Prove Set S Open: Using r & Triangle Inequality

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In summary: Oh, right. You're trying to use the ball convergence approach. Well, if every convergent sequence in the complement converges to a point in the complement, then the complement must be closed.
  • #1
dustbin
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Homework Statement



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

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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  • #2
dustbin said:

Homework Statement



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

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?

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.
 
  • #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?)?
 
  • #4
dustbin said:
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?)?

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.
 
  • #5
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?
 
  • #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!
 
  • #7
dustbin said:
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!

Do that. The sequential convergence approach sounds promising.
 
  • #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...
 
  • #9
dustbin said:
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...

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?
 

1. What is "Set S" in this context?

Set S refers to a set of points or values in a given space. In this context, it is used to represent a set of real numbers.

2. What does it mean for a set to be "open"?

In mathematics, a set is considered open if all of its points are contained within an open interval. In other words, there is no boundary or edge that excludes any points from the set.

3. What is the significance of using "r" in proving set S is open?

"r" represents the radius of an open interval. In order to prove that a set is open, it must be shown that for any point in the set, there exists an open interval with a radius r that contains the point. This means that the point is not an edge or boundary of the set, but rather within the set itself.

4. How does the Triangle Inequality play a role in proving set S is open?

The Triangle Inequality states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. In the context of proving a set is open, the Triangle Inequality is used to show that the distance between points in the set is always less than the radius r of an open interval. This ensures that no points are excluded from the set.

5. Why is it important to prove that a set is open?

Proving that a set is open is important in many areas of mathematics, including topology and analysis. It helps to define the boundaries and limits of a set, and can provide valuable insights into the properties and behavior of functions and equations within the set. Additionally, it allows for more accurate and precise calculations and proofs in various mathematical applications.

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