1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why is this set not closed?

  1. Feb 23, 2014 #1
    1. The problem statement, all variables and given/known data
    Z37YCsM.png


    2. Relevant equations
    A set is closed if it contains alll of its boundary points.

    A boundary point is a point where an open ball around that point has one point inside the ball that's in the set, and one point in the ball that's not in the set.


    3. The attempt at a solution
    As seen in the picture, I can create infinitely many open balls around the edges of the triangle, with the balls containing at least one point in the triangle (red point) and one point outside the triangle (blue point). Since the perimeter of the triangle has all the boundary points, this triangle should be closed, but it's not. Why isn't it, if it contains all its boundary points?
     
  2. jcsd
  3. Feb 23, 2014 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If you mean that the set in question is the interior of the triangle with all of the boundary edges included then it is closed. If you mean the interior of the triangle with one boundary edge omitted, (which drawing the dashed line usually means) then it's not closed. What do you mean?
     
  4. Feb 23, 2014 #3
    I mean the interior of the triangle with one edge omitted. Why is it not closed then, according to the fact that a set is closed if it contains all of its boundary points? Doesn't the figure, even with one edge omitted, contain all the boundary points of its edges?
     
  5. Feb 23, 2014 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It doesn't contain the edge you omitted. The edge contains boundary points (as your diagram shows) but the edge isn't in the set.
     
  6. Feb 23, 2014 #5
    So is it then not always true that a set is closed if it contains all of its boundary points?
     
  7. Feb 23, 2014 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    No, it's always true that a closed set contains all of its boundary points. Your set ISN'T closed because it doesn't contain all of its boundary points. The missing edge contains boundary points, the missing edge isn't in the set. Is it?
     
  8. Feb 23, 2014 #7
    But the missing edge is open, so there cannot be any points on that edge to be the center of a disc acting as a boundary point, can there?
     
  9. Feb 24, 2014 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Wouldn't that imply that all open sets are closed because they don't contain any edge points to act as the center of a disc acting as a boundary point?
     
  10. Feb 24, 2014 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    There are plenty of points on that "missing edge". They just aren't in your set.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Why is this set not closed?
  1. Closed set (Replies: 1)

  2. Closed set (Replies: 3)

  3. Closed sets (Replies: 8)

  4. Closed set (Replies: 11)

  5. Closed set (Replies: 14)

Loading...