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Help with simple proof - comlement of unions.

  1. Aug 8, 2008 #1
    I've been trained in applied mathematics, and I just started exploring pure mathematics (real analysis, to be exact). I'm already stuck at a very simple proof, and was wondering if anybody can help me out:

    http://img392.imageshack.us/img392/8971/proofeg2.png

    I understand it for the most part. However, I do not understand why the statement highlighted in yellow implies the statement that is highlighted in red. I copied that proof directly from Maxwell Rossenlicht's Introduction to Analysis.

    Any help is greatly appreciated.
     
  2. jcsd
  3. Aug 9, 2008 #2

    tiny-tim

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    Hi scorpion990! :smile:

    If x lies in at least one of A B C and D, then it lies in their union. :smile:

    (the red box isn't actually a statement … and if you begin it with "x lies in", that still only proves half the theorem :wink:)
     
  4. Aug 9, 2008 #3
    The red statement is supposed to start with "x is an element of...". And I do realize that this is only half the proof. I only bothered writing the part that gave me trouble.

    I understand that x is an element of the complement of at least of the sets in X, but I don't see how this automatically means that it lies in the union of all of the sets in X.
     
  5. Aug 9, 2008 #4

    tiny-tim

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    No … it lies in at least one complement, so it lies in the union of the complements of all of the sets in X. :smile:
     
  6. Aug 9, 2008 #5
    If [tex] x\notin X_j [/tex] for some j, then it is in the complement of X because the definition of a complement is [tex] A^c=\{x:x\notin A\} [/tex] thus we have that [tex] x\in X^c_j [/tex] for some j. As for any sets [tex] A \subset \bigcup A [/tex], that is if the union contains A then A is a subset of the union. Thus this implies that as [tex] X^c_j \subset \bigcup_{i \in I}X^c_i [/tex]. Thus again by definition then [tex] x \in \bigcup_{i \in I}X^c_i [/tex]

    Hope this helps
     
  7. Aug 9, 2008 #6
    Sorry guys =/
    Maybe I should stick to applied mathematics. I drew a picture to help gather my thoughts:
    http://img527.imageshack.us/img527/6457/setsyc4.png

    The first picture is, obviously, the intersection of all of the sets in X (which are all subsets of S). The middle picture is the complement of the intersection of all of the sets in X. This is what I started with. I argued, step by step, that x is an element of the complement of j. This is the third picture.

    For some reason, I still can't see why x can lie in the union of all of the complements. For example, it can't lie in (all of) the complement of the set labeled ii, because x can't lie in j.

    =/
     
  8. Aug 9, 2008 #7

    tiny-tim

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    Hi scorpion990! :smile:

    x lies in the green area in figure 3,

    which is part of the white area in figure 1.

    The white area in figure 1 is the union of the complements.

    If x lies in it, it must lie in the green-type area of at least one of of the sets, and vice versa. :smile:
     
  9. Aug 9, 2008 #8
    Okay. That makes more sense.
    I guess my problem is more fundamental than that.

    For example: Clearly, the set of integers is a subset of the set of real numbers. If x is an element of the set of integers, then it is an element of the set of real numbers. However, x cannot assume any value in the set of real numbers. In a way, I've actually "lost" information about the constraints on x by saying that it is an element of the set of real numbers, because it can only take on specific values in this set.

    If I understand correctly, this method of proof works in this way: I'm showing that if x is an element of the first set, then it must be an element of the second set. However, the second set may contain more elements than those which x is allowed to have. The second part of the proof (as somebody before mentioned), shows that if x is an element of the second set, then it must be an element of the first set, even though the first set may contain more elements than x is allowed to have. Alone, these two statements don't prove anything. But together, they prove that the two sets are equal since the first is a subset of the second, and the second is a subset of the first.

    Sorry if that's confusing.
     
  10. Aug 9, 2008 #9

    tiny-tim

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    Hi scorpion990! :smile:

    I'm not really following your last post.

    You've moved on from the "complements" part of the problem to the more fundamental question of how to prove that two definitions are equal.

    That, I follow.

    But you say:
    and yet you end with the apparently clear:
    and so I don't understand what the problem is. :confused:
     
  11. Aug 9, 2008 #10
    I was just thinking about sets incorrectly. For example:

    The set of integers is a subset of the set of real numbers. If x is an element of the set of integers, then it is also an element of the real numbers. Even though x belongs to the set of all real numbers, it can't necessarily equal any element in this set. For example, x can't equal pi, because I've previously stated that it must be an integer. All I've proven is that one set is a subset of the other.

    If my logic is wrong in any way, feel free to correct me.
     
  12. Aug 9, 2008 #11

    tiny-tim

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    Yes, that's right, that's all you've proved …

    but is anything worrying you about that? :smile:
     
  13. Aug 9, 2008 #12
    No :)
    It makes sense now. Thanks!!
     
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