The sum $\sum_{i \in I} \kappa_i$ and the product $\prod_{i \in I} \kappa_i$ of an indexed family of cardinals, defined under the Axiom of Choice

Assume the Axiom of Choice (The Axiom of Choice). Let $I$ be a set and $(\kappa_i)_{i \in I}$ a family of cardinals (Cardinal (initial ordinal) and cardinality), that is, a function on $I$ whose value at $i$ is the cardinal $\kappa_i$. Put

$\textstyle\bigsqcup_{i \in I} \kappa_i \;:=\; \bigcup_{i \in I} \big(\{i\} \times \kappa_i\big), \qquad \prod^{\mathrm{set}}_{i \in I} \kappa_i \;:=\; \{\, f : f \text{ is a function on } I \text{ with } f(i) \in \kappa_i \text{ for all } i \in I \,\},$

both sets by Replacement, Union and Power Set. The sum and product of the family are their cardinalities:

$\sum_{i \in I} \kappa_i \;:=\; \Big\lvert \textstyle\bigsqcup_{i \in I} \kappa_i \Big\rvert, \qquad \prod_{i \in I} \kappa_i \;:=\; \Big\lvert \prod^{\mathrm{set}}_{i \in I} \kappa_i \Big\rvert .$

Why the hypothesis is in the definition. Both right-hand sides are cardinalities of sets that ZF does not well-order. Under the Axiom of Choice every set is well-orderable (The well-ordering theorem) and both values exist (A set equinumerous with some ordinal has a least such ordinal, that ordinal is a cardinal, and equinumerous sets get the same one; no choice principle is used). Nothing else is being assumed: the two sets themselves are constructed in ZF, and the family $(\kappa_i)_{i \in I}$ is a function, so no representative is selected.

The finite cases are the operations already defined. Take $I = 2 = \{0,1\}$. Then $\bigsqcup_{i \in 2} \kappa_i = (\{0\} \times \kappa_0) \cup (\{1\} \times \kappa_1) = \kappa_0 \sqcup \kappa_1$ literally, so $\sum_{i \in 2} \kappa_i = \kappa_0 \oplus \kappa_1$ (Cardinal sum $\kappa \oplus \lambda$, product $\kappa \otimes \lambda$ and exponentiation $\kappa^{\lambda}$, and why they are written apart from the ordinal operations); and $f \mapsto (f(0), f(1))$ is a bijection from $\prod^{\mathrm{set}}_{i \in 2} \kappa_i$ onto $\kappa_0 \times \kappa_1$, with inverse $(a,b) \mapsto \{(0,a),(1,b)\}$, so $\prod_{i \in 2} \kappa_i = \kappa_0 \otimes \kappa_1$ by A set equinumerous with some ordinal has a least such ordinal, that ordinal is a cardinal, and equinumerous sets get the same one; no choice principle is used.

A constant family recovers $\otimes$ and exponentiation. If $\kappa_i = \kappa$ for every $i \in I$ and $\lambda = \lvert I \rvert$, then $\bigsqcup_{i \in I} \kappa = I \times \kappa$ and $\prod^{\mathrm{set}}_{i \in I} \kappa = {}^{I}\kappa$, so

$\sum_{i \in I} \kappa = \lambda \otimes \kappa, \qquad \prod_{i \in I} \kappa = \kappa^{\lambda},$

by the transport clause of Cardinal sum $\kappa \oplus \lambda$, product $\kappa \otimes \lambda$ and exponentiation $\kappa^{\lambda}$, and why they are written apart from the ordinal operations together with Disjoint union, cartesian product, function space and power set respect equinumerosity, and for ordinals $\alpha, \beta$ the sets $\alpha \sqcup \beta$ and $\alpha \times \beta$ carry explicit well-orders, so their cardinalities exist in ZF and Injection, surjection, bijection.

The product set is the set of choice functions. An element of $\prod^{\mathrm{set}}_{i \in I} \kappa_i$ picks one element of $\kappa_i$ for every $i$, which is exactly a choice function for the family (Choice function). So the assertion "the product set is nonempty when every $\kappa_i$ is nonempty" is the Axiom of Choice for that family, in the formulation recorded in The Axiom of Choice, and it is not an incidental consequence of the definition.

Why the sum tags its blocks. Without the tag $\{i\} \times \kappa_i$ the union $\bigcup_i \kappa_i$ would be a union of ordinals, which is the supremum of the family and not its sum: with $\kappa_i = 1$ for every $i \in \omega$ the untagged union is $1$, while the sum is $\aleph_0$, and the difference is exactly that the tagged blocks are disjoint. The tagging is the same device Cardinal sum $\kappa \oplus \lambda$, product $\kappa \otimes \lambda$ and exponentiation $\kappa^{\lambda}$, and why they are written apart from the ordinal operations uses for $\oplus$, applied to an arbitrary index set.

What is not defined here. Nothing is said about $\sum$ and $\prod$ over an index set for which the family has no cardinal values, and nothing is said in ZF alone. The theorem this definition exists for, König's theorem: assuming the Axiom of Choice, if $\kappa_i < \lambda_i$ for every $i \in I$ then $\sum_{i \in I} \kappa_i < \prod_{i \in I} \lambda_i$, carries the same hypothesis for the same reason.