Series with terms indexed by transfinite ordinals

Question

I learned this past year how to deal with summation of infinite series $a_0 + a_1 + a_2 + ...$ with indices running over the natural numbers. I've also learned about what comes "after" the natural numbers ($ \omega, \omega+1, \omega+2, ... $), and so I'm naturally wondering about sequences $$ a_0 + a_1 + a_2 + ... + a_\omega + a_{\omega+1} + a_{\omega+2} + ... $$ My guess would be that we should sum the "first" series (beginning at 0) and also sum the second series (beginning at $\omega$) and then add the two sums. Is it that simple? Are there any interesting results for a series like this? What about going further: $$ a_0 + a_1 +... a_\omega + a_{\omega+1} + ... a_{2\omega} +...a_{\omega^2}+...a_{\omega^{\omega^\omega}}+... a_{\epsilon_0}+... $$

Answer 1

Assuming that the series converges absolutely, e.g. if all the elements are positive, you can always rearrange the series to be indexed by $\omega$, and get it over with.

If that this is not the case, e.g. the first $\omega$ elements are $-\frac1n$, you probably want to add the requirement that for every limit ordinal, the sum does converges.

However, in that case, if the series is $\sum_{\xi<\gamma}x_\xi$ is indexed by some limit ordinal $\gamma$. We can take a cofinal $\omega$-sequence $\langle\gamma_n\mid n<\omega\rangle$ and replace the series by $a_n=\sum_{\xi<\gamma_n} x_\xi$, and by the assumption that these sum already make sense, we get that $\sum_{\xi<\gamma}x_n=\sum_{n<\omega}a_n$. So we reduce this to the countable case again.

(Note that a sum indexed by a successor ordinal will have to be reducible to a finite sum by "condensing" the sum at the last limit ordinal.)

And of course, you can't get to $\omega_1$ anyway, since any uncountable set of real numbers has an infinite sum.

So while the whole notion is interesting on paper, by cofinality arguments we can reduce it back to the usual notion of an infinite sum indexed by $\omega$.

Answer 2

Let $(x_n)_{n\in N}$ be a real sequence such that $\sum_{n=1}^{\infty} |x_n| <\infty$, We can re-index $(x_n)_{n\in N}$ in any manner we like and still get the same value for the sum of the series. That is , let $f:N\to S$ be a bijection. Let $y_{f(n)}=x_n$ for $n\in N.$

Let $A=\{s\in S:y_s\geq 0\}$ and $B=\{s\in S: y_s<0\}.$

Let $P$ be the lub of $\sum_{s\in A^*}y_s$ over all finite $A^*\subset A,$ and let $Q$ be the glb of $\sum_{s\in B^*}y_s$ over all finite $B^*\subset B.$ (With the usual convention that the sum over an empty set is $0$.)

$$\text {Then }\quad\sum_{n=1}^{\infty} x_n=P+Q.$$ And if $S=\omega +\omega$ we also have $$P+Q=\sum_{s\in \omega}y_s+\sum_{s\in (\omega +\omega)\backslash \omega}y_s.$$