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Simple proof question

  1. Jan 30, 2012 #1

    SixNein

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    Now, I was told that the above proof was valid by a professor. But I don't see how it could be valid as it is written. The only proof I can arguably see here is a proof that AUB[itex]\subseteq[/itex]BUA.

    From the way its written, case 1 shows that A[itex]\subseteq[/itex]BUA while case 2 shows that B[itex]\subseteq[/itex]BUA; therefore, the conclusion would be AUB[itex]\subseteq[/itex]BUA.

    Maybe I'm missing something here..?
     
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  3. Jan 30, 2012 #2

    micromass

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    You are right. But if we substitute A and B, then we also get a proof for the other inclusion. That is: a proof for the other inclusion follows from proving the first inclusion.
     
  4. Jan 30, 2012 #3

    SixNein

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    See I tired to point this out in class. The professor argued that my argument of
    A→B and B→A therefore A=B was a totally different proof. And some how, he accomplishes the same thing without using this because of something to do with his description of an "arbitrary x".
     
  5. Jan 30, 2012 #4

    micromass

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    OK, what about this:

    First we prove (as in the OP) that [itex]E\cup F\subseteq F\cup E[/itex] for ALL sets E and F. This is what the OP does, right??

    Now, we want to prove that [itex]A\cup B=B\cup A[/itex] for all sets A and B.
    Well
    [itex]\subseteq[/itex] follows if we take E=A and F=B.
    [itex]\supseteq[/itex] follows if we take E=B and F=A.
    So equality holds.
     
  6. Jan 30, 2012 #5

    SixNein

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    Let me ask you this:

    Would you agree that in case 1: he essentially showed that A⊆BUA?
    Would you also agree that in case 2: he essentially showed that B⊆BUA?

    He believed that they didn't.

    Why would he think that?

    At any rate, I agree with you here; however, he seemed to be making a different argument (during the discussion).
     
  7. Jan 30, 2012 #6

    micromass

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    I agree with you here.

    Formally, you indeed need to provide justification for both inclusions.

    However, I wasn't present at the discussion, so I can't really say what your professor was trying to say. All I can say is that I think you have a good understanding of this situation and that what you say is correct.
     
  8. Jan 30, 2012 #7

    SixNein

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    I just needed some extra eyes on it. I could have been wrong.

    The class is being taught out of the computer science department. I honestly don't think this would have been an issue in the mathematics department.

    ANyway, thanks for your time.
     
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