# Question about unconditional Schauder basis definition

I could not make sense of a step when proving two definitions for unconditional Schauder basis are equivalent:

1. For any scalar sequence $(\alpha_{n})$, if $\sum{\alpha_{n}x_{n}}$ converge, then it converge unconditionally.
2. There exist $C > 0$, finite, such that: \begin{align} \Bigg\|\sum_{i=0}^n\epsilon_i\alpha_ix_i\Bigg\| \le C\Bigg\|\sum_{i=0}^n\alpha_ix_i\Bigg\| \end{align} uniformly over $n$ and all sequences $\epsilon_n$ with $|\epsilon_n| \le 1$, and all scalar coefficients ${\alpha_n}$.

To prove 2 $\Rightarrow$ 1, supposing $\sum \alpha_ix_i$ converges, i.e. $\forall \ \ \epsilon > 0, \exists\ \ n = n(\epsilon)$, s.t. \begin{align} \Bigg\|\sum_{i>n}\alpha_ix_i\Bigg\| < \frac{\epsilon}{2C}. \end{align}

Given any finite $A \subset (n, \infty)\cap\mathbb{Z}$, by choosing the sign sequence of $\{\epsilon_n\}$ s.t. equals $1$ on $A$, equals $0$ on $B = \{1, 2, ..., n\}$, and equals $\pm 1$ constantly on $\mathbb{N}\backslash (A\cup B)$. Using 2, we obtain: \begin{align} \Bigg\|\sum_{i\in A}\alpha_ix_i\Bigg\| \le 2C\Bigg\|\sum_{i>n}\alpha_ix_i\Bigg\| < \epsilon \ \ \ \ (\textbf{I could NOT understand this step}) \end{align} I could understand the rest steps, i.e. $\|\sum_{i\in A}x_i\| \le \epsilon \Leftrightarrow \sum x_n$ converges unconditionally.

Could anyone help me to understand the step above I put in bold?

• Sorry about the formatting, I'll edit it so that the math format works Commented Oct 18, 2016 at 2:57