Prove Set S Open: Using r & Triangle Inequality

[dustbin]

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?

 

[Dick]

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.

 

[dustbin]

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?)?

 

[Dick]

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.

 

[Dick]

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?

 

[dustbin]

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!

 

[Dick]

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.

 

[dustbin]

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...

 

[Dick]

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?