The discussion revolves around the set S in \(\mathbb{R}^2\) defined as S = \{(x, y) \in \mathbb{R}^2 : x > y\}, with participants tasked to show that this set is open. The boundary of this set is questioned, particularly in relation to the line y = -x versus y = x.
Some participants have provided clarifications regarding the boundary of the set, while others express confusion about the radius r needed for the proof of openness. There is an ongoing exploration of the implications of the boundary definition and how to determine the radius.
Participants are grappling with the mathematical definitions and conditions necessary for proving the openness of the set, including potential typos in the problem statement and the correct formulation of the radius r.
It's not - the boundary is the line y = x. I suspect a typo.tuggler said: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 said:It's not - the boundary is the line y = x. I suspect a typo.
That equation doesn't look right to me. Why is it (x1, y2), and why doesn't y2 show up on the right?tuggler said: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\}
tuggler said: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?