Let $S\subset\mathbb{R}^n$ be a semi-algebraic set, and let $U\subset \mathbb{R}^n$ be an open neighbourhood of $S$ (non necessarily semi-algebraic).
Is it true that there exists an open semi-algebraic neighbourhood $V\subset U$ of $S$? Is it true if, in addition, we assume $S$ is locally closed?
Unfortunately this can fail. Take for instance $S = \Bbb R\times\{0\}\subset \Bbb R^2$ and $U =\{(x,y)\mid -e^x < y < e^x\}$. Then if there exists such a $V$, the complement $V^c$ is a closed semi-algebraic subset of $\Bbb R^2$ containing the graph of $y=e^x$ and therefore must contain some interval of the $x$-axis, contradiction.
