# Construct dense and disjoint sets of $\mathbb{R}^m$ so that every element of their Cartesian product has full rank

Or equivalently, can one construct sets $S_1 ,S_2 ,\dots ,S_n \subseteq \mathbb{R}^m$ so that

(i) the sets $S_i$ are dense and disjoint; and

(ii) if one picks from each set $S_i$ any element $u_i$, the vectors $u_1,u_2,\dots,u_n$ are linearly independent?

Note that a necessary condition is that $m\geq n$. Ideally I'd like $m=n$, but bigger $m$ is fine as well.

Yes. Let $t_{11},t_{12},\ldots,t_{nn}$ be $n^2$ mutually transcendental elements of $\mathbb{R}$. Let $$T_{ij} \;=\; \{q\, t_{ij} \mid q\in\mathbb{Q}\text{ and }q\ne 0\}$$ and let $$S_i \;=\; T_{i1}\times \cdots \times T_{in}$$ for each $i$. Clearly each $S_i$ is dense, and $S_1,\ldots,S_n$ are disjoint.
Now, suppose we pick vectors $v_1,\ldots,v_n$, where $v_i\in S_i$. I claim that $v_1,\ldots,v_n$ are linearly independent. To prove this, consider the matrix $M$ whose rows are $v_1,\ldots,v_n$. This $ij$th entry of this matrix has the form $$m_{ij} \;=\; q_{ij} t_{ij}$$ for some nonzero $q_{ij} \in \mathbb{Q}$. Then $$\det(M) \;=\; p(t_{11},t_{12},\ldots,t_{nn})$$ for some polynomial $p(x_{11},x_{12},\ldots,x_{nn})$ with rational coefficients. Since $t_{11},t_{12},\ldots,t_{nn}$ are mutually transcendental, this can only be zero if $p$ is the zero polynomial. But if we let $$u_{ij} \;=\; \begin{cases}1/q_{ii} & \text{if } i = j,\\ 0 & \text{if }i\ne j.\end{cases}$$ then $p(u_{11},u_{12},\ldots,u_{nn})$ is the determinant of the $n\times n$ identity matrix, which is $1$. We conclude that $p$ is not the zero polynomial, so $M$ has nonzero determinant.
• @KennyLJ "Mutually transcendental" means algebraically independent over $\mathbb{Q}$. – Jim Belk Nov 13 '13 at 22:22