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Let $\{A_{i}\}_{i \in \mathbb{Z}_{+}}$ be a countably infinite collection of sets. Suppose that every set of the collection contains two distinct elements. Let $B = ⨉_{i \in \mathbb{Z}_{+}}A_{i}$ be the Cartesian product of the collection. Prove that $B$ is numerically equivalent to $\mathbb{R}$, that is, $|B| = |\mathbb{R}|$.

(Extra challenge: Can one relax the condition that every set has two distinct elements? Can some sets in the collection be empty or have just one element? If yes, then how many?)

I understand that for $|B| = |\mathbb{R}|$, there has to exist a bijective function $f:B \to \mathbb{R}$. But I'm a bit confused on how to set up the proof given the wording that every set contains two distinct elements. Is this the same as the Cartesian product of two countably infinite sets?

From what I gather, $\{A_{i}\}_{i \in \mathbb{Z}_{+}} = \{(x_{1}, y_{1}), (x_{2}, y_{2}),...,(x_{i}, y_{i})\}$. Would an inductive proof showing each pair $(x_{i + 1}, y_{i + 1})$ exists in $B$ work here?

Also for the extra challenge, I would imagine sets can have just one element since the Cartesian product would still be defined by itself. No sets should be able to be empty. I've just started learning about this topic, so if someone with more knowledge could enlighten me on what I'm missing I would be very appreciative.

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An element of S = {0,1}^N, a denumberable product of a two point set is a binary sequence. So map S -> [0,1), to the binary expression of the sequence. There are countable many sequences ending in 11111.... that will be double mapped. Remove those for a bijection.

Replacing the two point sets with one point sets won't change the cardinality as long as infinitely many are two point; otherwise S is finite.

What you gather, as mentioned above, is a wrong expression for an element of S. They are sequences.

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