# 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. ### stoGuest

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

2. ### William ElliotGuest

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

3. ### Brian M. ScottGuest

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
4. ### William ElliotGuest

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
5. ### Brian M. ScottGuest

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
6. ### stoGuest

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
7. ### stoGuest

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
8. ### William ElliotGuest

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.

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
9. ### stoGuest

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
10. ### William ElliotGuest

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
11. ### stoGuest

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
12. ### William ElliotGuest

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
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
13. ### stoGuest

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. ### stoGuest

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
15. ### William ElliotGuest

Indeed. I may come back to a difficult topic, several days, even weeks or
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
16. ### stoGuest

circular angle measure?

sto, Dec 22, 2008