1. ## Sets

If B has the set {-5, -3, -1, 1} and D has the set {-1, 1}

Then what is D x B x D

2. How would you start?

3. Originally Posted by brumby_3
what is D x B x D
The binary Cartesian product operation is not associative. You need to specify either

D X (B X D)

or

(D X B) X D

4. Originally Posted by MoeBlee
The binary Cartesian product operation is not associative. You need to specify either

D X (B X D)

or

(D X B) X D
Wikipedia n-ary Cartesian product.

5. Then that is stipulating association to the left, that is:

(D X B) X D

So what problem is the original poster having in simply manually iterating the members?

By the way: Wikipedia contradicts itself between the article on Cartesian products and the article on tuples, unless their "variant" method for tuples is adopted. In my experience, for tuples, association to the left is more common (though, as I understand, in computer science, association to the right is often more practical).

6. Originally Posted by MoeBlee
By the way: Wikipedia contradicts itself between the article on Cartesian products and the article on tuples, unless their "variant" method for tuples is adopted. In my experience, for tuples, association to the left is more common (though, as I understand, in computer science, association to the right is often more practical).
For whatever it's worth, I don't see this as a contradiction. Under the subheading "Tuples as nested ordered pairs" here, there is, in my opinion, clear statement that the definition can be made for either direction. (Not all definitions are universally agreed upon.)

In general for sets A,B,C, there is a natural bijection between (A X B) X C and A X (B X C)... I suppose whether this is an isomorphism depends on what algebraic structure is imposed on the sets. (?)

7. Originally Posted by undefined
For whatever it's worth, I don't see this as a contradiction. Under the subheading "Tuples as nested ordered pairs" here, there is, in my opinion, clear statement that the definition can be made for either direction.
Yes, I said "unless".

Originally Posted by undefined
In general for sets A,B,C, there is a natural bijection between (A X B) X C and A X (B X C)... I suppose whether this is an isomorphism depends on what algebraic structure is imposed on the sets. (?)
That they biject and that a relation can be defined to provide an isomorphism is not in contention. But for purposes of listing the exact members of the set (such as the question given in this thread), it is is necessary to specify association to the left or association to the right.

8. Originally Posted by MoeBlee
Yes, I said "unless".
Well it seemed you were indicating the Wikipedia page should be corrected.. I guess it gets into the emphasis of words in your sentence structure, which doesn't seem worth discussing..

Originally Posted by MoeBlee
That they biject and that a relation can be defined to provide an isomorphism is not in contention. But for purposes of listing the exact members of the set (such as the question given in this thread), it is is necessary to specify association to the left or association to the right.
Not really, just use the convention used for R^3; list members of A X B X C as ordered triples (a,b,c) with a in A, b in B, c in C.

9. Originally Posted by undefined
Well it seemed you were indicating the Wikipedia page should be corrected
I don't use Wikipedia as a reference in mathematics. However, what I said in this instance is not that Wikipedia needs to be corrected, but rather what I said is as exactly as I wrote under the ordinary sense of the word "unless".

Originally Posted by undefined
list members of A X B X C as ordered triples (a,b,c) with a in A, b in B, c in C.
Not if ordered triples reduce to ordered tuples (such as even in the Wikipedia mention you lihnked to). Of course, one can take such objects to be actual sequences (actual FUNCTIONS) with, e.g., triples having the domain {0 1 2} = 3, or some other convention, and then that iterated Cartesian products are to be taken in this sense. But, again, that requires specification that that particular sense is being used. (Of course, in the case where the indexing set is INFINITE, the Cartesian products is indeed taken in the sense of a set of functions since it is the most obvious sense of what it would mean to iterate pairing "infinitely").

But In the sense earlier mentioned (even as you linked to it as a particular specification in Wikipedia), it is not the case that in general D X (B X D) = (D X B) X D.

10. Originally Posted by MoeBlee
I don't use Wikipedia as a reference in mathematics. However, what I said in this instance is not that Wikipedia needs to be corrected, but rather what I said is as exactly as I wrote under the ordinary sense of the word "unless".

Not if ordered triples reduce to ordered tuples (such as even in the Wikipedia you mentioned). Of course, one can take such objects to be actual sequences (actual FUNCTIONS) with, e.g., triples having the domain {0 1 2} = 3, or some other convention, and then that iterated Cartesian products are to be taken in this sense. But, again, that requires specification that that particular sense is being used. (Of course, in the case where the indexing set is INFINITE, the Cartesian products is indeed taken in the sense of a set of functions since there is no there it is the most obvious sense what it would mean to iterate pairing "infinitely").

In the sense earlier mentioned (even as you linked to it as a particular specification in Wikipedia), it is not the case that in general D X (B X D) = (D X B) X D.
This discussion is getting rather silly. I agree that it is not the case that in general D X (B X D) = (D X B) X D. But for listing out the elements of D X B X D, you can do so unambiguously by listing all ordered triples (d_1, b, d_2) with d_i in D and b in B.

Whether we define (d_1,b,d_2) as ((d_1,b),d_2) or (d_1,(b,d_2)) has absolutely no relevance to listing out all elements as (d_1,b,d_2).

Here, I'll just do it, since I really don't desire for this ridiculousness to continue

D X B X D =
{
(-1,-5,-1),
(-1,-5,1),
(-1,-3,-1),
(-1,-3,1),
(-1,-1,-1),
(-1,-1,1),
(-1,1,-1),
(-1,1,1),
(1,-5,-1),
(1,-5,1),
(1,-3,-1),
(1,-3,1),
(1,-1,-1),
(1,-1,1),
(1,1,-1),
(1,1,1)
}

11. Originally Posted by undefined
for listing out the elements of D X B X D, you can do so unambiguously by listing all ordered triples (d_1, b, d_2) with d_i in D and b in B.
But then you've abandoned the stipulation gven in your original link to Wikipedia.

Ordinarily, triples reduce to tuples. And, for example,

((-1 -5) -1) does not equal (-1 (-5 -1)) in the ordinary sense of a tuples as ordered pairs (such as in the Kuratowski definition).

Yes, IF we agree that triples are not reduced to tuples but rather that triples are literal functions on the domain {0 1 2}, then the notation '(-1 -5 -1)' stands for {(0 -1) (1 -5) (2 -1)}, and then IF we further agree that 'D X B X D' stands for {(x y z) | x in D & y in B & z in D} then there is no matter as to association.

I don't mind that you adopt whatever notational convention you want. But IF we take triples as tuples (which was the stipulation mentioned at the partiucular point in the Wikipedia reference you gave) and not as literal functions, then ((-1 -5) -1) does not equal (-1 (-5 -1)). And your OWN reference to Wikipedia bears that out.

Ordinarily this is a matter of agreeing on whatever convention we find suitable. But it deserves being clear as to which particular convention is being agreed upon.

And this is not always entirely trivial. For example:

To prove the existence of the set, for all n, of n-tuples generated from a non-empty set, we require the axiom schema of replacement. But to prove merely the existence of the set of finite sequences (literal functions) over a non-emtpy set does not require the axiom schema of replacement.

12. Well I debate whether or not to post this, but it seems worth mentioning...

Originally Posted by MoeBlee
Ordinarily, triples reduce to tuples.
Triples ARE tuples. A triple is another name for a 3-tuple, just as a pair is another name for a 2-tuple.

We learn in grade school that there are ordered pairs, ordered triples, (ordered quadruples, etc.) .. we need a name for these objects that have arbitrary n number of elements, to make our lives easier, and we call them n-tuples. (The term "ordered" is optional.)

So why would a triple reduce to a tuple, when a triple IS a tuple?

Once we have a clear understanding that tuples exist, we do not need to constantly go back to the definition of tuple (whichever definition we choose). In the same way, once we have constructed the naturals, we don't have to keep going back to the definition.

If we have a question, 1+1 = x, we will answer x = 2, we won't answer x = {0,1}. Moreover we won't answer x = {{},{{}}}. Even though technically speaking these are valid answers...

I believe it is standard to consider 3-ary Cartesian products A X B X C as sets of triples, but someone correct me if I'm wrong.

13. Originally Posted by undefined
Triples ARE tuples. A triple is another name for a 3-tuple, just as a pair is another name for a 2-tuple.
In context, I hope it would be clear that by 'tuple' I meant specifically ordered pairs. And, indeed, under the ordinary set theoretic definition, a triple does reduce to an ordered pair. My referring to triples as tuples in the sense of ORDERED PAIRS is right in line with the exact link you provided in Wikipedia, where it is mentioned: "If tuples are defined as nested ORDERED PAIRS [...]" [emphasis added], which is the context I even agreed to accept from you having given it by link.

Originally Posted by undefined
We learn in grade school that there are ordered pairs, ordered triples, (ordered quadruples, etc.) .. we need a name for these objects that have arbitrary n number of elements, to make our lives easier, and we call them n-tuples. (The term "ordered" is optional.)
And in adult mathematics we learn a specification more rigorous than that ordinarily conveyed in grade school.

Originally Posted by undefined
So why would a triple reduce to a tuple, when a triple IS a tuple?
In the context of my remarks, it should be clear that a triple reduces to a tuple in the sense that a triple is obtained by iterated ordered pairing, as opposed to ANOTHER, but related, notion of a triple as a function whose domain is {0 1 2}.

Originally Posted by undefined
Once we have a clear understanding that tuples exist, we do not need to constantly go back to the definition of tuple (whichever definition we choose). In the same way, once we have constructed the naturals, we don't have to keep going back to the definition.
That will depend on what specific question is at hand. If the question is what are the MEMBERS of the natural number 1 then we do have to refer to whatever set theoretic definition of '1' we are using.

Originally Posted by undefined
If we have a question, 1+1 = x, we will answer x = 2, we won't answer x = {0,1}. Moreover we won't answer x = {{},{{}}}. Even though technically speaking these are valid answers...
Actually, depending on context, {0 1} might very well be the notation we seek.

Originally Posted by undefined
I believe it is standard to consider 3-ary Cartesian products A X B X C as sets of triples, but someone correct me if I'm wrong.
Yes, of course. But answers to certain questiions may ride on what SENSE of 'triple ' is meant. There may be a good degree of variation from author to author. However, in most set theory contexts I have found, iterated Cartesian products group to the left. So, 'A X B X C' would be (A X B) X C, which is not in general the same set as A X (B X C), where (A X B) X C = {((x y) z) | x in A & x in B & x in C} and A X (B X C) = {(x (y z)) | x in A & x in B & x in C}.

As I said, I don't begrudge you taking whichever notational convention you like. However, the answer to the specific question of what are the MEMBERS of the set DOES depend on which of various conventions we do adopt, whether association to the left, association to the right, or an actual function.

14. Oh Gosh what has happened to this thread lol. What have I started? Thanks undefined, I happened to get the same answers as you specified after thinking about it some more last night. Haha I'm happy to see such a vigorous debate here. Thanks for all your contributions.

15. Originally Posted by brumby_3
Thanks undefined, I happened to get the same answers as you specified after thinking about it some more last night.
You're welcome!

Originally Posted by brumby_3
Oh Gosh what has happened to this thread lol. What have I started? ... Haha I'm happy to see such a vigorous debate here.
I know, it's a little out of hand.

Edit: Originally wrote out lengthy response to MoeBlee, but on second thought it's really not worth it. I'll just let MoeBlee have the last word. All that needed to be said has already been said. I believe there was some communication failure but there's no point discussing it further.