The set S = {(x, y) ∈ ℝ² : x > y} is confirmed to be an open set in ℝ², with the boundary defined by the line y = x, not y = -x. The discussion clarifies that for S to be open, it must satisfy the condition S = S°. A radius r can be determined using the formula r = (x₁ - y₁)/√2, where P = (x₁, y₁) is a point in S. This radius ensures that all points within the distance r from P remain in the set S.
PREREQUISITESStudents of mathematics, particularly those studying real analysis and topology, as well as educators looking for clarification on open sets and their boundaries.
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?