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Proof involving sets. NEED HELP

  1. Feb 26, 2009 #1
    Proof involving sets. NEED HELP!!!

    1. The problem statement, all variables and given/known data

    Prove directly "If A U B = A, then B is a subset of A." and also provide a proof by contrapositive of its converse.


    2. The attempt at a solution

    Here is what i did, but I don't know if it is right or not,

    Direct Proof: Assume A U B = A, then x ∈ (A U B) and x ∈ A. So it follows that B ∈ A = B is a subset of A.
    Contrapositive of Converse Proof: Assume that A U B ≠ A, then x ∈ (A U B) and x ∉ A. Then, B ∉ A and so B is not a subset of A.

    I don't think this is right. Could someone help me out please??
     
  2. jcsd
  3. Feb 27, 2009 #2

    J$C

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    Re: Proof involving sets. NEED HELP!!!

    Direct Proof: Assume A U B=A as you have. Then to show a set B is a subset of a set A the standard technique is to let x be in B, then show it is also in A. Notice if x is in B then it is clearly also in A U B and the conclusion follows from your initial assumption.

    Converse: If B is a subset of A then A U B=A
    Contrapositive of Converse: If A U B /neq A then B is not a subset of A

    Notice A U B /neq A but A is clearly a subset of A U B. So what's left to make that not equals is A U B is not a subset of A. That gives that there's an element in A U B that is not in A. Go from there.
     
  4. Feb 27, 2009 #3
    Re: Proof involving sets. NEED HELP!!!

    How about now...

    Directly -
    Assume A U B = A, then x ∈ B. Then x ∈ (A U B), and since A U B = A, x ∈ A. So B is a subset of A.

    Contrapositive of Converse -
    Assume that A U B ≠ A. Since A is a subset of A U B, there must be an x ∈ (A U B), such that x ∉ A, since A U B ≠ A. This means that there is a y ∈ B, such that y ∉ A. So B is not a subset of A.

    Is that a complete proof??
     
  5. Feb 27, 2009 #4

    lanedance

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    Homework Helper

    Re: Proof involving sets. NEED HELP!!!

    in contrapositive of converse

    though its implied, I think you should change it to A is a proper subset of AUB
     
  6. Feb 27, 2009 #5
    Re: Proof involving sets. NEED HELP!!!

    Allright. Thanks!! =D
     
  7. Feb 27, 2009 #6

    J$C

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    Re: Proof involving sets. NEED HELP!!!

    Looks good.
     
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