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)

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.

[...]

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

[...]

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...)?

This seems to be some kind of standard algebraic technique.

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 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.
 >

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.

 >>> One does not jump from the finite to the infinite without an
infinite boost.
 >>
 >> Irrelevant: he's not doing so.  You've completely misunderstood his
question.

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)

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?

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.