The set of finite "variations" of an unconditionally convergent series is pre-compact Proposition: If $\sum x_i$ is an unconditionally convergent series in a Banach space $X$, then $S=\{\sum_{i=1}^n \varepsilon_ix_i:n\in\mathbb N, \varepsilon_i=\pm1\}$ is pre-compact.
Proof:
1) $S'=\{\sum^\infty \varepsilon_ix_i:\varepsilon_i=\pm1\}$ is compact
2) $2\sum_{i=1}^n\varepsilon_ix_i=\sum_{i=1}^\infty\varepsilon_ix_i+\sum_{i=1}^\infty\varepsilon_i'x_i$
where $\varepsilon_i=\varepsilon_i'$ for $i=\overline{1,n}$
and $\varepsilon_i=-\varepsilon_i'$ for $i=\overline{n,\infty}$
3) Therefore, $S$ is pre-compact.
Problem: To prove that $S$ is pre-compact, it is enough to examine the limits of sequences in $S$ (since metric spaces are Hausdorff). Apparently, the proof does differently.
As far as I can interpret it, one can express any element in $S\over 2$ as a sum of two elements in $S'$. It does not follow, however, that $\overline {S\over 2}=S'$, which would nearly conclude the proof... How can I better understand this concept? Any help would be appreciated.

Apparentlly $$S\subset \frac{S'+S'}{2}$$ Also addition and multiplication are continuous in Banach spaces, therefore $\frac{S'+S'}{2}$ is compact. Therefore, $S$ is pre-compact as a subset of a compact set. q.e.d.
