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!

Prove that a set is open.

  1. Sep 27, 2007 #1
    1. The problem statement, all variables and given/known data

    In general, in R^n, what is the best way to approach the problem - a given set is open?
    The given set E is such that for all x,y that belong to the given set, d(x,y) < r.

    2. Relevant equations


    3. The attempt at a solution

    let x be the center of the sphere and y be any point such that d(x,y) < r. Now, let z be any boundary point such that d(x,z) = r.
    Also let d(y,z) < epsilon. We can make a neighborhood N with epsilon as radius and y as point such that all points of N are subset of the given set. In general we can construct a neighborhood N of smallest (of all possible neighborhoods with the same center) radius r ,
    such that N is a subset of E. Hence, all points of the given open set are internal points. Hence, the given set is open.

    Is it an okay proof? Or should I be proving that the complacent of the open set in a given universe is closed. Hence, the set is open?.

    I am somewhat new to the method of writing proofs, and so want to know that which is a better way to prove?
     
    Last edited: Sep 27, 2007
  2. jcsd
  3. Sep 27, 2007 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    In the most general situation, go back to the definition of "open". You didn't say what definition you are using (there are several) but most often the definition is "all points in the set are interior points. Of course the definition of "p is an interior point of set A" (in metric spaces which Rn is) is that there exist some neighborhood of p which is completely contained in A. I'm not entirely sure how your set is defined/

    2. Relevant equations


    3. The attempt at a solution

    let x be the center of the sphere and y be any point such that d(x,y) < r. Now, let z be any boundary point such that d(x,z) = r.
    Also let d(y,z) < epsilon. We can make a neighborhood N with epsilon as radius and y as point such that all points of N are subset of the given set. In general we can construct a neighborhood N of smallest (of all possible neighborhoods with the same center) radius r ,
    such that N is a subset of E. Hence, all points of the given open set are internal points. Hence, the given set is open.

    Is it an okay proof? Or should I be proving that the complacent of the open set in a given universe is closed. Hence, the set is open?.

    I am somewhat new to the method of writing proofs, and so want to know that which is a better way to prove?[/QUOTE]
    I have a friend who is a math professor. He tells me that it was when he was able to prove exactly the theorem you mention that he knew he could be and wanted to be a mathematician! I'm concerned that if I help you too much you might miss that thrill!

    Let y be a point in the set. One of the things you should think about is how close y is to the boundary set. I strongly recommend you draw a picture. In R2 of course. Draw a circle with center x and mark a point y in the circle. How large can a neighborhood about y be and still be in the set? Now mark a point z in that neighborhood. How far is z from x? Use the triangle inequality.
     
  4. Sep 27, 2007 #3
    Thanks for your reply. I did not want to state any assumptions in the problem because there could be several ways to approach. I wanted to leave it as an open ended question.... but I am beginning to find that thrill although I get stuck many times!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Prove that a set is open.
  1. Proving a set is open. (Replies: 5)

  2. Prove a set is open. (Replies: 8)

Loading...