Equivalent notions for convergence of an unordered sum over a countable set

Question

Let $(X,\|\cdot\|)$ be a normed space and let $A:=\{a_{k} \ | \ k\in\mathbb{N}\}$ be a countable subset of $X$. Define $\mathscr{F}:=\{F\subset A \ | \ F \ \text{is finite}\}$. We say that the net $(\sum_{a\in F}a)_{F\in\mathscr{F}}$ converges to $S\in X$ iff for every $\varepsilon>0$ there exists a $F_{0}\in\mathscr{F}$ such that $\|\sum_{a\in F}a-S\|<\varepsilon$ whenever $F_{0}\subset F\in\mathscr{F}$. In this case, we denote $\sum_{a\in A}a:=S$. I want to prove the following theorem:

THEOREM: The net $(\sum_{a\in F}a)_{F\in\mathscr{F}}$ converges if and only if for every bijection $b\colon\mathbb{N}\to\mathbb{N}$ the series $(\sum_{k=1}^{l}a_{b(k)})_{l=1}^{\infty}$ converges (in the usual sense).

I am able to prove the implication "$\implies$", but I got stuck on the converse.

MY ATTEMPT: Assume that for any bijection $b\colon\mathbb{N}\to\mathbb{N}$ the sequence $(\sum_{k=1}^{l}a_{b(k)})_{l=1}^{\infty}$ converges and let $\varepsilon>0$ be given. I want to prove that the net $(\sum_{a\in F}a)_{F\in\mathscr{F}}$ converges to $\sum_{k=1}^{\infty}a_{b(k)}$ for some bijection $b$. Given a bijection $b$, define $$\hat{b}\colon\mathbb{N}\to A,\qquad\hat{b}(k):=a_{b(k)}.$$ Note that $\hat{b}$ is also bijective. By assumption there exists a $L_{b}\in\mathbb{N}$ (depends on $b$) such that for every $l\geq L_{b}$, $$\left\|\sum_{k=1}^{l}a_{b(k)}-\sum_{k=1}^{\infty}a_{b(k)}\right\|=\left\|\sum_{a\in\hat{b}(\mathbb{N}_{\leq l})}a-\sum_{k=1}^{\infty}a_{b(k)}\right\|<\varepsilon.$$ Seemingly, the natural thing to do is to define the finite set $F_{0}:=\hat{b}(\mathbb{N}_{\leq L_{b}})$. Now, here is where I got stuck: I think that it is not true that $$F_{0}\subset F\in\mathscr{F}\implies\left\|\sum_{a\in F}a-\sum_{k=1}^{\infty}a_{h(k)}\right\|<\varepsilon.$$ For as far as I know, this is only true for sets $F$ of the form $F=\hat{b}(\mathbb{N}_{\leq l})$ for some $l\geq L_{b}$.

I think that I have to use a certain bijection $b$, but I have no idea where to start.

Any suggestions are more than welcome! Thanks in advance!

Answer 1

Here's the key lemma. Given a bijection $b:\mathbb{N}\to\mathbb{N}$, let $S(b)=\sum_{k=1}^\infty a_{b(k)}$. We will refer to a set of the form $\{b(k):k\leq l\}$ as an initial segment of $b$.

Lemma: If $S(b)$ exists for all $b$, then it is always the same: that is, $S(b)=S(c)$ for all $b,c$.

Proof: Suppose $S(b)\neq S(c)$ for some $b$ and $c$. We can then construct a sequence of finite subsets $$F_1\subset G_1\subset F_2\subset G_2\subset\dots$$ with the following properties:

• Each $F_n$ is an initial segment of $b$, and $\left\|\sum_{k\in F_n} a_k-S(b)\right\|<2^{-n}$
• Each $G_n$ is an initial segment of $c$, and $\left\|\sum_{k\in G_n} a_k-S(c)\right\|<2^{-n}$.

We can construct such sets by just building them one by one. Given $F_n$, we can extend it to a set which is an initial segment of $c$, and then we can extend that initial segment to a larger initial segment $G_n$ such that $\left\|\sum_{k\in G_n} a_k-S(c)\right\|<2^{-n}$. Similarly, given $G_n$, we can extend it to get $F_{n+1}$.

Now let $d:\mathbb{N}\to\mathbb{N}$ be a bijection which has both every $F_n$ and every $G_n$ as an initial segment. We then see that $\sum_{k=1}^\infty a_{d(k)}$ has subsequences of partial sums that approach both $S(b)$ and $S(c)$. Thus $S(d)$ does not exist, which is a contradiction.

Now we prove that if $S(b)$ always exists, then $\sum_{a\in A} a$ exists and is equal to the common value $S$ of $S(b)$ for all $b$. The proof is by a method similar to the proof of the Lemma. If the net does not converge to $S$, then there is some $\epsilon>0$ such that for any finite $F\subset\mathbb{N}$ there is a finite $F'\supseteq F$ such that $\left\|\sum_{k\in F'}a_k-S\right\|>\epsilon$; call such a set $F'$ bad. We construct a sequence of bad sets $$F_1\subseteq F_2\subseteq F_3\subseteq\dots$$ one by one whose union is $\mathbb{N}$. The construction is easy: given $F_n$, we know there is some bad set $F_{n+1}$ containing $F_n\cup\{n\}$. Since $n\in F_{n+1}$ for all $n$, we can be sure that the union of the $F_n$ is all of $\mathbb{N}$. Now let $b:\mathbb{N}\to\mathbb{N}$ be a bijection which has each $F_n$ as an initial segment. We then see that since each $F_n$ is bad, $S(b)$ cannot be equal to $S$, which is a contradiction.