Results 1 to 6 of 6

Math Help - Sets question (Proving)

  1. #1
    Junior Member
    Joined
    Feb 2011
    Posts
    56

    Sets question (Proving)

    Prove that

    (A U B)∩(B U C)∩(C U A) = (A ∩ B)U(A ∩ C)U(B ∩ C)

    I am so clueless on solving this... can anyone help me?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,969
    Thanks
    1011

    Re: Sets question (Proving)

    Try drawing a Venn Diagram.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,394
    Thanks
    1478
    Awards
    1

    Re: Sets question (Proving)

    Quote Originally Posted by MichaelLight View Post
    Prove that
    (A U B)∩(B U C)∩(C U A) = (A ∩ B)U(A ∩ C)U(B ∩ C)
    That is just a messy, messy proof. No one likes doing it.
    Here is a suggestion. Consider any point from the LHS. Show it must belong to the union on the RHS.
    Then reverse the order: RHS to LHS.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Re: Sets question (Proving)

    Quote Originally Posted by MichaelLight View Post
    Prove that

    (A U B)∩(B U C)∩(C U A) = (A ∩ B)U(A ∩ C)U(B ∩ C)

    I am so clueless on solving this... can anyone help me?
    It's known that:


    A\cap (B\cup C)=(A\cap B)\cup (A\cap C)

    and


    A\cup (B\cap C)=(A\cup B)\cap (A\cup C)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,417
    Thanks
    718

    Re: Sets question (Proving)

    Consider what intersections you get when you apply the first distributive law that Zarathustra gave above to the left-hand side

    (A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A).

    Each resulting intersection will have one set from each of the three sets of parentheses. Thus, there will be

    A ∩ B ∩ C,
    A ∩ B ∩ A,
    A ∩ C ∩ C,
    A ∩ C ∩ A,

    and so on. Note that repetitions can be removed because X ∩ X = X for any set X.

    Each intersection will have either two sets (A ∩ B or A ∩ C or B ∩ C) or all three sets (A ∩ B ∩ C). Also, each of A ∩ B, A ∩ C, and B ∩ C will be a part of the left-hand side. Finally, note that A ∩ B ∩ C ⊆ A ∩ B, and if X ⊆ Y, then X ∪ Y = Y, so A ∩ B ∩ C can be omitted from the final union.

    All together, this implies that the left-hand side simplifies to (A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,394
    Thanks
    1478
    Awards
    1

    Re: Sets question (Proving)

    Quote Originally Posted by MichaelLight View Post
    Prove that
    (A U B)∩(B U C)∩(C U A) = (A ∩ B)U(A ∩ C)U(B ∩ C)
    As I said above , I prefer a 'pick-a-point' proof.
    Suppose that t \in \left[ {(A \cup B) \cap (A \cup C) \cap (B \cup C)} \right]
    That means all of three are true:
    t \in (A \cup B)\; \Rightarrow \;t \in A \vee t \in B.~~\color{blue}[1]

    t \in (A \cup C)\; \Rightarrow \;t \in A \vee t \in C.~~\color{blue}[2]

    t \in (B \cup C)\; \Rightarrow \;t \in B \vee t \in C.~~\color{blue}[3]

    If t\in A\cap B\cap C the clearly it is RHS.

    So suppose that t\notin A. The by [1] & [2] we t\in B\wedge t\in C.
    So it belongs to the RHS.

    That is a sample of a different approach.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving Sets
    Posted in the Algebra Forum
    Replies: 2
    Last Post: July 8th 2011, 04:40 PM
  2. Proving cardinality of sets
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: April 29th 2011, 09:54 AM
  3. Proving Sets
    Posted in the Number Theory Forum
    Replies: 6
    Last Post: February 28th 2010, 01:22 AM
  4. Proving finite sets
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: September 21st 2009, 09:11 AM
  5. Proving sets unbounded
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: September 17th 2009, 11:30 AM

Search Tags


/mathhelpforum @mathhelpforum