Open Set Boundary: Proving S as the Half Plane with y = -x Line Boundary

[tuggler]

Homework Statement

Show that the set of \mathbb{R}^2 given by S = \{(x, y) \in \mathbb{R}^2 : x > y\} is open.

Homework Equations

The Attempt at a Solution

Why is S the half plane that has boundary given by the line y = -x?

 

[Mark44]

Homework Statement

Show that the set of \mathbb{R}^2 given by S = \{(x, y) \in \mathbb{R}^2 : x > y\} is open.

Homework Equations

The Attempt at a Solution

Why is S the half plane that has boundary given by the line y = -x?

It's not - the boundary is the line y = x. I suspect a typo.

 

[STEMucator]

It's not - the boundary is the line y = x. I suspect a typo.

This is what I also thought when I read the problem.

S is open if and only if S=S°.

 

[tuggler]

Thanks for the clarification. I am still having trouble with this problem.

I know I must determine a radius r such that \{(x_1, y_2)\in \mathbb{R}^2: \sqrt{|x-y|^2 +|x_1-y_1|^2}<r\}

but how can I find such an r?

What I did was:

I located the region of the plane where x\gt y then I place a point P=(x_1,y_1) in it where it is fairly close to the line y=x.

Now I moved to the left from the point P until I hit the line y=x, which I marked Q then this Q has coordinates (y_1,y_1).

But I can't find an r. The answer given is \frac{x_1 - y_1}{\sqrt{2}}.

Which I don't know how they got?

 

[Mark44]

Thanks for the clarification. I am still having trouble with this problem.

I know I must determine a radius r such that \{(x_1, y_2)\in \mathbb{R}^2: \sqrt{|x-y|^2 +|x_1-y_1|^2}<r\}

That equation doesn't look right to me. Why is it (x₁, y₂), and why doesn't y₂ show up on the right?

but how can I find such an r?

What I did was:

I located the region of the plane where x\gt y then I place a point P=(x_1,y_1) in it where it is fairly close to the line y=x.

Now I moved to the left from the point P until I hit the line y=x, which I marked Q then this Q has coordinates (y_1,y_1).

But I can't find an r. The answer given is \frac{x_1 - y_1}{\sqrt{2}}.

Which I don't know how they got?