Definition talk:Cartesian Product

For an arbitrary collection $\{X_i:i \in I\}$, we can define $\prod_{i\in I}X_i$ as follows:

$\displaystyle \prod_{i \in I}X_i = \left\{f:I\to\coprod_{i \in I}X_i \vert \forall i: f(i)\in X_i\right\}$

(hmm.. \middle does not work with MathJAX apparently; according to Google it will be in future releases)

The $\coprod$ signifies the disjoint union. It always exists.

Non-emptyness of this Cartesian product for all $I$ is equivalent to the Axiom of Choice. (Similarly, for countable $I$ is equivalent with Axiom of Countable Choice). --Lord_Farin 16:05, 26 October 2011 (CDT)

\middle doesn't work, no - but \mid does: it's $\mid$ and it has nice gaps either side.

But the house style calls for $:$ as $\mid$ can get confused with "absolute value" delimiters etc. in certain contexts. See Definition:Set Definition by Predicate. Colon or pipe are fairly evenly spread thru the literature so it's a reasonable option to go for. --prime mover 16:29, 26 October 2011 (CDT)

Actually, \middle corresponds to \left and \right to size for example the pipe in the middle of the above set (where I deemed using a colon instead confusing). --Lord_Farin 16:32, 26 October 2011 (CDT)

Having said that, there's already two colons in the set definition so pipe works better here. As you were ... --prime mover 16:29, 26 October 2011 (CDT)

Oh yes, and while I think of it, for Definition:Disjoint Union we've used $\sqcup$ and $\displaystyle \bigsqcup$, which can be argued as easier to distinguish from $\prod$ than $\coprod$. --prime mover 16:33, 26 October 2011 (CDT)

In any case, with Equivalence of Versions of Axiom of Choice up, it is about time this definition for arbitrary sets makes it to the actual page itself. --Lord_Farin 04:45, 30 January 2012 (EST)