Set in $\mathbb{R}^n$ that does not belong to unions of hyperplanes I want to find an explicit finite set $S$ in $\mathbb{R}^n$ 
such that $S$ does not belong to any union of $m$ hyperplanes of $\mathbb{R}^n$. 
Also I have a requirement for the size of $S$: $|S| = \text{poly} (n,m)$. 
 A: If you pick $n+1$ points by random from a uniform distribution over a set with positive $n$-dimensional volume, the probability that all of them lie in the same hyperplane is $0$. Therefore, if you pick $mn+1$ points by random, then with probability $1$ no $n+1$ points lie on the same hyperplane and you need $m+1$ hyperplanes to cover the set.
So such a set with $mn+1$ certainly exists, but you wanted an explicit set.
I suspect that letting the $i$th coordinate of the $j$th point be $\sqrt[p_{nj+i}]{p_{nj+i}}$ (where $p_k$ is the $k$th prime) works, but perhaps someone better in number theory can prove that.

Actually, as Ivan Neretin says in a comment, $\sqrt{p_{nj+i}}$ is actually enough. Here's the outline of a proof: Our claim is that
$$
\begin{pmatrix}
\sqrt{p_{1,1}} - \sqrt{p_{0,1}} \\
\vdots \\
\sqrt{p_{1,n}} - \sqrt{p_{0,n}}
\end{pmatrix}, \cdots ,
\begin{pmatrix}
\sqrt{p_{n,1}} - \sqrt{p_{0,1}} \\
\vdots \\
\sqrt{p_{n,n}} - \sqrt{p_{0,n}} 
\end{pmatrix}
$$
are linearly independent if all $p_{i,j}$ are distinct primes. Wlog, this means that the last vector can't be expressed as linear combination of the others. 
Assume for induction that 
$$
\begin{pmatrix}
\sqrt{p_{1,1}} - \sqrt{p_{0,1}} \\
\vdots \\
\sqrt{p_{1,n-1}} - \sqrt{p_{0,n-1}}
\end{pmatrix}, \cdots ,
\begin{pmatrix}
\sqrt{p_{n-1,1}} - \sqrt{p_{0,1}} \\
\vdots \\
\sqrt{p_{n-1,n-1}} - \sqrt{p_{0,n-1}} 
\end{pmatrix}
$$
are linearly independent. Then
$$
\begin{pmatrix}
\sqrt{p_{n,1}} - \sqrt{p_{0,1}} \\
\vdots \\
\sqrt{p_{n,n-1}} - \sqrt{p_{0,n-1}} 
\end{pmatrix}
$$
can be represented as a unique linear combination of those vectors with coefficients $c_1,\dots c_{n-1}$. But $\sqrt{p_{n,n}} - \sqrt{p_{0,n}}$ can't be expressed as 
$$
c_1 (\sqrt{p_{1,n}} - \sqrt{p_{0,n}}) + \cdots +
c_{n-1} (\sqrt{p_{n-1,n}} - \sqrt{p_{0,n}}),
$$
because the LHS must be in $\mathbb{Q}[p_{0,1},\dots,p_{0,n},p_{1,1},\dots,p_{1,n},\dots,p_{n,1},\dots,p_{n,n-1}]$ and $\sqrt{p_{n,n}} - \sqrt{p_{0,n}}$ is not.
A: A partial answer: let $S_0$ denote the set
$$
S_0 = \{e_i : 1 \leq i \leq n\}
$$
where $e_1,\dots,e_n$ denotes the canonical basis of $\Bbb R^n$ (for instance, $e_2 = (0,1,0,\dots,0)$).  Define
$$
S_k = \{x + (k,k,\dots,k) : x \in S_0\}.
$$
That is, $S_k$ is the set $S_0$ "shifted over" in a direction orthogonal to the hyperplane that contains them $k$ units.  For $m \leq n$, the set
$$
S = S_0 \cup S_1 \cup \cdots \cup S_m
$$
requires (I believe) $m$ hyperplanes, and contains $mn$ points.  I suspect, however, that this set can be captured in a smaller number of hyperplanes when $m$ is sufficiently large.  Rotating the sets $S_k$ about the $(1,1,\dots,1)$ axis might be a way to fix this.  Perhaps it suffices to rotate by a constant angle that is not a rational multiple of $\pi$.
