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Abstract Algebra, Sets Proof

  1. Jan 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Question 1. Let U be a universal set, A and B two subsets of U.
    (1) Show that
    B ⊆ A ∪ (B ∩ A^c).
    (2) A = B if and only if there exists a subset X of U such that A ∪ X = B ∪ X and X\A^c = X\B^c.

    3. The attempt at a solution

    My attempt at a solution is as follows:

    Part 1: Showing B ⊆ A ∪ (B ∩ A^c)

    (A∪B)∩(A∪A^c)
    (A∪B)∩([itex]\bigcup[/itex])

    Since A∪B are both in universe, it serves that B ⊆ A ∪ (B ∩ A^c).

    Part 2
    A = B if and only if there exists a subset X of U such that A ∪ X = B ∪ X and X\A^c = X\B^c.

    So I claimed double inclusion proof here, letting X[itex]\in[/itex]A

    Case 1: X[itex]\in[/itex]X

    X[itex]\in[/itex]A[itex]\cup[/itex]X[itex]\Rightarrow[/itex]X[itex]\in[/itex]B[itex]\cup[/itex]X[itex]\Rightarrow[/itex]X[itex]\in[/itex]B

    Case 2: X[itex]\notin[/itex]X[itex]\Rightarrow[/itex]X[itex]\in[/itex]X\A^c[itex]\Rightarrow[/itex]X[itex]\in[/itex]X\B^c[itex]\Rightarrow[/itex]X[itex]\in[/itex]X or X[itex]\in[/itex]B^c but X[itex]\notin[/itex]X so X[itex]\in[/itex]B

    So I think the problem with this question is that I am not fully understanding the concept of the property known as the difference.
     
  2. jcsd
  3. Jan 25, 2013 #2

    jbunniii

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    I don't understand your proof. May I suggest something along these lines: if [itex]b \in B[/itex], and [itex]b \not\in A[/itex], then [itex]b \in B \cap A^c[/itex]. Therefore...
    You may find it helpful to use the following identity: [itex]A \setminus B = A \cap B^c[/itex] for any sets [itex]A, B \subset U[/itex].
     
  4. Jan 25, 2013 #3

    When A and B in this case are already compliments, that makes no difference? I can still switch using that identity?
     
  5. Jan 25, 2013 #4

    jbunniii

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    Yes, the identity is valid in all cases. If you have something like [itex]X \setminus A^c[/itex] then that equals [itex]X \cap(A^c)^c[/itex] = [itex]X \cap A[/itex].
     
  6. Jan 26, 2013 #5
    Hi jbunni, could you check out my second attempt at question part 1:




    So B⊆A∪(B∩A^c)

    To show this, we can say that when x∈B, implies there exists an x∈A∪(B∩A^c)

    We will let x∈B,

    Case 1: x∈ A

    x∈ A--->x∈ A or x∈(B∩A^c). So clearly, x∈A.

    Case 2: x∉A

    x∈A or x∈(B∩A^c). We have declared x∉A, so x∈B and x∈A^c. Now we note that X∈A^c is = to X∉A.

    Thus proving that B⊆A∪(B∩A^c)
     
  7. Jan 26, 2013 #6

    jbunniii

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    I think you have the right idea, but I would word it somewhat differently. See if this sounds cleaner to you:

    Let [itex]x \in B[/itex]. We consider two cases:

    Case 1: [itex]x \in A[/itex]. In this case [itex]x \in A \cup (B \cap A^c)[/itex] because [itex]A \subset A \cup (B \cap A^c)[/itex].

    Case 2: [itex]x \not\in A[/itex]. Then [itex]x \in (B \cap A^c)[/itex]. Therefore [itex]x \in A \cup (B \cap A^c)[/itex], because [itex](B \cap A^c) \subset A \cup (B \cap A^c)[/itex]

    In both cases, we have established that [itex]x \in B[/itex] implies [itex]x \in A \cup (B \cap A^c)[/itex]. This shows that [itex]B \subset A \cup (B \cap A^c)[/itex].
     
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