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Simfish

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Simfish

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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]

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]

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- #3

Hurkyl

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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.

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

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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- #4

HallsofIvy

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Order is important but the "parentheses" don't- the Cartesian product is "associative" but not "commutative".

- #5

CRGreathouse

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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?Order is important but the "parentheses" don't- the Cartesian product is "associative" but not "commutative".

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