Is Cartesian Product associative?

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December 1st 2010, 11:49 AM
Matt Westwood

Is Cartesian Product associative?

Let $A, B, C$ be sets.

Is it the case that $A \times (B \times C) = (A \times B) \times C$
where $(A \times B)$ denotes Cartesian Product?

That is, $A \times B = \{(a, b): a \in A, b \in B\}$ .

I have J.A. Green's "Sets and Groups" (1965, Routledge & Kegan Paul) saying:
"Show that the sets $A \times (B \times C)$ and $(A \times B} \times C$ are never the same" (Chapter One Exercise 8), on the one hand.

On the other hand, I have W.E. Deskins: Abstract Algebra (1964, Dover) saying:
"Theorem 1.9. $(a, b, c) = (d, e, f)$ if and only if $a=d, b=e, c=f$ ."

The definition of $(a, b, c)$ in Deskins is given as $(a, (b, c))$ , i.e. as a construction of ordered pairs.

This latter sort of seems to imply that $A \times (B \times C) = (A \times B) \times C$ , but no mention of it is made in that book.

T.S. Blyth's "Set Theory and Abstract Algebra" (Longman, 1975) gives:
$E \times F \times G = \{(x, y, z): x \in E, y \in F, z \in G\}$
which sort of seems to imply that in this context associativity does hold.

I've seen this set as an exercise in several places, but never have I seen a definitive proof one way or another. It hinges on whether $(a, (b, c)) = (a, b, c) = ((a, b), c)$ which you instinctively feel ought to be true, but I can't come up with a convincing argument either way, unless you take literally the expression of $(a, b)$ as equal to $\{\{a\}, \{a, b\}\}$ (according to Wiener and Kuratowski) which is in the final instance a convenient way of obtaining an ordered pair in axiomatic set theory (e.g. ZF). And Green's "Sets and Groups" (mentioned above) is pretty assertive when it comes to exercise 1.9, and no mention has been made in that work of the Wiener/Kuratowski definition.

What's the current thinking on this result?
December 1st 2010, 12:30 PM
DrSteve

From a set theoretic point of view, every object is a set which means that any true definition of an ordered pair should be equivalent to the last definition you have in terms of sets. In this case there are no issues.

If associativity doesn't hold, then it still almost holds in the sense that there is a natural bijection between the cartesian products.

I'm not sure if anyone currently uses the nonassociative notion, but I certainly wouldn't.
December 1st 2010, 12:37 PM
emakarov

When one talks about associativity, one assumes that the product is a binary, not ternary, operation. Therefore, if Blyth's definition $E \times F \times G = \{(x, y, z): x \in E, y \in F, z \in G\}$ involves ordered triples, it probably defines a ternary operation and is irrelevant to the question of associativity.

$(A\times B)\times C$ consists of elements of the form $((a, b), c)$ , while $A\times (B\times C)$ consists of elements of the form $(a, (b, c))$ . At least when A, B, C themselves don't contain pairs, two elements from $(A\times B)\times C$ and $A\times (B\times C)$ cannot be equal.

As was pointed out, there is a natural bijection that maps ((a, b), c) to (a, (b, c)).
December 1st 2010, 12:55 PM
Matt Westwood

*smacks forehead* D'oh. Of course. Thanks bro's.