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Boolean Rings: System of Switches.

  1. Jul 2, 2013 #1
    Okay, So I just got done reading and re-reading Chapter one of Introductory Real Analysis by Andrei Kolmogorov and S.V Fomin, I have to say it is a lot easier to read than the symbolic logic book I have. But to the point, There is this section which talks about Boolean Rings and it defines them just as you would define any Boolean Algebra by the intersection of any sets must be in the Algebra(Ring) and the symmetric difference between any two sets must be in the Algebra(Ring).

    My problem begins not with the definition, but with the concept of the Boolean Lattice ##BL_n## and n is at least 1. So when you create this Boolean Lattice and it's similar to a Boolean Algebra's/(Ring's) Power Set, right? Also, If it is the power set of The Boolean Algebra(Ring), then would it mean that the ##BL_n## is a semi ring of the Boolean Algebra(Ring), or what? I mean the definition of a semi ring is that, if you take any two subsets of The ring, lets call it ##\mathcal{X}## and take their intersection then, this intersection must be in ##\mathcal{X}##. Furthermore, that ##\mathcal{X}## contains the empty set and all sets can be represented as a finite, or countably infinite cover of that set. Next natural question, is the semi ring ##\mathcal{X}## a powerset?
     
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  3. Jul 2, 2013 #2

    micromass

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    I'm not really sure what you're asking.

    First, you talk about ##BL_n##, what is the definition of this? And what is a semi ring?
     
  4. Jul 2, 2013 #3
    The Boolean lattice? Oh, it is graphs who has a vertex set of subsets of ##\{1,2,3,\cdots , n\}## and the rest of the structure is similar to that of a Boolean Algebra. I am asking two questions, my first one is Is this ##BL_n## similar to the power set of a Boolean ring, and would this make a semi ring, since ##BL_n## contains the null set? I guess my second question is just a restatement of the first but with the definition of a semiring :/
     
  5. Jul 2, 2013 #4

    micromass

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    You define a Boolean lattice as a graph?? No, that can't be right. Do you have a reference?
     
  6. Jul 2, 2013 #5
    Yeah, Graph Theory by J.A Bondy, but I am only trying to equate the Boolean semiring to ##BL_n## and if I can do that I believe that it should be a powerset, if and only if the semiring is a power set and ##BL_n## is a powerset. Maybe, I am not asking the right question. I apologize for any confusion :(
     
  7. Jul 2, 2013 #6

    micromass

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  8. Jul 3, 2013 #7
    You are right I see it now, it does say the set of all subsets of the vertex set, I guess I have been reading that wrong, but still leaves the second part, is the semi ring a power set too? I am probably not right in saying that it is, but since it contains the null set it is kind of hard for me not to think that it is, I just don't want to be making a huge mistake and have to turn back D:
     
  9. Jul 3, 2013 #8

    micromass

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    For the last time, what do you mean with a semi ring?
     
  10. Jul 3, 2013 #9
    Okay, in the book it defines ##\mathcal{X}## as the system of semi rings (of sets) if
    i.) ##\emptyset \in \mathcal{X}##
    ii.) ##\bigcap_{n=1}^{m}A_{n} \in \mathcal{X}## when ever ##A_{n} \in \mathcal{X}##
    iii.) When ##A \in \mathcal{X}## and ##A_{1} \subset A##, then ##A## can be represented as a finite union of ##A = \bigcup_{n=1}^{m}A_{n}## and then it begins to list some examples :/
     
  11. Jul 3, 2013 #10

    micromass

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    Your definition has some errors, but now I finally know what kind of semiring you mean.

    So, you want to ask whether a semiring is a power set? The answer is no. You can very easily construct semirings which are not the entire power set.
    For example, take ##\mathcal{X}\subseteq \mathcal{P}(\{0,1\})## and take ##\mathcal{X} = \{ \emptyset, \{0\}, \{1\} \}##. This is a semiring, but not the power set.
     
  12. Jul 3, 2013 #11
    Okay I get what you are saying , I think... So, there are some instances in which the semiring acts as a power set, but in general it need not be a power set. Okay, thank you, that cleared it up.
     
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