The convergence of unordered sum $X$ is an arbitrary normed vector space. $A=\{x_i\in X\mid i\in J\}$ is an indexed set. $J$ contains the indices and is uncountably infinite. Let $\mathcal F=\{F\mid F\subseteq J, F \text{ is finite}\}$.
The following definition of the convergence of unordered sum is taken from https://www.math.ucdavis.edu/~hunter/m201b_old/sums.pdf. We claim that the unordered sum $\sum \limits_{i\in J}x_i$ converges if and only if $$\exists x\in X,\ \forall \varepsilon>0,\ \exists F_\varepsilon \in \mathcal F,\ \forall F_\varepsilon \subseteq F\in \mathcal F,\ \left\|\sum_{i\in F} x_i-x\right\|\leq \varepsilon.$$ Suppose the unordered sum $\sum \limits_{i\in J} x_i$ converges. Let $I\subseteq J$. Does $\sum \limits_{i\in I}x_i$ converge? (Note that it may converge to a different point) How to prove it? Thanks!
 A: If $X$ an incomplete normed space and $x_n\in X$ are such that $\sum\limits_{n=1}^\infty \|x_n\| <\infty$ but the series has no limit just take $I=\mathbb Z \setminus \lbrace 0 \rbrace$ and $x_{-n}=-x_n$. The the series $\sum\limits_{n\in I} x_n$ converges with limit $0$ but the subseries $\sum\limits_{n\in\mathbb N}x_n$ does not converge. For Banach spaces the situation is opposite because you have the Cauchy criterion.
A: I found a simple counter-example for this question. The answer is no. 
Define a vector space $X=\mathbb{Q}$ over the field $\mathbb{Q}$. $\mathbb{Q}$ is the collection of rational numbers. $X$ is equipped with the canonical distance $|\cdot|$. Now, we construct two nonnegative increasing sequences of rational numbers, $\{a_n\}_{n=1}^\infty$ and $\{b_n\}_{n=1}^\infty$. We let $a_n\rightarrow \sqrt{2}$, $b_n\rightarrow 2-\sqrt{2}$. Define $c_n=a_{n+1}-a_n$, $d_n=b_{n+1}-b_n$. So, $c_n\geq 0, d_n\geq 0$. So, $\sum \limits_{n=1}^\infty c_n=\sqrt{2}$, $\sum \limits_{n=1}^\infty d_n=2-\sqrt{2}$. Let $e_{2n-1}=c_n, e_{2n}=d_n$. We know $\sum \limits_{n=0}^\infty e_n=2$.
From the above construction, we know if we let $A=\{e_n|n\in \mathbb{N}\}$, $\sum \limits_{n\in \mathbb{N}}e_n$ converges to $2$. But $2\mathbb{N}-1\subset \mathbb{N}$, $\sum \limits_{n\in 2\mathbb{N}-1}e_n=\sqrt{2}$, which doesn't converge in $X$. So, it is a contradiction.
