My friend and I are having a disagreement over whether the number of terms in the following series is countable or uncountable:
$$\sum_{i=1}^\infty a_i + \sum_{i=1}^\infty\sum_{j=1}^\infty a_{ij}+ \sum_{i=1}^\infty\sum_{j=1}^\infty\sum_{k=1}^\infty a_{ijk} + \ldots$$
i.e. the sum goes to have an infinite number of "sigmas".
My argument is that each of the sums has a countable number of terms, and there are a countable number of sums (indexed by the number of "sigmas") in the total sum, and hence the total number of terms is the cardinality of a countable union of countable sets, which is countable.
My friend's argument is that given any irrational number in $[0,1]$, he can find a corresponding term in the sum with the subscripts $i, j, k, \ldots$ coming from the decimal expansion of the irrational number, and vice versa (though not all irrational number won't be mapped and vice versa, the subset being mapped to is still uncountable). So the number of terms is uncountable.
We think that the answer to the question boils down to whether the cardinality of $\lim_{k\to\infty}\mathbb N^k$ is equal to $\mathbb N^\infty$ (i.e. an infinite Cartesian product) or not. Is this statement true, and is it applicable to the bigger question of the cardinality of the number of terms in the series?
Answer
Note that $\Bbb N^k$ is countable, for every finite $k$. Therefore the limit is $\aleph_0$, because the sequence is essentially constant. On the other hand, $\Bbb{N^N}$ is uncountable.
Cardinal exponentiation is not continuous.
(Note by the way, that $\infty$ from calculus is not quite compatible with infinite products in cardinal arithmetic.)
No comments:
Post a Comment