Results 1 to 6 of 6

Math Help - Determine whether the equations hold (equality, inclusion etc.)

  1. #1
    Member
    Joined
    Feb 2010
    Posts
    146

    Determine whether the equations hold (equality, inclusion etc.)

    Let A, B, and A_ \alpha denote subsets of a space X. Determine whether the following equations hold, if an equality fails, determine whether one of the inclusions \subset or \supset holds.

    (1) \overline{AintersectB}= \overline{A} \cap \overline{B}

    My Work: A \capB \subsetA,B. Thus A \capB \subset \overline{A}, \overline{B} and therefore A \capB \subset \overline{A} \cap \overline{B}. Thus \overline{AintersectB} \subset \overline{A} \cap \overline{B}

    Is this enough? Or do I need further information explaining it, that the part I always lack in my proofs.

    (2) \overline{the intersection of A_alpha}= \bigcap \overline{A}_ \alpha

    My Work: \overline{the intersection of A_alpha} \subset \bigcupA_ \overline{alpha}. I only need to show \subset right because I only need to show what inclusion works if equality fails? i'm having the most trouble with this one because i'm very unsure about it.

    (3) \overline{A-B}= \overline{A}- \overline{B}.

    My Work: \overline{A-B} \cup \overline{B}= \overline{A-B UNION B}= \overline{A UNION B} \supset \overline{A} \Rightarrow \overline{A-B} \supset \overline{A}- \overline{B}. ...A:=(0,2) B:=(1,2) \overline{A-B}:=[0,1] \overline{A}- \overline{B}:=[0,1). Is this correct? and/or enough? or again do I need more explanation.

    ***I am sorry in advanced for my bad latex skills I'm learning and tried very hard this time to make it as clear as I could with what i know.***
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Sep 2010
    Posts
    86
    What you did is not really a proof, in my opinion. Make sure to use the definitions of the particular items you are using. For (1):

    If x \in \overline{A\cap B} then (i) x \in A\cap B \vee (ii) \forall r > 0: N_r(x)\cap(A\cap B)\setminus\{x\} \ne \emptyset.

    If (i): \overline{A}\cap\overline{B} \supset A\cap B \ni x.
    If (ii): N_r(x)\cap A\setminus\{x\} \ne \emptyset \wedge N_r(x)\cap B\setminus\{x\} \ne \emptyset \Rightarrow x\in \overline{A} \wedge x \in \overline{B} \Rightarrow x \in \overline{A}\cap\overline{B}

    Thus: \overline{A\cap B} \subset \overline{A}\cap \overline{B}

    If x \in \overline{A}\cap\overline{B} then (i) x \in A\cap B \vee (ii) \forall r > 0: N_r(x)\cap A\setminus\{x\} \ne \emptyset \wedge N_r(x)\cap B\setminus\{x\}.

    If (i): \overline{A\cap B} \supset A\cap B \ni x.
    If (ii): You must ask yourself, can there be limit points of both A and B that are no limit points of A\cap B. The answer is no again since: \emptyset \ne N_r(x)\cap A\setminus\{x\} \cap N_r(x)\cap B\setminus\{x\} = N_r(x)\cap (A\cap B\setminus\{x\}) \Rightarrow x \in \overline{A\cap B}

    Thus: \overline{A\cap B} \supset \overline{A}\cap \overline{B} and therefore \overline{A\cap B} = \overline{A}\cap \overline{B}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2010
    Posts
    86
    If x is a limit point of \cap A_{\alpha} then this limit point must have neighbourhoods that allways contain points of this intersection. Since we look at an intersection, all A_{\alpha} must contain the same elements and therefore this limit point is a limit point of any A_{\alpha}. It follows that \overline{\cap A_{\alpha}} \subset \cap \overline{A_{\alpha}}.

    On the other side. Imagine A_{\alpha} = (0,\frac{1}{\alpha}) with \alpha \in \mathbb{N}. Then \cap A_{\alpha} = \emptyset. But the closure of each A_{\alpha} contains the element zero. Thus the intersection of all closures will contain 0. Thus the intersection of all closures is not empty. Thus \overline{\cap A_{\alpha}} \not\supset \cap \overline{A_{\alpha}} in the general case.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2010
    Posts
    86
    For (3) I belive you are wrong. Since \overline{A - B} \supset \overline{A} - \overline{B} but not necessarily the other way round. Think of an counter example. The proof for the former only contains steps I showed you before so try to solve this yourself first.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Feb 2010
    Posts
    146
    Quote Originally Posted by raphw View Post
    For (3) I belive you are wrong. Since \overline{A - B} \supset \overline{A} - \overline{B} but not necessarily the other way round. Think of an counter example. The proof for the former only contains steps I showed you before so try to solve this yourself first.
    hmm.. \overline{R-Q}=R \neq \overline{R}- \overline{Q}=(Empty Set)?

    R and Q as in the reals and the rationals
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Sep 2010
    Posts
    86
    Yep. Good example! You can still prove that the subset constraint I posted above holds, however.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: December 17th 2011, 12:48 PM
  2. Why does this congruence hold?
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: April 3rd 2011, 04:16 PM
  3. determine the range of ln equations
    Posted in the Advanced Math Topics Forum
    Replies: 4
    Last Post: January 24th 2011, 01:46 PM
  4. Replies: 4
    Last Post: November 21st 2010, 11:35 AM
  5. Replies: 4
    Last Post: August 25th 2008, 04:28 AM

Search Tags


/mathhelpforum @mathhelpforum