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A set equality proof without elements, I broke my brain?

  1. Oct 15, 2013 #1
    1. The problem statement, all variables and given/known data

    b(B) = cls(B) \ Int(B)

    where b(B) is the boundary, cls is the closure, and int is the interior of set B.

    This was not hard for me to prove by picking elements and showing that the sets were contained in one another. However, I decided it would be fun to try to derive it by just going by definitions, my interpretations of definitions, DeMorgan's laws, things about sets, etc. While I'm not sure entirely what's wrong, or where I broke down, something must just not make sense here.

    2. Relevant equations



    3. The attempt at a solution

    So for

    [itex]\overline{B}[/itex]

    I defined this to be "the complement of the union of all open sets disjoint from B" in order to make it fit nicely with the type of manipulation I had in mind. I thought this definition may have been the problem but Office_Shredder confirmed its accuracy in this thread.

    I defined

    [itex]Int{B}[/itex]

    To be the union of all open subsets of B. This is the textbook definition.


    And, I defined

    [itex]b(B)[/itex]

    To be the intersection of all closed sets containing B. This is also the textbook definition.


    So my goal is to just reason directly that

    [itex](X \setminus (\cup O_{i} | O_{i} \cap B = \oslash, O_{i} is open)) \setminus (\cup Q_{i} | Q_{i} \subseteq B, Q_{i} is open)[/itex]

    is the boundary of B.



    1.)

    Using some ideas with complements, and changing these to "absolute complements" for the sake of simplicity, I think that this is equivalent to that:

    [itex](\cup O_{i} | O_{i} \cap B = \oslash, O_{i} is open)^{c} \cap (\cup Q_{i} | Q_{i} \subseteq B, Q_{i} is open)^{c}[/itex]

    And then, being an intersection of complements, is a complement of unions, so I have that this is (removing the characteristics of the sets in the unions, it's ugly, pretend they are there)

    [itex]X \setminus ((\cup O_{i}) \cup (\cup Q_{i}))[/itex]

    So now, I think that the union of unions is just a normal union, so I merged them into one, and merge their conditions to an or. Then, I pulled the new compound union out of the complement, making it an intersection, so their conditions are now an and. Also, instead of O and Q, call everything O now.

    [itex]\cap X \setminus O_{i} | O_{i} \cup B = \oslash AND O_{i} \subseteq B[/itex]


    Which is exciting because I now have an intersection of closed sets. The b(B) is an intersection of closed sets containing B.

    So, with the condition
    [itex]O_{i} \cup B = \oslash[/itex]
    and
    [itex]O_{i} \subseteq B[/itex]
    the B-containment should follow for each X \ Oi, but it doesn't as far as I can tell. In fact, it means that every Oi is empty, so every closed set X\Oi is just X!

    What went wrong?

    Thanks!
     
    Last edited: Oct 15, 2013
  2. jcsd
  3. Oct 16, 2013 #2

    haruspex

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    I think you have made a mistake in the logical inversion to reach that.
     
  4. Oct 16, 2013 #3
    Should I negate each condition in addition to making the statement a conjunction? Doing that doesn't seem to get me the right logical implication of X\Oi containing B.

    It's hard for me to tell here. I didn't think it was an inversion at all. I thought it just became an intersection.

    Thanks

    EDIT woops, if you are referring to the union symbol for the disjoint condition, that is just a typo.
     
    Last edited: Oct 16, 2013
  5. Oct 16, 2013 #4
    I found the problem. My boundary definition makes no sense at all! Proof followed very quickly with the right definition!
     
    Last edited: Oct 16, 2013
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