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This is really bugging me. On a final exam I had an extra credit question that said to prove that the countable Cartesian Product of countable sets is countable. I know this is true for a finite product and I thought I proved it using induction. But on the way home I thought up a counter example:

Ignoring the decimal point, every real number can be expressed as a string of characters 0-9 with the first number non-zero. Therefore the real numbers can be put in 1-1 correspondence with a subset of

$\mathbb{Z}_{10}-\{0\}\times\mathbb{Z}_{10}\times\mathbb{Z}_{10}\times\mathbb{Z}_{10}\times\mathbb{Z}_{10}\times...$

which is a countable Cartesian product of finite sets. Yet, wouldn't this provide a counterexample since a countable set cannot have an uncountable subset?

CantorStudent
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    I'm pretty sure that in fact your counter example is true, and that the countable cartesian product can be uncountable- see: https://math.stackexchange.com/questions/500849/infinite-cartesian-product-of-countable-sets-is-uncountable. – SEWillB May 12 '17 at 20:08
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    However, what should be noted is the potential ambiguity in the term countable and countably infinite. The reason I said 'may be uncountable' above is that all the sets are ${0}$ then the countable cartesian product is obviously countable... If all sets in the product are countably infinite then the product itself is uncountable! – SEWillB May 12 '17 at 20:11
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    Also note that proving this result by induction only shows that essentially $\Bbb{N}^k$ is countable $\forall k \in \Bbb{N}$ which is not the same as what your title asks! – SEWillB May 12 '17 at 20:14
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    For what it's worth, a very common way of topologically thinking of the irrational numbers is as the product of a countably infinite number of copies of the natural numbers. See here, for example. – Dave L. Renfro May 12 '17 at 20:27
  • Thank you for the comments and the link. – CantorStudent May 12 '17 at 20:49

1 Answers1

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No, a Cartesian product of countably infinite, countable sets (with at least two elements in each) is uncountable. Let $S_i|i\in \mathbb{N}$ be the countably many countable sets. Then the Cartesian product of them is just the set of all sequences $(s_1,s_2,s_3,\dots ,s_i,\dots)$ with $s_i \in S_i$ for each $i\in \mathbb{N}$. You can apply Cantor's diagonal argument to see that this set is not countable.

Arpan1729
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  • OK, I think I understand now. Let's say each set has exactly 2 elements. Then the nth product would have $2^n$ elements. So infinitely many of them would have 2 to the power of $\aleph_0$ elements, making it uncountable. – CantorStudent May 12 '17 at 20:40