Vtyjkyuk

I came up with this theorem since the proof of $I$ is infinite, $A_k$ is countably infinite, and $A_i$ is countable for all $i neq k$. Is $prodlimits_iin IA_i$ countable? is realated to the rearrangement of $A_i$.

While I'm quite sure about the bijection, I'm not sure if my proof for its injection and surjection contains any error. I'm very happy If you can verify that part. Thank you so much!

The Cartesian product of a family $(A_imid iin I)$ is defined as $$prodlimits_iin IA_i=f:Itobigcup A_imid f(i)in A_i text for all i in I$$

Theorem: Let $h:I to I$ be a bijection. Then $prodlimits_iin IA_i$ and $prodlimits_iin IA_h(i)$ are equinumerous.

Proof: Consider

$$begin
arrayl
G : & prodlimits_iin IA_i
& longrightarrow & prodlimits_iin IA_h(i) \
& phi & longmapsto & G(phi) endarray$$

where $phi:I to bigcup_iin IA_i$ such that $phi(i) in A_i$ for all $i in I$.

$G(phi)$ is defined by $G(phi)(i)=phi circ h(i)$ for all $i in I$.

1. $G(phi) in prodlimits_iin IA_h(i)$

Because $h:I to I$ is bijective, $h(i) in I$ for all $i in I$. Thus $phi circ h(i) in A_h(i)$ since $phi in prodlimits_iin IA_i$. It follows that $G(phi)(i) in A_h(i)$ for all $i in I$. Hence $G(phi) in prodlimits_iin IA_h(i)$.

2. $G$ is surjective
   

For $psi in prodlimits_iin IA_h(i)$, define $phi$ by $phi(i)=psi circ h^-1(i)$ where $h^-1$ is the inverse mapping of $h$.

Then $psi:I to bigcup_i in IA_h(i)$ such that $psi(i) in A_h(i)$ for all $i in I$.

By definition, $phi(i)=psi circ h^-1(i) in A_h circ h^-1(i)=A_i$ for all $i in I$. Thus $phi in prodlimits_iin IA_i$.

By definition, $G(phi)(i) = phi circ h(i) = (psi circ h^-1) circ h(i) = psi circ (h^-1 circ h) (i) = psi(i)$ for all $i in I$.

Thus $G(phi)=psi$. Hence $G$ is surjective.

3. $G$ is injective
   

For $phi_1,phi_2 in prodlimits_iin IA_i$ and $G(phi_1)=G(phi_2)$. Then $G(phi_1)(i)=G(phi_2)(i)$ for all $i in I$, then $phi_1 circ h(i) = phi_2 circ h(i)$ for all $i in I$.

Since $h:I to I$ is bijective, $phi_1 (i) =phi_1 circ h(i')$ and $phi_2 (i) =phi_2 circ h(i')$ for a unique $i' in I$. Since $phi_1 circ h(i') = phi_2 circ h(i')$ for all $i' in I$, $phi_1 (i) = phi_2 (i)$ for all $i in I$.

Thus $phi_1= phi_2$. Hence $G$ is injective.

I agree with the proof in essence.

If you're being this detailed, also check the well-definedness of your maps: check that $G(phi) in prod_i in I A_h(i)$ (using the definition of the Cartesian product that you also gave, please do use a capital C in honour of Descartes...) and also in the proof of surjectivity that the $phi$ you define is in $prod_i in I A_i$. It's a minor quibble though.

Also you could spell out why $phi_1(h(i)) = phi_2(h(i))$ for all $i$ implies $phi_1(i) = phi_2(i)$ for all $i$ (namely because all $i in I$ are of the form $h(i')$ for some $i' in I$ etc.). I'd say if you're going detailed (i.e. all the gory details) also do these details.