Sum of a series indexed by ordinals

If $\mu$ is an ordinal, how can we formalize that

$$\sum_{\lambda<\mu}x_{\lambda}=z$$

When $\mu=\omega$, this is just the usual infinite series, the partial sums converge to $z$. What is the definition for higher ordinals?

I don't think this is quite the same as an uncountable sum, it is a sum over an well-ordered index. We can define it as a limit points of finite sums, but I was wondering whether there is a "better" (possibly equivalent) definition that takes advantage of the extra structure (well-oredring of the index). I suspect this will make sense only if countably many elements are non-zero.

As for the context, I guess the most general one could be that of topological vector spaces, so that we have the vector operations and limits.

• Usually, it would be defined in terms of "unordered sums", which is essentially the same as saying that $\sum_{\lambda \in A} x_\lambda$ is the Lebesgue integral of $x$ over $A$, where $A$ is viewed as a discrete space with the counting measure. When $x_\lambda$ is nonnegative, this is the same as the supremum of the sums over finite subsets of $A$. – Carl Mummert Oct 3 '16 at 17:33
• I've voted to close as a duplicate reading the question as just asking for a definition. If there is some more specific issue related to ordinals in particular, please edit the question, and it's easy to re-open it if there is a different question to answer. – Carl Mummert Oct 3 '16 at 17:35
• @CarlMummert I disagree with your duplicate closure. I think the question asks for a definition which takes into account the well-ordering. It seems natural for example to define the partial sum for each limit ordinal $\lambda < \mu$ as the limit of the partial sums up to that point, if this limit exists. Much more interesting than when you just take the sum over an index set instead of an index ordinal. – 6005 Oct 3 '16 at 20:10
• @6005: I am somewhat skeptical, but I will vote to reopen since the question has been edited, and because I wasn't really trying to close the question unilaterally when I voted to close as a duplicate. The question asks for "the definition", which to me would indicate a pre-existing definition rather than speculation about what the definition could be. If the latter is what is desired, the question is too broad. – Carl Mummert Oct 3 '16 at 21:28
• @CarlMummert See my answer for why I think this is an interesting question, and not a duplicate of the earlier post; the answers to the earlier post didn't address conditionally convergent series at all. – Mitchell Spector Oct 4 '16 at 0:06

I'm assuming here that the $x_{\lambda}$ are real numbers. (Complex numbers would be fine too -- that doesn't matter. This was written before the edit mentioning topological vector spaces in general, and I haven't thought about it at that level of generality.)

In the first part of this answer, we'll see how to define convergence of a sum indexed by a countable ordinal.

In the second part of this answer, we'll see that there is no way to define the convergence of any well-ordered sum with uncountably many non-zero terms (it's easy to eliminate the case where all the non-zero terms are positive reals, but in fact we'll rule out any non-zero values at all). This will be done via an axiomatization of the notion of convergence of a well-ordered sum.

COUNTABLE WELL-ORDERED SUMS

In this part, we'll handle well-ordered sums where all but countably many of the terms are $0.$

For countable ordinals $\mu,$ the sum can be defined by transfinite induction, as follows:

$\sum_{\alpha<0}x_{\alpha}=0,$

$\sum_{\alpha<\mu}x_{\alpha}=(\sum_{\alpha<\beta}x_{\alpha})+x_\beta,$ if $\mu=\beta+1,$

and, for $\mu$ a countable limit ordinal,

$\sum_{\alpha<\mu}x_{\alpha}=z$ iff for every increasing function $f\colon\omega\to\mu$ which is cofinal in $\mu,$ $z=\lim_{n\to\infty}\sum_{\alpha<f(n)}x_{\alpha}.$

Here's how to extend this to uncountable ordinals $\mu\!:$

$\sum_{\alpha<\mu}x_{\alpha}=z$ iff the set $A=\lbrace \alpha < \mu \mid x_\alpha \ne 0\rbrace$ is countable and $z=\sum_{\alpha\in A}x_\alpha,$ where that last sum means $\sum_{\alpha < (\text{order type of }A)}x_{\text{the }\alpha^{\text{th}}\text{ member of }A}.$

By the way, just taking the supremum of all finite sums is the same thing for absolutely convergent series, but not for conditionally convergent series.

UNCOUNTABLE WELL-ORDERED SUMS

We'll see that there is no way to extend this usefully to any sequence with uncountably many non-zero terms.

This goes beyond the easily observed fact that there's no way to assign a finite sum to any uncountable series of positive numbers. We'll eliminate the possibility even of conditionally convergent series with uncountably many non-zero terms; so we can't have any convergent series with uncountably many non-zero terms, even if some are negative or have a non-zero imaginary part.

Let $S$ be either $\mathbb{R}$ or $\mathbb{C},$ and write ${}^\mu S$ for the set of all functions from $\mu$ to $S.$

We'll make only the following assumptions about what "convergence" means:

1. For each ordinal $\mu,$ we have a subset $E_\mu$ of ${}^\mu S$ and a function $e_\mu \colon E_\mu \to S.$ We say that the sum $\sum_{\alpha\lt\mu}x_\alpha$ converges iff $(x_\alpha)_{\alpha\lt\mu}\in E_\mu.$ If $\sum_{\alpha\lt\mu}x_\alpha$ converges, then we'll call $e_\mu((x_\alpha)_{\alpha\lt\mu})$ the value that the sum converges to, and we'll write $\sum_{\alpha\lt\mu} x_\alpha$ to mean $e_\mu((x_\alpha)_{\alpha\lt\mu}).$

2. The sum of the empty sequence is $0.$

3. For every $\mu,$ $\sum_{\alpha<\mu+1}x_\alpha$ converges iff $\sum_{\alpha<\mu}x_\alpha$ converges, and $\sum_{\alpha<\mu+1}x_\alpha=(\sum_{\alpha<\mu}x_\alpha)+x_\mu.$

4. If $\mu$ is a limit ordinal, then $\sum_{\alpha\lt\mu}$ converges to a value $L$ iff for every $\varepsilon\gt 0,$ there exists an ordinal $\gamma\lt\mu$ such that for all $\beta,$ if $\gamma\lt\beta\lt\mu,$ then $\lvert L-\sum_{\alpha\lt\beta}x_\mu \rvert \lt \varepsilon.$

5. If a sum $\sum_{\alpha\lt\mu} x_\alpha$ converges, then $\sum_{\alpha\lt\gamma} x_\alpha$ converges for every $\gamma\lt\mu.$

6. If a sum $\sum_{\alpha\lt\mu} x_\alpha$ converges and if $A=\lbrace \alpha < \mu \mid x_\alpha \ne 0\rbrace,$ then $\sum_{\alpha\in A} x_\alpha$ converges to the same value as $\sum_{\alpha\lt\mu} x_\alpha.$

Our definition of convergence for countable series satisfies the above properties, and in fact is the only definition of convergence for countable series satisfying those properties.

Suppose we have an uncountable sequence $\sum_{\alpha\lt\mu} x_\alpha$ whose sum converges (where "converges" has some meaning satisfying conditions 1-6 above). We'll show that all but countably many $x_\alpha$ must equal $0.$

Assume that, to the contrary, uncountably many $x_\alpha$ are non-zero. Let $y_\alpha$ be the $\alpha^{\text{th}}$ non-zero element in the sequence $\langle x_\alpha \mid \alpha \lt \mu \rangle.$ Then, by conditions 5 and 6, $\sum_{\alpha\lt\omega_1} y_\alpha$ converges; let its value be $L.$

By condition 4, for every positive rational $q,$ there exists $\gamma_q\lt\omega_1$ such that for all countable $\beta\gt\gamma_q,$ $\lvert L-\sum_{\alpha\lt\beta}y_\alpha \rvert \lt q.$

Since the set of rationals is countable, there is a countable ordinal $\gamma$ greater than all the $\gamma_q$ for $q$ rational and positive. But then any $\beta \ge \gamma$ satisfies $\sum_{\alpha\lt\beta}y_\alpha=L,$ since $\lvert L-\sum_{\alpha\lt\beta}y_\alpha \rvert$ is less than every positive rational number. Condition 3 then implies that $$y_\gamma=(\sum_{\alpha\lt\gamma+1}y_\alpha)-(\sum_{\alpha\lt\gamma}y_\alpha)=L-L=0,$$ contradicting the fact that all the $y_\alpha$ are non-zero.

What is the context? In the setting of analysis, $\sum_{i\in\mathbb N}x_i$ is defined as usual; other than that, the infinite sum $\sum_{i\in I}x_i$ is defined only when $\sum_i|x_i|$ is defined (so, we do not have a proper theory of conditionally convergent series). Assume then that the $x_i$ are non-negative, in which case $\sum_i x_i$ is defined as the unique limit of the set of sums $\sum_{i\in J}x_i$, where $J$ varies over all finite subsets of $I$. (So, if the set has no limit points, or more than one, the limit of the series does not exist.) Note that all that matters here is the size of $I$, not that it is given as an ordinal.

In other contexts, it makes sense to consider ordered sums, and then the fact that $\mu$ is an ordinal (rather than an arbitrary index set) matters.

• Why should $\sum_{x\in I}x_i$ be defined only if the $x_i$ are nonnegative? This is the same as saying that only nonnegative functions are Lebesgue-integrable. A necessary condition for $\sum_{x\in I}x_i$ to exist is that $\sum_{i\in I}|x_i|$ exists, but certainly $x_i\ge0$ is not at all required. – egreg Oct 4 '16 at 17:36
• @egreg Yes, of course. I wrote too fast. – Andrés E. Caicedo Oct 4 '16 at 17:42