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$\sum^{\aleph_0}_{\mu=\infty}\mu=\aleph_0$? If this is true, however, then $\sum^{\aleph_0}_{\mu=0}\mu=\aleph_1$?

Jeremy Upsal
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yayab
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1 Answers1

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As far as I'm aware, the expressions $\sum_{\mu=\infty}^{\aleph_0} \mu$ and $\sum_{\mu=0}^{\aleph_0} \mu$ don't have definitions. If an expression hasn't been defined, then it doesn't have a value.

In general, there's no definition of a summation that has $\aleph_0$ as one of its bounds.

Tanner Swett
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    Actually there is, but it's usually cardinal summation, and so the first one with bound $\infty$ doesn't make sense, however, as a cardinal sum, $\displaystyle\sum_{\mu<\aleph_0}\mu = \aleph_0$ – Maxime Ramzi Mar 06 '17 at 06:55
  • @Max Actually that makes sense even as an ordinal sum, as long as the index set over which we're summing is well-ordered. – Noah Schweber Mar 07 '17 at 18:13
  • @Noah : yep I understood what you meant afterwards and thus deleted my comment – Maxime Ramzi Mar 07 '17 at 18:47