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Open set

  1. Sep 23, 2013 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations



    3. The attempt at a solution

    Why is S the half plane that has boundary given by the line [tex]y = -x?[/tex]
     
  2. jcsd
  3. Sep 23, 2013 #2

    Mark44

    Staff: Mentor

    It's not - the boundary is the line y = x. I suspect a typo.
     
  4. Sep 23, 2013 #3

    Zondrina

    User Avatar
    Homework Helper

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

    S is open if and only if S=S°.
     
    Last edited: Sep 23, 2013
  5. Sep 23, 2013 #4
    Thanks for the clarification. I am still having trouble with this problem.

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

    but how can I find such an r?

    What I did was:

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

    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 cant find an r. The answer given is [tex]\frac{x_1 - y_1}{\sqrt{2}}.[/tex]

    Which I don't know how they got?
     
  6. Sep 24, 2013 #5

    Mark44

    Staff: Mentor

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