Span of Infinite Vandermonde matrix?

Q1: Does there exist $x \in (0,1)^\mathbb{N}$ such that for any decreasing $v \in (0,1)^\mathbb{N}$ that converges to $0$ (meaning, $v_{n+1} < v_n$ and $v_n \rightarrow 0$), there is $a \in (0,1)^\mathbb{N}$ with: $$\begin{array}{lll} \sum a_n & =& v_0\\ \sum a_n x_n& = &v_1\\ \sum a_n x^2_n & = &v_2\\ \mbox{etc.}&& \end{array}$$

Equivalently, think of an infinite Vandermonde matrix with coefficients $x \in (0,1)^\mathbb{N}$
$$V(x) = \left( \begin{array}{llllll} 1 & 1 & 1 & \cdots & 1 & \cdots\\ x_1 & x_2 & x_3 & \cdots & x_n & \cdots\\ x^2_1 & x^2_2& x^2_3 & \cdots & x^2_n & \cdots\\ \vdots &\vdots&\vdots & \ddots &\vdots & \vdots\\ x_1^m & x^m_2 &x^m_3 & \cdots & x^m_n & \cdots\\ \vdots &\vdots&\vdots & \ddots &\vdots & \vdots \end{array} \right)$$ (where superscrits represent exponents and not just row position) and ask whether there is $x \in (0,1)^\mathbb{N}$ such that for any decreasing and convergent (to $0$) $v \in (0, 1)^\mathbb{N}$, there is $a \in (0, 1)^\mathbb{N}$ satisfying $V(x) \cdot a= v$.

Intuitively, $x$ cannot be chosen arbitrarily. I suspect that a positive answer requires $x$ to be such that $\{x_n \in (0,1) : n\in\mathbb{N}\}$ is dense in $(0,1)$.

As always, thank you all for your help.

• Question 2 is independent of Question 1. If you know that $\mathbb{Q}$ is dense in $\mathbb{R}$ and that there is a surjective map $\phi : \mathbb{N}\to\mathbb{Q}$ then you will easily find an answer. – Nathanael Skrepek Jul 3 '17 at 13:32
• @supinf For Q1, I have tried to gain intuition with the finite case (which seems to be standard, by Vandermonde determinant itself). But the infinite dimensional part throws me off. I have considered $x_n = 1 - b^n$ for $b \in (0,1)$ but it's not it. – Alexny Jul 3 '17 at 13:42
• @NathanaelSkrepek I have edited my question in response to your comment, which wasn't clear to me. – Alexny Jul 4 '17 at 13:26

The answer to Q1 is no: If there is $a,x\in (0,1)^{\mathbb N}$ such that $$\sum a_n = \nu_0, \ \sum a_nx_n = \nu_1$$ then this implies $\nu_1<\nu_0$ if $a\ne0$: Since $x_n\in(0,1)$, it holds for all $a_n>0$: $a_nx_n < a_n$. Summing up, yields $\nu_1<\nu_0$.