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By using the following lemmas:

  1. A countable union of countable sets is countable.

  2. Cartesian products of integers are countable.

Wouldn't it be possible to prove the countability of the reals between 0 and 1, by partitioning them into countable sets each one uniquely identified by a Cartesian pair (m, n)?

Each set (m, n) contains all reals that have their first m digits sum equal to n and all their other digits zero.

By the second lemma above there is a countable such (m,n) that cover the interval [1,0]. And each such set is also countable since it contains a finite amount of numbers. Therefore by using the first lemma above wouldn't that prove that the the reals between 1 and 0 are countable?

Appreciate any feedback on the above on where I went wrong; I suspect it might be in how I used the second lemma to encode the above sets and show there is countable number of such sets.

many thanks,

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    When do you count the real numbers whose decimal expansion has no zeros, say, $\frac{1}{9}$?2012-06-24
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    Since $\mathbb{R}$ in uncountable, the answer must be negative :-). I suspect that arbitrary cartesian products of countable sets need not be countable...2012-06-24
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    *And each such set is also countable since it contains a finite amount of numbers*... Why is that so? The set of *reals that have the first m digits that sum up to n* is certainly not finite (and is in fact uncountable).2012-06-24
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    Thank you for the comments, I have modified the third paragraph again. I hope this is not bad etiquette here. Also, I don't have any doubts that the proof is wrong, just trying to figure out how.2012-06-24
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    You have not addressed Asaf's objection: the subset of real numbers described in the third paragraph is only those whose decimal expansion ends in a string of infinitely many zeros. That leaves out a lot of real numbers.2012-06-24

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