions of open/closed set/ball and the fact that metric spaces induce a topology.
     
  10. Sep 21, 2008 #9

    Dick

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    How does that prove it?
     
  11. Sep 21, 2008 #10
    Those are just some facts that I can use.
     
  12. Sep 21, 2008 #11

    Dick

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    Ah, gotcha. Now why can't you use calculus or sequential closure, exactly? I think you are limiting yourself unfairly to the point where you can't prove it at all. Those properties are too vague to cover 1=xy.
     
  13. Sep 21, 2008 #12
    I'm assuming this problem is from the section called Open and Closed Sets since some of the problems before and after it involve proving that some set is open or closed. The next section is called Convergent Sequences, so I'm assuming the author is expecting me to use only those notions prior to this section to solve the stated problem. This doesn't include calculus since the aim of the book is to develop calculus rigorously--and there is always the danger of circular arguments.

    Anywho, I don't even understand how the sequential argument would work: Let B be the complement of A. I can show that B is closed by showing that any sequence of points in B converges to a point in B. But how is this true for the sequence (1,1), (2,1/2), (3,1/3)..., which goes to [itex](\infty,0) \not\in B[/itex]?
     
  14. Sep 22, 2008 #13
    Also, the derivative of (a-x)2 + (b-1/x)2 is -2a + 2x + 2b/x2 - 2/x3. I don't know how to find the zeros of that.
     
  15. Sep 22, 2008 #14

    Dick

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    You only have to show that the limits of CONVERGENT sequences are in B. I.e. if (a_i,1/a_i) converges to (b,c) can you show b=1/c? It's the same as the proof f(x)=1/x is continuous except at 0.
     
  16. Sep 22, 2008 #15
    Oops. That's right. Suppose (ai, 1/ai) converges to (b, c). Then ai converges to b and so 1/ai converges to 1/b, hence c=1/b.

    That was easy. What about finding the minima?
     
  17. Sep 22, 2008 #16

    Dick

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    I guess I can't think of a good way to compute or estimate the minimum distance. But now we don't have to, right?
     
  18. Sep 22, 2008 #17
    No. We don't have to. It would be interesting though.
     
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