Cartesian product of cartesian products

  • Thread starter Simfish
  • Start date
  • #1
Simfish
Gold Member
818
2

Main Question or Discussion Point

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)?
 

Answers and Replies

  • #2
225
0
I believe these are in fact ordered pairs. So order does matter.

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

As for the first thing say [itex]A[/itex] contains elements of the form [itex](x,y)[/itex] then [itex]A\times B = \{ ((x,y), b) | (x,y) \in A, b \in B \}[/itex]
 
Last edited:
  • #3
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
17
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:
[itex]\alpha((a, b), c) = (a, (b, c))[/itex]


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.
 
Last edited:
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,770
911
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)?
Order is important but the "parentheses" don't- the Cartesian product is "associative" but not "commutative".
 
  • #5
CRGreathouse
Science Advisor
Homework Helper
2,820
0
Order is important but the "parentheses" don't- the Cartesian product is "associative" but not "commutative".
I don't know. Using the standard (a, b) := {{a}, {a, b}}, [itex](a, (b, c))\neq((a, b), c)[/itex]. 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?
 

Related Threads for: Cartesian product of cartesian products

  • Last Post
Replies
9
Views
10K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
10
Views
10K
  • Last Post
Replies
4
Views
633
  • Last Post
Replies
8
Views
4K
Replies
3
Views
4K
Replies
3
Views
2K
Top