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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?

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    $\begingroup$ I would suggest writing $\bigcup_{k\in\mathbb{N}}\mathbb{N}^k$ instead using limits. Also, the symbol $\infty$ denotes a different concept than the cardinal infinities, like $ℵ_0$, $2^{ℵ_0}$, etc. In other words, you are wondering how do cardinalities of $\bigcup_{k\in\mathbb{N}}\mathbb{N}^k$ and $\mathbb{N}^\mathbb{N}$ compare. They do differ, that is the former is $ℵ_0$ (same as natural numbers), while the latter is $2^{ℵ_0}$ (same as real numbers). $\endgroup$ – dtldarek Apr 16 '14 at 11:08
  • $\begingroup$ @AsafKaragila You are right, fixed now. $\endgroup$ – dtldarek Apr 16 '14 at 11:08
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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.)

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Your friend's argument relies on changing the $a_i$, $a_{ij}$, $a_{ijk}$, etc. for each irrational number. Even if these are chosen from some fixed countable collection (such as the rationals), there are uncountably many different ways to choose these infinitely many terms, so this has no bearing on the number of terms.

Your reasoning is correct. For each group of $\sum$s the number of terms appearing in that group is countable (because $\mathbb{N}$ is countable, and finite products of countable sets are countable). The totality of terms is then the union of these groups, and is countable as union of countably many countable sets is countable.

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