Results 1 to 4 of 4

Math Help - "Closed" set in a metric space

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    404

    Unhappy "Closed" set in a metric space

    Suppose that (x_n) is a sequence in a metric space X such that lim x_n = a exists. Prove that {x_n: n E N} U {a} is a closed subset of X.

    =================================

    Let B(r,a)={x E X: d(x,a) < r} denote the open ball of radius r about a.
    Definition: Let D be a subset of X. By definition, D is open iff for all a E D, there exists r>0 such that B(r,a) is contained in D.
    Definition: Let F be a subset of X. F is called closed iff whenever (x_n) is a sequence in F which converges to a E X, then a E F. (i.e. F contains all limit points of sequences in F)
    Theorem: F is closed in X iff the complement of F is open.

    I know the definitions, but I just don't know out how to construct the proofs formally. I am not sure how to prove that some set is "closed"...I can't find any similar examples in my textbook.

    May someone kindly help me out?
    Any help is appreciated!



    [note: also under discussion in math links forum]
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Accidental post, see next post.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by kingwinner View Post
    Suppose that (x_n) is a sequence in a metric space X such that lim x_n = a exists. Prove that {x_n: n E N} U {a} is a closed subset of X.

    =================================

    Let B(r,a)={x E X: d(x,a) < r} denote the open ball of radius r about a.
    Definition: Let D be a subset of X. By definition, D is open iff for all a E D, there exists r>0 such that B(r,a) is contained in D.
    Definition: Let F be a subset of X. F is called closed iff whenever (x_n) is a sequence in F which converges to a E X, then a E F. (i.e. F contains all limit points of sequences in F)
    Theorem: F is closed in X iff the complement of F is open.

    I know the definitions, but I just don't know out how to construct the proofs formally. I am not sure how to prove that some set is "closed"...I can't find any similar examples in my textbook.

    May someone kindly help me out?
    Any help is appreciated!



    [note: also under discussion in math links forum]
    Let E be the set in question. Consider E' and let e\in E'. Let \delta=\frac{d(a,e)}{2}>0 and consider B_{\delta}(e),B_{\delta}(a) these are disjoint open sets about e,a respectively. Since x_n\to a we have that B_{\delta}(a) contains all but finitely many points of \{x_n\} say x_1,\cdots,x_m. Let B_{\delta_k}(e) be the open ball of radius \frac{d(e,x_k)}{2} and consider O=B_{\delta}(e)\cap\bigcap_{k=1}^{m}B_{\delta_m}(e  ), clearly O is an open set containing e disjoint form E. The conclusion follows.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,383
    Thanks
    1474
    Awards
    1
    In a metric space (X,d) you should know that
    D(A;x)=\inf\{d(a,x):a\in A\} is the distance the point x is from the set A.
    Moreover, you should know that A closure (A;x)=0\}" alt="\overline{A}=\{x(A;x)=0\}" />
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Question about a "closed walk" in a graph
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: July 5th 2011, 11:07 AM
  2. Replies: 2
    Last Post: June 4th 2011, 12:11 PM
  3. Replies: 2
    Last Post: April 24th 2011, 07:01 AM
  4. Replies: 1
    Last Post: October 25th 2010, 04:45 AM
  5. Replies: 1
    Last Post: June 4th 2010, 10:26 PM

Search Tags


/mathhelpforum @mathhelpforum