What is the cardinality of an infinite product of sets? We have $(A_i)_{i\in I}$ a family of sets. We can suppose that the axiom of choice is true. I am trying to prove that for every $\kappa$ infinite cardinal, we have $card(\prod_{i \in I} A_i)&lt;\kappa'$ for a certain $\kappa '$ cardinal (to be determined) if and only if $\forall i\in I,card(A_i)&lt;\kappa$ and $card(I)&lt;\kappa$. How can one prove this ? Is this even true ?
The cardinality of the product of a family of sets is easily proved to be the $\sup$ of the cardinalities of the sets in the family (provided they are infinite of course) or the cardinality of $I$, whatever is bigger.

If $\kappa$ is a regular cardinal, then you can set $\kappa' = \kappa$, because by definition no sequence of less than $\kappa$ cardinals all smaller that $\kappa$ can sum $\kappa$.

If on the other side $\kappa$ is singular (that is, its cofinality is strictly less than $\kappa$) then you can have that $\sup(card(A_i)) = \kappa$ if $I$ is the cofinality of $\kappa$. so you can set $\kappa'$ to be the successor of $\kappa$.

I'll leave you to fill out the details.

Edit:

Thinking a bit more, it seems that I am completely wrong!

One of the reasons is that (provided enough $A_i$ are at least $2$) then

$$
\left| \prod_{i\in I} A_i\right| ≥ \left| \mathscr{P}(I) \right|
$$

and that number may be anything...
The cardinality of the product of a family of sets is easily proved to be the $\sup$ of the cardinalities of the sets in the family (provided they are infinite of course) or the cardinality of $I$, whatever is bigger.

If $\kappa$ is a regular cardinal, then you can set $\kappa' = \kappa$, because by definition no sequence of less than $\kappa$ cardinals all smaller that $\kappa$ can sum $\kappa$.

If on the other side $\kappa$ is singular (that is, its cofinality is strictly less than $\kappa$) then you can have that $\sup(card(A_i)) = \kappa$ if $I$ is the cofinality of $\kappa$. so you can set $\kappa'$ to be the successor of $\kappa$.

I'll leave you to fill out the details.

Edit:

Thinking a bit more, it seems that I am completely wrong!

One of the reasons is that (provided enough $A_i$ are at least $2$) then

$$
\left| \prod_{i\in I} A_i\right| ≥ \left| \mathscr{P}(I) \right|
$$

and that number may be anything...