Results 1 to 4 of 4

Math Help - Cluster Point

  1. #1
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093

    Cluster Point

    let A be a subset of a space (X,T) and let y\in A-A' A' is the derive set (the set that contains the clusters points ) show that {y} is open in the subspace (A,T_A)

    I will try to show that A-{y} is closed by prove that \overline{A-\left(y\right)}=A-\left(y\right)
    (y)={y} \overline{A-\left(y\right)} is in T_A not in (X,T) and it is enough to prove that y\notin \overline{A-\left(y\right)}

    Suppose that y\in \overline{A-(y)} \Rightarrow \forall u_a \in T_A \text{   with } \text{          } y\in u_a\ :  u\cap A \ne \phi

    but
    y\in A-A'\Rightarrow y\in A\cap (X-A') so

    y\in A\; and\; y\in (X-A')

    y\in (X-A')\Rightarrow y\notin A' \Rightarrow \exists u\in T : u\cap A=\phi

    here where I stuck can anyone help me ??

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jun 2009
    Posts
    113
    You will obtain the result from the existence of an open subset U in X containing y such that U\cap A=y. But if such an open set does not exist, we would have a contradiction with y\in A\setminus A'.
    Last edited by Enrique2; August 13th 2009 at 10:55 AM. Reason: english ortography!!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by Enrique2 View Post
    You will obtain the result from the existence of an open subset U in X containing y such that U\cap A=y. But if such an open set does not exist, we would have a contradiction with y\in A\setminus A'.
    Thanks very much

    I think I can also solve it

    Quote Originally Posted by Amer View Post
    let A be a subset of a space (X,T) and let y\in A-A' A' is the derive set (the set that contains the clusters points ) show that {y} is open in the subspace (A,T_A)

    I will try to show that A-{y} is closed by prove that \overline{A-\left(y\right)}=A-\left(y\right)
    (y)={y} \overline{A-\left(y\right)} is in T_A not in (X,T) and it is enough to prove that y\notin \overline{A-\left(y\right)}

    Suppose that y\in \overline{A-(y)} \Rightarrow \forall u_a \in T_A \text{ with } \text{ } y\in u_a\ : u_a\cap A-(y) \ne \phi

    but
    y\in A-A'\Rightarrow y\in A\cap (X-A') so

    y\in A\; and\; y\in (X-A')

    y\in (X-A')\Rightarrow y\notin A' \Rightarrow \exists u_0\in T : u_0\cap A-(y)=\phi

    let v=u_0 \cap A and y\in v so v\in T_A and v\cap A-(y)=\phi contradiction

    I think it is now correct right ??

    Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jun 2009
    Posts
    113
    Yes, you are right. In fact, proving that the complementary of y in A is closed is the same as showing that y is open. It seems that you are used to prove closedness instead of opennes.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Show that x is also a cluster point of E...
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: March 1st 2010, 04:19 AM
  2. Cluster point of a finite sequence
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: October 22nd 2009, 09:57 AM
  3. !Cluster Point <=> Convergent
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: July 24th 2009, 08:53 AM
  4. Cluster point:
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 16th 2009, 01:11 PM
  5. Unique Cluster Point => Convergent?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 13th 2008, 03:21 PM

Search Tags


/mathhelpforum @mathhelpforum