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Prove a set is open.

  1. Jan 16, 2013 #1
    1. The problem statement, all variables and given/known data

    I want to prove that the set S={(x,y) in R^2 : x^2 > y} is open.

    3. The attempt at a solution

    Given any (x,y) in S, we must choose an r>0 so that we may show (a,b) in Br(x,y) (*The open ball of radius r centered at (x,y)*) implies (a,b) in S. Doing so, we have shown that for all (x,y) in S there exists r>0 s.t. Br(x,y) is contained in S... and thus S is open.

    I am stuck on choosing r. I have chosen r=x^2-y>0 and some variations of this assignment, but cannot get a^2>b using the triangle inequality (or reverse tri. inequality). Would anyone care to provide a hint as to what I am missing?
     
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  3. Jan 16, 2013 #2

    Dick

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    There are too many xy's around. Pick your ball center to be (x0,y0). Then you want to pick an r that is less than the distance from (x0,y0) to the curve y=x^2.
     
  4. Jan 16, 2013 #3
    Ah, yes! I did not think to do so...

    So, given any (x0,y0) in S, the distance from this point to the curve y=x^2 is obviously d=√[ (x0-x)^2+(y0-y)^2 ] = √[ (x0-x)^2+(y0-x^2) ]. I tried choosing some r's based on this, but no luck. Then I tried taking the derivative of d to solve for the shortest distance between (x0,y0) and the curve x=y^2. I end up with a cubic polynomial, which I'm not really sure helps... I ended up with 2x^3+x(1-2y0)-x0 =0. How can I solve for the real root in order to get a value for d that only involves x0, y0 (without using the cubic equation?)?
     
  5. Jan 16, 2013 #4

    Dick

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    Good point. You probably don't want to actually try to solve it. But if you look at the expression D(x)=(x0-x)^2+(y0-x^2)^2 you see that that is nonnegative for any x and it's equal to zero only if x=x0 and x^2=y0 (in other words (x0,y0) is on the parabola, which it's not). Since D(x) also approaches infinity as x goes to +/- infinity it must have a positive minimum for some value of x. Actually for any closed curve and a point not on the curve you know the distance between the point and the curve is positive, but that might be something you haven't proved yet.
     
  6. Jan 16, 2013 #5

    Dick

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    An even more direct way to prove it is that if you define f(x,y)=x^2-y then f is continuous and if you know that means that the inverse image of an open set is open you could go that route. Or do you HAVE to use balls?
     
  7. Jan 16, 2013 #6
    Sadly, we can only balls... Or prove the complement is closed by showing every convergent sequence in the comp. Converges to a point is the comp.

    I will continue playing with it and report back!
     
  8. Jan 16, 2013 #7

    Dick

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    Do that. The sequential convergence approach sounds promising.
     
  9. Jan 17, 2013 #8
    Hm. I am having no luck with proving that every convergent sequence within the complement converges to a point in the complement...

    I feel that there should be a way to prove it using balls, since we only just went over the convergent sequences theorem and we're supposed to use balls on the homework...
     
  10. Jan 17, 2013 #9

    Dick

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    There IS a way to do it using balls. We've basically covered that. It would be nice if you could just solve the cubic but that's not really necessary. I gave you an argument to show that there is a positive minimum distance between the point and the curve. Call it r. Use that for the ball. Using sequences should be a little neater. What's the problem with it?
     
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