let $(X\_i)\_{i \in I}$ be an infinite family of sets with $|X\_i| \geq 2$. we define an equivalence relation on $X = \prod\_{i \in I} X\_i$ by $x \sim y \Leftrightarrow \{i : x\_i \neq y\_i\}$ is finite. what is the cardinality of $X/\sim$? we may endow the $X\_i$ with group structures and write this set as $\prod\_{i \in I} X\_i / \oplus\_{i \in I} X\_i$.


it is easy to see that $\min\_i |X\_i| \leq |X/\sim|$ and $|X| \leq |X/\sim| \* \max\_i |X\_i|$. in particular, if $|X\_i|$ is constant, $|X/\sim| = |X|$.


if all the $X\_i$ are finite, it can be shown $|X/\sim|=2^{|I|}$. the same equation holds, if every $X\_i$ satisfies $\aleph\_0 \leq |X\_i| \leq |I|$ (I'll add the proofs if needed).


the general case struggles me.



cardinality of product modulo direct sum
The cardinality of the reduced product is always the same as that of the product, modulo omitting finitely many unusually large $X\_i$'s. (Even when $I$ is finite, in which case all $X\_i$'s should be omitted.)


Arrange the sets $X\_i$ in nondecreasing order of size in a wellordered sequence $(X\_\alpha)\_{\alpha<\tau}$. We may assume that $\tau$ is a limit ordinal. Otherwise, the last coordinate $\tau-1$ (like any single coordinate) contributes nothing to the reduced product. Omit $X\_{\tau-1}$ and repeat as long as necessary.


I will show that then $|X| = |X/{\sim}|$ where $X = \prod\_{\alpha<\tau} X\_\alpha$.


Let $\kappa = \sup\_{\alpha<\tau} |X\_\alpha|$. Since each ${\sim}$-equivalence class has size $|\tau|\cdot\kappa$ (see note) we have 


$|X| \leq |X/{\sim}|\cdot|\tau|\cdot\kappa = \max(|X/{\sim}|,|\tau|,\kappa).$ 


Since $|X| \geq 2^{|\tau|} > |\tau|$, we conclude that either $|X/{\sim}| = |X|$ or $|X/{\sim}| \leq |X| \leq \kappa$.


Since $\tau$ is a limit ordinal, the diagonal embedding $d:\kappa \to \prod\_{\alpha<\tau} |X\_\alpha|$, where 


$d\_\alpha(\xi) = \begin{cases} \xi & when\ \xi < |X\_\alpha| \\ 0 & otherwise,\end{cases}$


shows that $\kappa \leq |X/{\sim}|$. So, in the case $|X/{\sim}| \leq |X| \leq \kappa$, we in fact have $|X/{\sim}| = |X| = \kappa$.


Note: The elements ${\sim}$-equivalent to a given $x \in X$ are obtained by selecting finitely many new values from the sets $X\_\alpha-\{x\_\alpha\}$ to replace the corresponding value of the sequence $x$. There are at least $\sum\_{\alpha<\tau} |X\_\alpha-\{x\_\alpha\}|$ and no more than $\left(\sum\_{\alpha<\tau} |X\_\alpha|\right)^{<\omega}$ ways of doing this. Since $\tau$ is infinite and the $X\_\alpha$'s all have two or more elements, these two bounds are equal to $\sum\_{\alpha<\tau} |X\_\alpha| = |\tau|\cdot\kappa$.


