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Let $X\subseteq\mathbb{R}^k$ be a non-closed, unbounded semi-algebraic subset. Then it seems to me that Proposition 5.49 on p. 189 of the book Algorithms in Real Algebraic Geometry still holds true for $X$:

(Conic structure theorem at infinity) There exists an $r\in \mathbb{R}$ such that for all $r'\geq r$, there is a semi-algebraic deformation retraction from $X$ to $X_{r'}:=X\cap \overline{B_k(0,r')}$ and a semi-algebraic deformation retraction from $X_{r'}$ to $X_{r}$.

In the book the Proposition above is stated only under the additional assumption that $X$ is closed.

Tentative proof Consider the inclusions $\mathbb{R}^k\subseteq S^k\subseteq \mathbb{R}^{k+1}$. The map $\iota:x\mapsto \frac{x}{||x||^2}$ for $x\in \mathbb{R}^k$, extends to a semi-algebraic automorphism $\iota:S^k\rightarrow S^k$ of the $k$-dimensional sphere. Therefore, $\iota(X)\cap\mathbb{R}^k\subseteq\mathbb{R}^{k}$ is a semi-algebraic set such that $0\in \iota(X)$ is a non-isolated point of $\iota(X)$ (because $X$ was unbounded by assumption). Finally, one may apply the local conic structure theorem (Prop. 5.48 of the same book) to finish the argument.

Question: Is the proof above correct ?

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  • $\begingroup$ A proof of this result follows from Corollary 9.3.7 of the book "real algebraic geometry" of Bochnak, Coste and Roy. $\endgroup$ Jan 12, 2015 at 12:01

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