# Proving that product of mutually disjoint sets is equipotent

The question I am trying to prove is as follows:

Let $\{B_i\}_{i \in I}$ and $\{C_i\}_{i \in I}$ be families of mutually disjoint sets. If $B_i \approx C_i$ for each $i \in I$, prove that $\prod_{i \in I}B_i \approx \prod_{i \in I}C_i$.

My attempt:

Since $B_1 \approx C_1$, and $B_2 \approx C_2$,

this implies that $B_1 \times B_2 \approx C_1 \times C_2$

and, $B_1 \times B_2 \approx C_1 \times C_2$ and $B_3 \approx C_3$

this implies that $B_1 \times B_2 \times B_3 \approx C_1 \times C_2 \times C_3$

Repeating this process for all $i \in I$, we get the desired result.

Is this proof correct. Any comments would be appreciated.

• While this 'seems true' it doesn't suffice. You only handle the case that $I$ is finite. Commented Jun 12, 2017 at 15:48
• @Stefan Would I have to separate the proof into 2 cases (i.e. $I$ being finite and infinte) or is there a generalized direct proof for this? Commented Jun 12, 2017 at 15:49
• You can handle the finite and infinite case in one go - see my hint below. Commented Jun 12, 2017 at 15:50

Hint: Fix, for each $i \in I$, a bijection $f_i \colon B_i \to C_i$. Now find a way to combine all of these bijections into a single bijection
$$f \colon \prod_{i \in I} B_i \to \prod_{i \in I} C_i.$$
• Thank you for your hint. Should I use the fact that if $f_1:B_1 \to C_1$ and $f_2:B_2 \to C_2$ are bijective, then $f:B_1 \times B_2 \to C_1 \times C_2$ is bijective and repeat the process ith times? but isn't that exactly what I did? What extra should I do? Commented Jun 12, 2017 at 16:00
• Get away from the finite case... Where could you map an infinite tupels $(x_i \mid i \in I)$ to - using all the $f_i$ at once? Commented Jun 12, 2017 at 16:03
• I am still having trouble.. so the infinite tupels $(x_i : i \in I)$ will be mapped to $f(x_1, x_2, x_3,...) \in \prod C_i$. Is this all I should explain? Commented Jun 12, 2017 at 16:17
• No, you need to describe $f$. Take $f((x_i \mid i \in I)) := (f_i(x_i) \mid i \in I)$. Can you show that this is well-defined and bijective? Commented Jun 12, 2017 at 17:43
• For showing that $f$ is well-defined, would it suffice to show that 1. $f$ is an relation from $X$ to $Y$ 2. The domain of $f$ is $X$, every element in $X$ is related to some element of $Y$ 3.No element of $X$ is related to more than one element of $Y$ Commented Jun 14, 2017 at 13:33