Generalizing the usual defining or addition and multiplication of two cardinal numbers, one can define the sum and the product of the entire indexed family $(\kappa_i)_{i \in I}$ of cardinals:
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$\sum_{i \in I} \kappa_i = |\bigsqcup_{i \in I} \kappa_i|$
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$\prod_{i \in I} \kappa_i = |\prod_{i \in I} \kappa_i|$, that is, the cardinality of the Cartesian product of the family $(\kappa_i)_{i \in I}$.
The question is, how much the results of such addition and multiplication depend on a choice of an index set $I$? In the finite case, for cardinal number $\kappa,\lambda,\mu$ and $\nu$ we have
$$\kappa\cdot\lambda\cdot\mu\cdot\nu = \kappa\cdot\mu\cdot\lambda\cdot\nu = \nu\cdot\lambda\cdot\kappa\cdot\mu,$$
etc. It follows the the usual commutativity laws of addition and multiplication of cardinals. But what about infinite case?
For example, it is true that
for an two indexed families $(\kappa_i)_{i \in I}$ and $(\mu_j)_{j \in J}$ of cardinal numbers, if $|I| = |J|$ and for any $i \in I$ there is $j \in J$ so that $\kappa_i = \mu_j$, then the sums and products of these families agree?
If not, what is true? I guess I'm seeking the generalized commutatitivy law for such sums and products, something more than simply considering the rearranged family $(\mu_i)_{i \in I}$ for a permutation $\sigma\colon I\to I$ such that for any $i \in I$ we have $\mu_i = \kappa_{\sigma(i)}$.
P.S. I tag this as set-theory rather than elementary-set-theory because the very definition of a cardinal number depends on the axiom of choice and the info for the set-theory tag advises tagging questions pertaining the axiom of choice as such.
Best Answer
"For any $i\in I$ there is $j\in J$ so that $\kappa_i=\mu_j$" is certainly not enough. For example, we could take $I=J=\mathbb N$ and $$ \kappa_i = \begin{cases} 1& \text{if }i=42 \\ 2 &\text{otherwise} \end{cases} $$ $$ \lambda_j = \begin{cases} 2& \text{if }j=42 \\ 1 &\text{otherwise} \end{cases} $$ Then $\prod_i \kappa_i = 2^{\aleph_0}$ but $\prod_j \lambda_j = 2$.
If you have an actual bijection $f:I\to J$ such that $\kappa_i = \mu_{f(i)}$ it is a different matter. More than that I don't think you can get.