The set of "variations" of an unconditionally convergent series is compact Proposition: If $\sum x_i$ is an unconditionally convergent series in a Banach space $X$, then $S=\{\sum \varepsilon_ix_i:\varepsilon_i=\pm1\}$ is compact.
Proof:
1) $\{-1,1\}^{\mathbb N}$ is compact in the pointwise topology.
2) $f:\{\varepsilon_i\}\to\sum\varepsilon_ix_i$ is a continuous function.
3) Im$f=S$
Therefore, $S$ is compact. q.e.d.
Problem: The problem is with the continuity of the proposed function.
It is easy to show that for any unconditionally convergent series $\sum x_i$, in a Banach space
$$(\forall\varepsilon>0)
(\exists n_0\in\mathbb N)
(\forall m\geq n\geq n_0)
(||\sum_{i=n}^m \varepsilon_ix_i||<\varepsilon)$$
for any choice of $\{\varepsilon_i=\pm1\}$
Does it directly follow that $f$ is continuous? How come?
Any help would be appreciated.
 A: Every projection
$$
\pi_i:\{-1,1\}^\mathbb{N}\to\{-1,1\}:\{\varepsilon_j\}\mapsto\varepsilon_i
$$
is continuous by definition of topology of $\{-1,1\}^\mathbb{N}$. Every map
$$
m_i:\{-1,1\}\to X:\varepsilon_i\mapsto \varepsilon_i x_i
$$
is continuous because topology of $\{-1,1\}$ is discrete. So the map $m_i\pi_i$ is continuous for each $i\in\mathbb{N}$. Hence the map
$$
f_n:\{-1,1\}^\mathbb{N}\to X:\{\varepsilon_j\}\mapsto \sum\limits_{i=1}^n m_i\pi_i
$$
is continuous as finite sum of continuous functions. As you noted in your question
$$
\forall\varepsilon>0\quad\exists n_0\in\mathbb{N}\quad\forall m\geq n\geq n_0\quad\forall\{\varepsilon_i\}\in\{-1,1\}^\mathbb{N}\implies\left\Vert \sum\limits_{i=n}^m \varepsilon_i x_i\right\Vert<\varepsilon
$$
This is nothing more that the statement that the set of continuous fnction $\{f_n:n\in\mathbb{N}\}$ converges uniformly on $\{-1,1\}^\mathbb{N}$. Since uniform limit of continuous functions is continuous we see that
$$
f:\{-1,1\}^\mathbb{N}\to X:\{\varepsilon_j\}\mapsto\sum\limits_{i=1}^\infty\varepsilon_i x_i
$$
is continuous.
