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The question in the title is naively stated, so let be make it more precise: Let $\sum_{n\in\alpha}a_n$ be an ordinal-indexed sequence of real numbers such that $a_n>0$ for each $n\in\alpha$, where $\alpha$ is an ordinal number. What is the smallest $\alpha$ which guarantees that $\sum_{n\in\alpha}a_n$ diverges? Since bijections correspond to rearrangements of the sum, and since the $a_n$ are positive, the sum is either absolutely convergent or diverges to $+\infty$ regardless of the order, so it follows that $\alpha$ is a cardinal number. My intuition tells me that $\alpha=\omega_1$, but I can only prove that $\alpha\le\frak{c}^+$, as follows:

Let $\sum_{n\in\frak{c}^+}a_n$ be a sum of positive reals with $\frak{c}^+$ terms, and let $s_\beta=\sum_{n\in \beta}a_n$ be the sequence of partial sums, so that $s_{\beta+1}=s_\beta+a_\beta$. Then if $\beta<\gamma$, $s_\beta<s_\beta+a_\beta=s_{\beta+1}\le s_\gamma$, so in particular, the $\{s_\beta\}$ are all distinct, and $|\{s_\beta\}|=\frak{c}^+$. If $\sum_{n\in\frak{c}^+}a_n=A$ is finite, then every partial sum is less than $A$, so $\{s_\beta\}\subseteq [0,A]\subseteq\mathbb{R}$, so $\frak{c}^+=|\{s_\beta\}|\le|\mathbb{R}|=\frak{c}$, a contradiction. Thus $\sum_{n\in\frak{c}^+}a_n$ is not finite.

As indicated above, $\alpha\ge\omega_1$ is obvious because $\alpha$ is a cardinal, and $\sum_{n\in\omega}2^{-n}=2$ is finite, so $\alpha>\omega$. Can anyone prove that $\alpha\le\frak{c}$ or that $\alpha=\frak{c}^+$? I can't imagine any set of positive numbers indexed by reals whose sum could possibly be finite, so I lean strongly toward $\alpha\le\frak{c}$, but I don't know how to prove it.

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    $\begingroup$ Probably I am using Choice. The answer is $\omega_1$. For take our collection of numbers, and classify them as being $\gt 1$, between $1/2$ and $1$, between $1/3$ and $1/2$, and so on. If we have $\omega_1$ numbers, one class must have infinitely many numbers. $\endgroup$ – André Nicolas Apr 25 '13 at 3:22
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Following André's idea, let me add a small further refinement to Mario's answer: Work in $\mathsf{ZF}$. If $X$ is a set, and $a_x>0$ for each $x\in X$, as usual we define the series $\sum_{x\in X}a_x$ as $$\sup\biggl\{\sum_{x\in X'}a_x\mid X'\subset X\mbox{ is finite}\biggr\}.$$ If the series converges, then we can write $X$ as a countable union of finite sets: $X=\bigcup_{n\in\omega}A_n$, where each $A_n$ is finite. Conversely, for any such set $X$ there are $a_x>0$ for $x\in X$ such that $\sum_{x\in X}a_x$ converges.

Note that the requirement on $X$ is in general strictly weaker than being countable. For example, a Russell set is a set $X$ that can be written in the form $X=\bigcup_n X_n$, where each $X_n$ has size two, the $X_n$ are pairwise disjoint, and no infinite subfamily of the $X_n$ admits a choice function, that is, for any infinite $I\subseteq\omega$, we have that $\prod_{n\in I}X_n$ is empty. It is consistent with $\mathsf{ZF}$ that these sets exist. (The name, of course, comes from Russell's remark that given infinitely many pairs of indistinguishable socks, there is no way to pick a sock from each pair. Curiously, Russell's anecdote actually involved boots rather than socks.)

Suppose first that $X=\bigcup_n A_n$. For each $n$, let $m_n=|A_n|$, and define $\displaystyle a_x=\frac1{m_n 2^n}$ for all $x\in A_n$. We have that $\sum_{x\in X}a_x=2$.

Conversely, suppose that $\sum_{x\in X}a_x$ converges. The point is that, for each positive integer $n$, the set $A_n=\left\{x\in X\mid a_x\in\left[\frac1n,\frac1{n-1}\right)\right\}$ is finite (where $1/0$ is interpreted as $+\infty$). This is because if $|A_n|\ge m$, then $$S=\sum_{x\in X}a_x\ge\sum_{x\in A_n}a_x\ge \frac{m}{n+1},$$ so $S$ diverges if $A_n$ is infinite. But for each $x\in X$, since $a_x>0$, then $x\in A_n$ for some (unique) $n\in\omega$, and it follows that $X=\bigcup_n A_n$.

In fact, for each $n\in\omega$, let $B_n=\{a_x\mid x\in A_n\}$. Note that $B_n$ is the image of a finite set, so it is finite. Moreover, since $B_n$ is a set of reals, it comes equipped with a natural enumeration. For each $x\in X$, let $n_x=|\{y\in X\mid a_y=a_x\}|$. Let $C_n=\bigcup_{x\in A_n}\{a_x\}\times n_x$. Note that $C_n$ is a finite set, and is linearly ordered lexicographically, using the natural ordering of $B_n$ and of the number $n_x=\{m\in\omega\mid m<n_x\}$. It follows that $C=\bigcup_n C_n$ is countable (without any appeal to the axiom of choice). Each $c\in C$ has the form $(a_x,m)$ for some $x\in X$ and some $m<n_x$, and we can define $b_c=a_x$. The point of this is that $\sum_{x\in X}a_x=\sum_{c\in C}b_c$, which is to say that, even if there are convergent series of positive reals indexed by uncountable sets $X$, the fact that $X$ is uncountable is superfluous, as the series can be explicitly rewritten as a countable series.

Finally, let me point out that, if we insist that $X$ is well-ordered (as in Mario's question) then it is countable, as the $A_n$ are naturally ordered by the well-order they inherit from $X$, so $X$ is a countable union of explicitly counted sets. (This shows that if $X$ is an ordinal, then $X<\omega_1$. Note that it is consistent with $\mathsf{ZF}$ that $\omega_1$ is a countable union of countable sets.) Since any countable ordinal can be written as a countable union of finite sets (in fact, of singletons), it follows that $\alpha$, as defined in the question, is indeed $\omega_1$.

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  • $\begingroup$ I think that Russell sets are required to be Dedekind-finite, which is a slightly stronger requirement than having no choice function on the product of all the pairs (it's equivalent, though, to requiring that the product of any infinite subset of the pairs is empty). $\endgroup$ – Asaf Karagila Apr 25 '13 at 6:57
  • $\begingroup$ Yes, thank you, that's what I meant to write. $\endgroup$ – Andrés E. Caicedo Apr 25 '13 at 7:00
  • $\begingroup$ Russell must have had some uncomfortable boots if they were indistinguishable. On topic, I'm slightly confused by your construction of $C$. It would seem that you have reordered an uncountable set to make a countable one, but that doesn't make any sense. Surely you could just read it backwards to have a bijection into $X$ from $C$, thus proving the countability of $X$ as well. $\endgroup$ – Mario Carneiro Apr 25 '13 at 7:17
  • $\begingroup$ On Russell: He actually acknowledge years later the silliness of his chosen example. He attributed the curious boots to a "millionaire excentrique." $\endgroup$ – Andrés E. Caicedo Apr 25 '13 at 7:33
  • $\begingroup$ On $X$ and $C$: Suppose $X=\bigcup_m A_m$ is a Russell set, with each $A_m$ of size two, and that for each $m$ we have $a_x=a_y$ for $x,y\in A_m$. Call this common value $d_m$, so each $B_m=\{d_m\}$ is a singleton, and $n_x=n_y=2$ for $x,y\in A_m$. Then each $C_m$ is also of size 2, with $C_m=\{d_m\}\times 2=\{a_x\}\times 2=\{a_y\}\times 2$. The crucial difference is that we have, uniformly in $m$, an ordering of $C_m$: $(d_m,0)$ comes before $(d_m,1)$. (Cont.) $\endgroup$ – Andrés E. Caicedo Apr 25 '13 at 7:43
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Given any uncountable set $\cal S$ of positive reals, there must be some $n\in{\Bbb Z}_{>0}$ such that uncountably many members of $\cal S$ exceed $n^{-1}$. Therefore, the sum of $\cal S$ must diverge to $+\infty$. It follows that $\alpha\le\omega_1$, and it is easy to construct a convergent sum $\sum_{n<\beta} a_n$ for all $\beta<\omega_1$. So, $\alpha=\omega_1$.

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  • $\begingroup$ I don't follow the argument in your first sentence. $\endgroup$ – Mario Carneiro Apr 25 '13 at 3:30
  • $\begingroup$ If, for all $n$, only countably many members of $\cal S$ exceeded $n^{-1}$, then $\cal S$ would be a countable union of countable sets, so it would be countable. $\endgroup$ – David Moews Apr 25 '13 at 3:32
  • $\begingroup$ Not to pile on, but I don't follow the argument in the third sentence. $\endgroup$ – Charles Apr 25 '13 at 3:34
  • $\begingroup$ $\omega_1$ is uncountable $\ $ ∴ $\ $ $\sum_{n<\omega_1} a_n=+\infty$ regardless of $a_n$ $\ $ ∴ $\ $ $\alpha\le \omega_1$. $\endgroup$ – David Moews Apr 26 '13 at 1:42
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This is a refinement of André Nicolas's comment.

Suppose there exists a sequence of reals $a_n$ of length $\omega_1$ such that $\sum_{n\in\omega_1}a_n$ is finite. For each $m\in\omega$, define a sequence $i_m(n)$ such that $i_m(n)$ is the smallest ordinal greater than $i_m(n-1)$ such that $a_{i_m(n)}\in[(m+1)^{-1},m^{-1})$ (and terminating when there is no such integer). Each sequence is finite (let the length be $l_m$), because if for some $m$ the sequence $\{i_m(n)\}_{n\in\omega}$ was infinite, we would have $\sum_{n\in\omega_1}a_n\ge\sum_{n\in\omega}a_{i_m(n)}\ge\sum_{n\in\omega}(m+1)^{-1}=\infty$, in contradiction. By ordering the $i_m(n-1)$ lexicographically with $n$ first, then $m$, we obtain a bijection from the set of all $i_m(n)$ to $\omega$, and since $\omega$ is not equinumerous with $\omega_1$, we are forced to conclude that there exists some $\beta\in\omega_1$ such that $\beta\ne i_m(n)$ for any $m\in\omega,n\in l_m$. Now $a_\beta\in[(m+1)^{-1},m^{-1})$ for some $m$, so $\beta$ must be in the set $\{i_m(n)\}_{n\in l_m}$, a contradiction. Thus the sum must be infinite, and $\alpha=\omega_1$.

Note that AC was unnecessary for the proof, since the $a_n$ are well-ordered from the start.

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  • $\begingroup$ A very substantial refinement! $\endgroup$ – André Nicolas Apr 25 '13 at 4:16

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