Summability vs Unconditional Convergence

I'm reading that summability and unconditional convergence are the same. (At least if the index set is countable, so unconditional convergence makes sense.)

Then unconditional convergence is nothing else than convergence of the net of all rearranged partial sums. Thus uniqueness of the limit becomes fairly easy as the limit of every net in a Hausdorff space is unique.

But the usual proof deals with separating functionals on Banach spaces, which is a much more serious technique. So I think there must be a hook in my reasoning. Can somebody resolve this issue? Thanks a lot!

Best Regards, Freeze_S

For countable families $\{ x_n : n \in \mathbb{N}\}$ in a Banach (or Fréchet) space, summability and unconditional convergence are the same.

Let $\mathscr{F} = \{ F \subset \mathbb{N} : \operatorname{card}F < \infty\}$ the directed (by inclusion) set of finite subsets of $\mathbb{N}$, and for $F\in\mathscr{F}$, let $s_F = \sum\limits_{n\in F} x_n$.

The family is (by definition) summable if and only if the net $(s_F)_{F\in \mathscr{F}}$ converges (to $\lambda$, say).

Let the family be summable. Every bijection $\sigma \colon \mathbb{N}\to \mathbb{N}$ induces a map $\sigma^\ast \colon \mathbb{N}\to \mathscr{F}$ via $\sigma^\ast(n) = \{ \sigma(k) : 0 \leqslant k \leqslant n\}$, and $\{\sigma^\ast(n) : n \in \mathbb{N}\}$ is cofinal in $\mathscr{F}$, so $s^\sigma \colon n \mapsto s_{\sigma^\ast(n)}$ is a subnet of $(s_F)_{F\in \mathscr{F}}$ and hence also converges to $\lambda$. That is the unconditional convergence of $\sum\limits_{n=0}^\infty x_n$.

For the converse, we argue indirectly. Suppose the family is not summable. We have to show that it isn't unconditionally convergent.

Consider the sequence of partial sums

$$s_n = \sum_{k=0}^n x_k.$$

If that sequence doesn't converge, we're done, otherwise let $\mu$ be its limit. That the family isn't summable to $\mu$ means that there is an $\varepsilon > 0$, such that for every $F_1 \in \mathscr{F}$, there is a $c(F_1)\in \mathscr{F}$ with $F_1 \subset c(F_1)$ and

$$\left\lVert\sum_{n\in c(F_1)} x_n - \mu \right\rVert \geqslant \varepsilon.$$

We can now construct a bijection $\sigma \colon \mathbb{N}\to\mathbb{N}$ such that

$$s^\sigma_n = \sum_{k=0}^n x_{\sigma(k)}$$

does not converge to $\mu$, which shows that $\{x_n\}$ is not unconditionally convergent.

Let $n_1 = \min \{n\in \mathbb{N} : \lVert s_n - \mu\rVert \leqslant \varepsilon/2\}$. Whenever $n_k$ is defined, we let $F_k = \{0,\dotsc,n_k\}$. For $k \leqslant n_1$, let $\sigma(k) = k$. Let $m_1 = \operatorname{card} c(F_1)$, and $\sigma(n_1+1),\dotsc,\sigma(m_1)$ the elements of $c(F_1)\setminus F_1$ in ascending order. Let $r_1 = \max c(F_1)$ and $n_2 = \min \{n > r_1 : \lVert s_n-\mu\rVert \leqslant \varepsilon/2\}$, and $\sigma(m_1+1),\dotsc,\sigma(n_2)$ the elements of $F_2\setminus c(F_1)$ in ascending order.

We continue in this way, when $n_k$ and $\sigma\lvert_{F_k}$ are determined, we let $m_k = \operatorname{card} c(F_k)$, $\sigma(n_k+1),\dotsc,\sigma(m_k)$ the elements of $c(F_k)\setminus F_k$ in ascending order, $r_k = \max c(F_k)$, $n_{k+1} = \min \{ n > r_k : \lVert s_n - \mu\rVert \leqslant \varepsilon/2\}$, and $\sigma(m_k+1),\dotsc,\sigma(n_{k+1})$ the elements of $F_{k+1}\setminus c(F_k)$ in ascending order.

The construction yields $n_k < m_k \leqslant r_k < n_{k+1}$ for all $k$, and

$$\left\lVert \sum_{k=0}^{m_k} x_{\sigma(k)} - \mu\right\rVert \geqslant \varepsilon,$$

so $(s^\sigma_n)$ does not converge to $\mu$ (it does not converge at all, in fact).