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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 $\mathbb{R}$.2012-05-08
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    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](http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem).2012-05-08
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    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.2012-05-08

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