Results 1 to 3 of 3

Math Help - Closed and Open Sets in R^d

  1. #1
    Newbie
    Joined
    Sep 2011
    Posts
    6

    Closed and Open Sets in R^d

    This is the question:
    Let A be an open set and B a closed set. If B ⊂ A, prove that A \ B is
    open. If A ⊂ B, prove that B \ A is closed.


    Right before this we have a theorem stated as below:
    In R^d,
    (a) the union of an arbitrary collection of open sets is open;
    (b) the intersection of any finite collection of open sets is open;
    (c) the intersection of an arbitrary collection of closed sets is closed;
    (d) the union of any finite collection of closed sets is closed.

    So in each case, I think we can just say A\B=A∩B^c (B^c means B complement), and since both A and B^c are open (by assumption and then by definition of a closed set where if B is closed, B-complement is open) we use part (b) of the previous theorem, A\B is open. But in this proof, I didn't use the assumption that B⊂A, so I know it isn't right. Same goes for the second part of the question. PLEASE HELP!!!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,706
    Thanks
    1637
    Awards
    1

    Re: Closed and Open Sets in R^d

    Quote Originally Posted by xsavedkt View Post
    This is the question:
    Let A be an open set and B a closed set. If B ⊂ A, prove that A \ B is open. If A ⊂ B, prove that B \ A is closed.
    Right before this we have a theorem stated as below:
    In R^d,
    (a) the union of an arbitrary collection of open sets is open;
    (b) the intersection of any finite collection of open sets is open;
    (c) the intersection of an arbitrary collection of closed sets is closed;
    (d) the union of any finite collection of closed sets is closed.

    So in each case, I think we can just say A\B=A∩B^c (B^c means B complement), and since both A and B^c are open (by assumption and then by definition of a closed set where if B is closed, B-complement is open) we use part (b) of the previous theorem, A\B is open. But in this proof, I didn't use the assumption that B⊂A, so I know it isn't right. Same goes for the second part of the question.
    Actually, what you did is correct!
    The subset condition is irrelevant.
    This true because R^n \;\& \,\emptyset are both open and closed.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2011
    Posts
    6

    Talking Re: Closed and Open Sets in R^d

    Quote Originally Posted by Plato View Post
    Actually, what you did is correct!
    The subset condition is irrelevant.
    This true because R^n \;\& \,\emptyset are both open and closed.
    Ah, This makes sense!!! Thanks for your help! I guess I thought since the question indicated the "if"s that I've got to be doing something wrong, haha! Thank you so much for your help though!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Metric spaces, open sets, and closed sets
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: March 16th 2011, 05:17 PM
  2. Open vs. closed sets
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: September 14th 2009, 04:33 AM
  3. Open vs. closed sets
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: September 13th 2009, 12:25 AM
  4. Open and closed sets #3
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 2nd 2008, 05:25 PM
  5. Open and Closed Sets
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: May 28th 2008, 09:29 AM

Search Tags


/mathhelpforum @mathhelpforum