# Cardinality of a product of countable many sets

I'm working on this problem that involves the collections of sets. I'm not really sure how to approach this problem. I understand that to prove that something is numerically equivalent one must show that there is a bijection. Any help would be appreciated.

Let $\{A_i\}_{i \in \mathbb{Z_+}}$ be a countable collection of sets. Let $B = \displaystyle \prod_{i\in \mathbb{Z_+}}A_i$ be the Cartesian product of the collection. Prove that if every set of the collection $\{A_i\}_{i\in \mathbb{Z_+}}$ contains two distinct elements, then $B$ is numerically equivalent to $\mathbb{R}$, that is, $|B|=|\mathbb{R}|$

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Are you assuming the Axiom of Choice? And what are your hypothesis on the sizes of the $A_i$? As stated, the result is false: just take $A_1$ to be a set of cardinality strictly larger than that of $\mathbb{R}$ and all others to have at least two elements. That set contains a set of size $|A_1|$, which is strictly larger than that of $\mathbb{R}$. –  Arturo Magidin May 8 '12 at 4:59
I assume that you mean that each of the sets $A_i$ contains exactly two elements, as otherwise $B$ could be too large. It's very difficult to specify an actual bijection. It's much easier to exhibit injections in both directions and then appeal to the Cantor-Schröder-Bernstein Theorem. –  Brian M. Scott May 8 '12 at 5:02
If you know about cardinals, $$2^{\aleph_0} =(2^{\aleph_0})^{\aleph_0},$$ so if $2\leq |A_i|\leq 2^{\aleph_0}$ for each $i$, then the result follows. –  Arturo Magidin May 8 '12 at 5:04

Do you know how to prove that $\mathbb{R}$ is numerically equivalent to $\mathcal{P}(\mathbb{N})$? Show that $\mathbb{R}$ is numerically equivalent to $(0,1)$, then show (using binary representation; careful with the numbers with dual representation) that there is an embedding $(0,1)\hookrightarrow \mathcal{P}(\mathbb{N})$. Then show that there is an embedding $\mathcal{P}(\mathbb{N})\hookrightarrow (0,1)$, say by looking at decimal representations of numbers that only use two digits, neither of them $0$ or $1$.
Now, if each $A_i$ has exactly two elements, do you see a connection between $\prod A_i$ and $\mathcal{P}(\mathbb{N})$?