Show that if $\{x_1,...,x_n\}$ are elements of a normed space $X$ over field $F$ such that $0$ belongs to the closure of
$A= \left\{a_1x_1+a_2x_2+...+a_nx_n | a_i\in F, \prod_{i=1}^n (a_i-1)=0\right\}$.
then $\{x_1,...,x_n\}$ are linearly dependent.
To me, it's clear that if $0$ belongs to the set above then obviously $\{x_1,...,x_n\}$ are linearly dependent. Where I'm lost is where $0\in \bar{A}$ comes in. Let $\{y_n\}$ be a sequence of elements of $A$. This means that for every $\varepsilon>0$. There exists an $N\in\mathbb{N}$ such that whenever $m>N$. We have that $\|y_m-0\|<\varepsilon$.
So $\|y_m\|<\varepsilon$. Where $y_m=b_1x_1+...+b_nx_n$ for some $b_1,...,b_n\in F$ and $\prod_{i=1}^n (b_i-1)=0$.
How does this tell me that the set is linearly dependent?
Let $$X_i=\{(a_1,\dots,a_n)\in F^n\mid a_i=1\}$$ and $$X=\bigcup_{i=1}^nX_i.$$
For every $N\in\Bbb N,$ there exists $b(N)\in X$ such that $$\left\|\sum_{i=1}^nb(N)_ix_i\right\|<\frac1N.$$
For at least one of the $i$'s, $b(N)$ belongs to $X_i$ infinitely often, so that wlog (up to taking a subsequence and permuting the indices), we may assume $$\forall N\in\Bbb N\quad b(N)_1=1.$$
Then, $$\begin{align}x_1&=-\lim_{N\to\infty}\sum_{i=2}^nb(N)_ix_i\\&\in\overline{\operatorname{span}(x_2,\dots,x_n)}\\&=\operatorname{span}(x_2,\dots,x_n).\end{align}$$