Ring of Cartesian products A X B X C = Ring of Cartesian products(A X B) X C ????

Discussion in 'Undergraduate Math' started by sto, Dec 1, 2008.

  1. sto

    sto Guest

    If R1 and R2 are (Boolean) rings, then I can show that the set of all
    finite, disjoint unions of Cartesian products A1 X A2, A1 in R1, A2 in
    R2, is a (Boolean) ring. Then by induction it is simple to show that if
    R1, ..., Rn are rings then the set of all finite, disjoint unions of
    Cartesian products

    (...(A1 X A2) X A3 ) ... X An), A1 in R1, ..., An in Rn

    is a ring. What I don't understand is how one then makes the leap to
    inferring that the set of all finite, disjoint union of Cartesian
    products of the form A1 X A2 X ... X An is also a ring (in fact the
    "same" ring?) Clearly there exists a bijection between say the ordered
    triplet ((x1,x2),x3) and the ordered triplet (x1,x2,x3) but I don't see
    that this by itself is enough to show the necessary equivalence without
    using non-rigorous, hand waving arguments.

    Can anyone fill in the missing piece of logic here? (I've never taken an
    algebra class)

    Thanks,
     
    sto, Dec 1, 2008
    #1
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  2. One does not jump from the finite to the infinite without an infinite boost.
    Let { Rj | j in I } be a collection of rings indexed by some index set I.

    Define R = prod_j Rj as the infinite product of the Rj's.
    and define the operations + and * pointwise. That is
    if x = prod_j xj in R, then define
    x + y = prod_j xj + prod_j yj = prod_j (xj + yj)
    and similar with *

    Show R with the so defined operations is a ring.
    This is a standard algebraic construction.

    prod_j Rj = { f:I -> \/_j Rj | for all j in I, f(j) in Rj }
     
    William Elliot, Dec 2, 2008
    #2
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  3. On Mon, 1 Dec 2008 22:22:12 -0800, William Elliot
    Irrelevant: he's not doing so. You've completely
    misunderstood his question.

    [...]

    Brian
     
    Brian M. Scott, Dec 2, 2008
    #3
  4. Huh? Do you mean { (a,b) | a in R1, b in R2 } is a ring? Yes it is.
    I don't know what you're saying. Are you asking if
    { ((a,b),c) | a in R1, b in R2, c in R3) }
    =
    { (a,(b,c)) | a in R1, b in R2, c in R3) }

    No, they aren't equal, they're isomorphic,
    ie algebraic equivalent as rings.

    Otherwise use the function definition of tuples.
    AxBxC = { f:{ 1,2,3 } -> A\/B\/C | f(1) in A, f(2) in B, f(3) in C }
    (x,y,z) = { (1,x), (2,y), (3,z) }
    (x1,.. x_n) = { (j,xj) | j = 1,.. n
    Do you understand what he's saying? The construction I suggested still
    remains a way to avoid induction. What do you think he's saying?
     
    William Elliot, Dec 2, 2008
    #4
  5. On Tue, 2 Dec 2008 00:49:59 -0800, William Elliot
    No, he doesn't.

    [...]
    Yes. With luck I might make time today to post an answer,
    if no one beats me to it.

    Brian
     
    Brian M. Scott, Dec 2, 2008
    #5
  6. sto

    sto Guest

    Thanks.

    What I mean is let's say I construct a proof (which I did) that given
    two rings R1 and R2 the class of finite, disjoint unions of ordered
    pairs (a,b), a in R1, b in R2 is a ring. Then I want to generalize that
    proof to the class of finite, disjoint unions of ordered n-tuples
    (a,b,c,d...), a in R1, b in R2, c in R3... Here n is an integer, so the
    n-tuple is finite.

    It is simple to show by induction that the n-tuples (((a,b),c),d),...)
    are a ring, but these are not the same as the n-tuples (a,b,c,d...) How
    do you get from the induction proof for (((a,b),c),d)....) to stating
    that the same theorem is valid for (a,b,c,d...)?
    I think this is related to the answer to my question. After randomly
    digging through the library for some time, I've now settled on a book on
    abstract algebra which talks about homomorphisms and isomorphisms on
    rings. It seems to me that if I can show that there exists a bijection
    between (((a,b),c),d,.....) and (a,b,c,d...) that preserves set union
    and set difference then that would somehow be sufficient to imply that
    both sets "obey the same algebra", and I can then infer that
    (a,b,c,d...) is a ring directly from the fact that I have proved that
    (((a,b),c),d...) is a ring? Or maybe not.

    This seems to be some kind of standard algebraic technique.


    I haven't looked at the infinite case yet. Here n is strictly finite.
     
    sto, Dec 2, 2008
    #6
  7. sto

    sto Guest

    Forget that last post. After looking at some algebra books I see my
    question has to be stated in a completely different manner. When I
    refer to the term "ring" what I really should be saying is a ring of
    subsets of given set. I define a ring R_ as a nonempty class of subsets
    of a given set X which is closed under the formation of unions and
    differences. (I think this becomes a ring in the usual algebraic sense
    if we define an operation * to be set intersection and an operation + to
    be symmetric difference.) So the way I should have phrased my question
    is that I first prove the following theorem:

    Thm 1: If R1_ and R2_ are rings of subsets of the sets W1 and W2
    respectively, then the class R_ of all finite, disjoint unions of
    rectangles of the form A1 x A2, where A1 in R1_ and A2 in R2_, is a ring.

    Having proved this, it then becomes easily to generalize by induction to
    the following result:

    Thm 2:If R1_, R2_, ..., Rn_ are rings of subsets of the sets W1, W2,
    ...., Wn, then the class R_ of all finite, disjoint unions of rectangles
    of the form (...(A1 x A2) x A3) ..... x An-1) x An), is a ring.

    But that is not the final result I want. The result I want is that the
    class of all finite disjoint union of rectangles of the form A1 x A2 x
    ..... x An is a ring. This set is not the same as (...(A1 x A2) x A3)
    ...... x An-1) x An). How do I get from Thm 2 to the final result I want?

    As far as I know, I could construct the bijection

    p: (...(A1 x A2) x A3) ..... x An-1) x An) -> A1 x A2 x .... x An

    where p( (a1, a2), a3), ... , an) ) = (a1, a2, ... , an)

    and realize that it preserves set intersection and set difference (and
    everything else for that matter). But how does the fact that this
    bijection has these properties actually imply that the rectangles A1 x
    A2 x ..... x An and (...(A1 x A2) x A3) ..... x An-1) x An) share the
    property of being a ring?
     
    sto, Dec 3, 2008
    #7
  8. When A subset R1xR2, A is a set of ordered pairs.
    When A,B subset R1xR2 and A /\ B = nulset, then A \/ B is a disjoint union
    of ordered pairs?

    When a,b in R1, u,v in R2 and (a,u) /\ (b,v) = nulset
    then (a,u) \/ (b,v) is a disjoint union of ordered pairs?
    I suppose.
    What kind of rings are your talking about?

    A Boolean ring is an ordinary algebraic ring with the added
    condition that all elements are idempotent. The elements of
    a Boolean ring don't have to be subsets of some set.
    Every Boolean ring embeds into a product of Z_2's.

    Another sense of ring is a ring over a set S which
    is a subset of P(S) that's closed under unions and
    relative complements. Sigma rings over S are rings
    over S that are closed under countable unions.

    ----
     
    William Elliot, Dec 3, 2008
    #8
  9. sto

    sto Guest

    I was talking about a ring R_ of subsets of a given set W, where a ring
    is a class of subsets of W that is closed under unions and set
    differences. I now realize there are all sorts of other rings in
    algebra that have nothing to do with sets. So my question has to be
    completely re-stated as follows:
     
    sto, Dec 3, 2008
    #9
  10. R = { AxB | A in R1, B in R2 } ?

    No, R is generated by { AxB | A in R1, B in R2 }.
    That it's true for each finite instance, does not imply it's true for any
    infinite instances. What if instead of induction, you proved the finite
    case directly? Does experience proving the two case give insight how to
    directly prove the three case and from that to clues how to prove the
    n-case. Ok, assume you've done that. Now try extending the method to
    countable infinite first. Does it go through? Is there a hang up? If
    there is, then you may have found how to construct a counter example.
    Alternatively you may need to use limits or some sense of density to get
    around the hang up.

    A counter example I attempted was to generate an infinite disjoint
    collection of subsets. That failed because in finite cases attempts to
    produce collections of disjoint subsets seemingly remained constant with
    the number of products.
    I would certainly think it would.
    Let f:(G,+) -> (H,+) be a surjection with
    for all x,y in G, f(x + y) = f(x) + f(y).

    If G is a group, must H be a group? Certainly.
    If a,b in H, then some x,y in G with f(x) = a, f(y) = b.
    Then a + b = f(x + y) is in H.

    If a in H, then some x in G with f(x) = a and
    f(0) + a = f(0 + x) = f(x) = a. Thus f(0) is left identity of H.
    Etc.
     
    William Elliot, Dec 4, 2008
    #10
  11. sto

    sto Guest

    William Elliot wrote:

    THanks.

    After flipping through the group theory book (and a lot of struggle) I
    think I have this figured out.
    What I meant is R generated by {AxB | A in R1, B in R2}. By now I've
    also managed to prove that {AxB} is a semiring that generates R.

    True, for the finite n-case my life this entire month would probably
    have been a lot easier by just proving the theorem directly. But I
    still needed to understand precisely what the confusing "associativity"
    of the Cartesian product really represented mathematically. And there
    was a whole ensemble of related theorems to prove; I could prove maybe
    one or two directly for the n-case, but induction seemed like the only
    simple way to prove all of them for the n-case.

    Ok, assume you've done that. Now try extending the method
    For now I am completely skipping the infinite case, just because of lack
    of time. I may be forced to revisit it if I need to work with infinite
    sequences of random variables.

    I think this is the key piece of information that I was missing. I
    proved a number of related theorems by induction for various classes of
    subsets of (...((W1 x W2) x W3) x W4)...x Wn). Then for each class of
    subsets of (...((W1 x W2) x W3) x W4)...x Wn) for which a proof by
    induction had been carried out, I proved that there exists a bijection
    onto a class of subsets of W1 x W2 x ... x Wn that preserves the entire
    set algebra and therefore for which the theorem is still valid.

    I am still thinking about the relationship between integrals in the two
    spaces, and in precisely what way they can be considered to be equivalent.
     
    sto, Dec 17, 2008
    #11
  12. Good, glad to have been of help. Note that after two days of no response,
    I remove the tread and there after delete all posts that aren't in the
    newer threads. Though I barely remembered the problem somehow,
    especially in view of the low volume of this usegroup, I looked twice at
    the title for it's interest. Luckily I pause to think about it and then
    recalled, perhaps this is a delayed reply. It was. To assure that
    delayed replies don't get discarded, remove the 'Re:' from the subject
    line as I've done with this reply to indicate a renewed thread.

    Having looked at your replies, I see nothing for me to comment upon since
    for the most part it appears to be you thinking out loud and getting a
    handle on the problem. Has it become your koan? ;-)
     
    William Elliot, Dec 18, 2008
    #12
  13. sto

    sto Guest

    To be sure I actually understood this, I had to work through all the
    "trivial" details that nobody pays attention to, which is what took so
    long. In the end, one man's measure space is another man's koan.
     
    sto, Dec 18, 2008
    #13
  14. sto

    sto Guest

    To be sure I actually understood this, I had to work through all the
    "trivial" details that nobody pays attention to, which is what took so
    long. In the end, one man's measure space is another man's koan.
     
    sto, Dec 18, 2008
    #14
  15. Indeed. I may come back to a difficult topic, several days, even weeks or
    months since the last comments.
    Happily it's your problem and not mine.

    Riddle of the day. What's the mean measure of a vicious circle?
     
    William Elliot, Dec 19, 2008
    #15
  16. sto

    sto Guest

    circular angle measure?
     
    sto, Dec 22, 2008
    #16
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