Talk:Cardinality of Cartesian Product

When we say "$\left|{S \times T}\right| = \left|{S}\right| \times \left|{T}\right|$", is the second $\times$ also the Cartesian product? If it is, please delete this comment. If it's not, maybe we should specify we're using $\times$ in two different ways? --GFauxPas 11:24, 18 November 2011 (CST)
 * Seriously, I never gave it a thought that this would pose the slightest problem. $S$ and $T$ are sets, so $\times$ is the Cartesian product, as is specified. On the other hand $|S|$ and $|T|$, defined as they are of being the cardinality of $S$ and $T$ (that is, the number of elements in $S$ and $T$) are numbers. I didn't think it would be necessary to specify the second use of $\times$ as being ordinary boring multiplication on numbers, especially as it becomes obvious in the body of the proof. Do you think it's worth while explaining exactly what $|S| \times |T|$ really means. --prime mover 14:00, 18 November 2011 (CST)
 * I don't know, because I've seen several different definitions of cardinality, and I don't know which one is "correct"
 * Definition: The cardinality of a set is a natural number etc. etc.
 * Definition: The cardinality of a set is the unique property common to all sets that have a one-to-one relation between them. That is, the property of being conumerous to the sets as the following:
 * A set has one element iff $x \in X \land y \in X \implies x = y$
 * A set has two elements iff $x \in X \land y \in X \land z \in X \implies x = y \lor x = z$, etc. (Don't know how this works with infinite sets) --GFauxPas 14:08, 18 November 2011 (CST)
 * They're the same thing. The page you can get to by pressing the link under "cardinality" explains well enough for the purpose of this page. I've specified that the sets need to be finite, that should sort it out. --prime mover 14:10, 18 November 2011 (CST)
 * Okay now that the set is finite then for sure it means the product of two numbers, and I don't have questions anymore. You understood my question better than how I said it, thanks. --GFauxPas 14:12, 18 November 2011 (CST)
 * The really subtle thing is that this result does work for infinite sets, just that we haven't got round to specifying the arithmetic of transfinite cardinals. (E.g. if $S$ is countable but $T$ is uncountable, then $S \times T$ is uncountable, and so on.) But my set theory "sucks real bad", as they say in the States. --prime mover 14:14, 18 November 2011 (CST)