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In answering the question Why do we classify infinities in so many symbols and ideas?, William's answer asserted that summing over an uncountable index set necessarily results in an infinite sum. I am curious about whether this is true and, if so, if it is possible to characterise the transfinite nature of the resultant sums over uncountable index sets.

Is it possible to conjecture, e.g., that sums over index sets of $\aleph_{n}$ result in sums of cardinality $\aleph_{n}$? At most $\aleph_{n}$? Between $\aleph_{n-1}$ and $\aleph_{n}$?

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  • $\begingroup$ The notion of infinity that occurs as an infinite sum is NOT THE SAME as the infinities that occur as infinite cardinal numbers in set theory. $\endgroup$ – Dave L. Renfro Apr 28 '16 at 18:31
  • $\begingroup$ You can sum over an uncountable index set as long as only countably many values of the indexed family are nonzero. $\endgroup$ – BrianO Apr 28 '16 at 18:34
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Summing over an uncountable index results in an infinite sum if uncountably many terms are non-zero.

To see this, we prove the contrapositive: that if a sum over an uncountable index is finite implies that at most countably many terms are non-zero.

Proof. Let $\sum_{\alpha \in A} x_\alpha = L$. Let $S_n = \{\alpha \in A \mid x_\alpha > 1/n\}$. Then

$$L = \sum_{\alpha \in A} x_\alpha > \sum_{\alpha \in S_n} 1/n = \frac{|S_n|}{n}$$

So $| S_n| < nL$.

Let $S = \{\alpha \in A \mid x_\alpha > 0\}$. The $S$ is the countable union of each $S_n$, which are in turn each at most countable, so the result is at most countable.

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  • $\begingroup$ You meant "contrapositive" rather than "converse". $\endgroup$ – Andreas Blass Apr 28 '16 at 19:15
  • $\begingroup$ @AndreasBlass i do, thanks! $\endgroup$ – MCT Apr 28 '16 at 21:42

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