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Math Help - Set theory: One more quick proof (low level)

  1. #1
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    Set theory: One more quick proof (low level)

    A,B,C,D are sets.

    Prove that if C is contained in A and D is contained in B, then C∩ D is contained in A∩ B.

    Would proof by contrapositive be appropriate?

    I.e. show that if the first intersection is not contained in the second, then C not contained in A and so on or is a different method better? I just have trouble formally showing these things.
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  2. #2
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    Quote Originally Posted by zhupolongjoe View Post
    A,B,C,D are sets.

    Prove that if C is contained in A and D is contained in B, then C∩ D is contained in A∩ B.

    Would proof by contrapositive be appropriate?

    I.e. show that if the first intersection is not contained in the second, then C not contained in A and so on or is a different method better? I just have trouble formally showing these things.
    That sounds like an excellent idea! If C∩D is NOT contained in A∩B, there exist x in C∩D that is not in A∩D and there for x is in C and x is in D. Be sure you don't say "because x is not in A∩D, it is in neither A nor D"- that is not true.
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  3. #3
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    Ok thanks, is this an acceptable proof:

    Let x be any element.

    Then There exists (x that belongs to C∩D) and (x does not belong to A∩ B)

    So x belongs to C and x belongs to D

    If x belongs to C, since C is contained in A, then x belongs to A.
    If x belongs to D, since D is contained in B, then x belongs to B.
    So x belongs to A intersect B, a contradiction.

    Then the original statement is true.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by zhupolongjoe View Post
    Ok thanks, is this an acceptable proof:

    Let x be any element.
    any element? no, that won't work. that statement is too vague to be of any use. you mean let x be any element of ...
    Then There exists (x that belongs to C∩D) and (x does not belong to A∩ B)
    you never stated what you are assuming for you to get to this

    So x belongs to C and x belongs to D
    ok

    If x belongs to C, since C is contained in A, then x belongs to A.
    If x belongs to D, since D is contained in B, then x belongs to B.
    So x belongs to A intersect B, a contradiction.
    when were we given that C is contained in A and D is contained in B? we are using the contrapositive right? that means, we cannot use these

    sorry, but this requires a complete do over.

    start by saying what you are assuming. begin, for example, as follows:


    Proof:
    Assume the C \cap D \not \subseteq A \cap B. then there is some element x \in C \cap D that is not in A \cap B. But that means ...
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