- #1
chipotleaway
- 174
- 0
Cartesian product of indexed family of sets
The definition of a Cartesian product of an indexed family of sets [itex](X_i)_{i\in I}[/itex] is [itex]\Pi_{i\in I}X_i=\left\{f:I \rightarrow \bigcup_{i \in I} \right\}[/itex]
So if I understand correctly, it's a function that maps every index [itex]i[/itex] to an element [itex]f(i)[/itex] such that [itex]f(i) \in X_i[/itex]…my question is, is there supposed to be a notion of 'order' implied in the definition here?
The 'usual' Cartesian product is [itex]\Pi^n_{i=1}X_i=\left\{ (x_i)^n_{i=1}: x_i \in X_i \right\}[/itex] and here, there seems to be some notion of 'order' cause the first element is in [itex]X_1[/itex], second in [itex]X_2[/itex] etc.
But for the first, if the index set is I={1,2,3}, then couldn't we have have [itex]X_1 \times X_2 \times X_3[/itex] or [itex]X_2\times X_1\times X_3[/itex] (any reordering of indices in the index set)? Then the definition of the Cartesian product could be [itex]((f(1), f(2), f(3))[/itex] or [itex](f(2), f(1), f(3))[/itex]...
The definition of a Cartesian product of an indexed family of sets [itex](X_i)_{i\in I}[/itex] is [itex]\Pi_{i\in I}X_i=\left\{f:I \rightarrow \bigcup_{i \in I} \right\}[/itex]
So if I understand correctly, it's a function that maps every index [itex]i[/itex] to an element [itex]f(i)[/itex] such that [itex]f(i) \in X_i[/itex]…my question is, is there supposed to be a notion of 'order' implied in the definition here?
The 'usual' Cartesian product is [itex]\Pi^n_{i=1}X_i=\left\{ (x_i)^n_{i=1}: x_i \in X_i \right\}[/itex] and here, there seems to be some notion of 'order' cause the first element is in [itex]X_1[/itex], second in [itex]X_2[/itex] etc.
But for the first, if the index set is I={1,2,3}, then couldn't we have have [itex]X_1 \times X_2 \times X_3[/itex] or [itex]X_2\times X_1\times X_3[/itex] (any reordering of indices in the index set)? Then the definition of the Cartesian product could be [itex]((f(1), f(2), f(3))[/itex] or [itex](f(2), f(1), f(3))[/itex]...
Last edited: