Intereting Posts

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Riemann Zeta function – number of zeros
Choose parameters to make a harmonic function
Verifying $|F(r)| \geq \frac{1}{1-r}\log(\frac{1}{1-r}) $ and $|F(re^{i \theta})| \geq c_{q/r}\frac{1}{1-r}\log({\log(\frac{1}{1-r})})$
Finding a closed form expression for $\sum_{i=1}^{n-1}\csc{\frac{i\pi}{n}}$
Convergence of a sequence with assumption that exponential subsequences converge?
Distributivity of ordinal arithmetic
For Infinite Cardinals does $A > B \Rightarrow A^C > B^C$?
Baire space homeomorphic to irrationals
Showing a subset of the torus is dense
show that $\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}=\frac {(2n)!\pi}{2^{2n}(n!)^2}$
Pullbacks and pushouts in the category of graphs
What does insolvability of the quintic mean exactly?
Topologist's Sine Curve not a regular submanifold of $\mathbb{R^2}$?
Does every number not ending with zero have a multiple without zero digits at all?

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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- I read that ordinal numbers relate to length, while cardinal numbers relate to size. How do 'length' and 'size' differ?
- Basic examples of ordinals
- Ordinals - motivation and rigor at the same time

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- How to find $n$'th term of the sequence $3, 7, 12, 18, 25, \ldots$?
- uniform convergence of sequence of functions $f_n(x)=x e^{-nx^2} , x\in{\mathbb{R}}$.
- How is this series in denominator converted to a series in numerator?
- Does $\sum\limits_{n=1}^{\infty}\frac{1}{P_n\ln(P_n)}$ converge to the golden ratio?
- Elementary proof that $a_n \to a \implies a_n^r \to a^r$ for $r \in \mathbb{Q}$
- Prob. 17, Chap. 3 in Baby Rudin: For $\alpha > 1$, how to obtain these inequalities from this recurrence relation?
- “Closed” form for $\sum \frac{1}{n^n}$
- What is wrong with the sum of these two series?
- Prove that if $\sum_{n=1}^\infty a_{2n}$ and $\sum_{n=1}^\infty a_{2n-1}$ converge then $\sum_{n=1}^\infty a_n$ converges.

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 $\omega_1$.

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, $\alpha=\omega_1$.

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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