Cartesian product of cartesian products

Gold Member
So if we have sets A X B where A is (a_n, b_m) and C X D where C is (c_o, d_p), the cartesian product of the sets is (A X B) X (C X D) [(a_n, b_m, c_o, d_p)]. Is this correct? And thus, do parenthesis matter at all in Cartesian products? What about order? Is (a_n, b_m) equivalent to (b_m, a_n)?

I believe these are in fact ordered pairs. So order does matter.

If we use the definition that an ordered pair $(x,y) = \{\{x\}, \{x,y\}\}$ then clearly $(y,x) = \{\{y\}, \{x,y\}\} \neq (x,y)$

As for the first thing say $A$ contains elements of the form $(x,y)$ then $A\times B = \{ ((x,y), b) | (x,y) \in A, b \in B \}$

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Hurkyl
Staff Emeritus
Gold Member
The cartesian product is a symmetric monoidal operation (both on sets, and on elements) -- that means it has an identity, is associative, and is commutative... but only up to a natural isomorphism.

An identity set is any set with a single element. I'll call such a set '1'.
For all sets A, B, C, we have:
An isomorphism 1xA --> A ('left identity')
An isomorphism A --> Ax1 ('right identity')
An isomorphism (AxB)xC --> Ax(BxC) (the 'associator')
An isomorphism AxB --> BxA (the 'braiding')

For example, the associator is the function:
$\alpha((a, b), c) = (a, (b, c))$

None of these isomorphisms need be equalities. In fact, for the usual set-theoretic model of the ordered pair, none of these will be equalities. (I think the only exception is when the empty set is involved)

If we have a function A --> B, then we have functions
AxC --> BxC
CxA --> CxB

And each of these natural isomorphisms 'commute' with applying a function. i.e. the two compositions

AxC --> BxC --> CxB
and
AxC --> CxA --> CxB

both yield the same function AxC --> CxB.

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HallsofIvy
I don't know. Using the standard (a, b) := {{a}, {a, b}}, $(a, (b, c))\neq((a, b), c)$. There is a natural bijection between them, but then again there's a natural bijection between (a, b) and (b, a) too -- so why do you say that they associate but not commute?