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Iosif Pinelis
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$\newcommand\R{\mathbb R}$This is not true in general. E.g., let $p(s):=(\arctan s,0,\dots,0)$. Then $$\bigcup_{s\in\R}N_s=(-\pi/2,\pi/2)\times\R^{n-1}\ne\R^n.$$


After the above answer was posted, the OP has changed the question by adding the condition that $p(s)$ be pointwise polynomial, thus invalidating the above answer. However, this does not help: for any $n\ge2$, let $p(s):=(s_+^2,(1-s)_+^2,0,\dots,0)$, where $u_+:=\max(0,u)$. Then $p'(s)\ne0$ for any real $s$, whereas $$U:=\bigcup_{s\in\R}N_s\subseteq\{(x_1,x_2,\dots,x_n)\in\R^n\colon x_1\ge0\text{ or }x_2\ge0\},$$ so that $\ne\R^n$.

Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229