Cardinality of a product of countable many sets

Question

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, $|B|=|\mathbb{R}|$

Answer 1

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 $\mathcal{P}(\mathbb{N})$?

Answer 2

Consider each set as {0,1} Map the elements of the product onto the closed unit interval by mapping each to a binary number. For example, [0,1,0,1,0,1,..] maps to .01010101... [I'm assuming the Axiom of Choice.]

The image is uncountable with countably many duplications, hence B must have the same cardinality.