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Let $S\subset\mathbb{R}^n$, with $n\geq 3$, such that for any hyperplane $L$ one has $L\cap S$ closed, semialgebraic, and convex. Is it true that $S$ itself is semialgebraic?

A colleague explained to me that closedness is necessary, as one can take an open ball in $\mathbb{R}^n$ and paste a sufficiently nasty curve into its boundary.

It is probably possible to weaken the convexity assumption to something like "no isolated points", and still have a meaningful question.

PS. the is basically a question recently asked by Lev Birbrair and Aris Daniilidis.

PPS. The if $n=2$ then for any convex set $S$ the intersection with a line is semialgebraic, so the answer is obvious "no" in this case.


One way, suggested by Birbrair, to make question more interesting is to assume in addition that $S$ is defined in an $o$-minimal structure. This would preclude examples like in the my own answer below.

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  • $\begingroup$ well, that one has an easy answer. $\endgroup$ – Dima Pasechnik Dec 1 '17 at 12:36
  • $\begingroup$ This one may have an easy negative answer too . . . Can we make $S = \cap_{k=1}^\infty S_k$ with each $S_k$ convex and semialgebraic (possibly even a half-space), $S$ not an intersection of finitely many $S_k$, but any $L\cap S$ equal to the intersection of finitely many $L \cap S_k$? $\endgroup$ – Noam D. Elkies Dec 3 '17 at 3:52
  • $\begingroup$ Any condition on $L$ ? $\endgroup$ – Dima Pasechnik Dec 3 '17 at 11:34
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    $\begingroup$ @NoamD.Elkies Cutting infinitely many tiny caps of fast decreasing sizes off the unit ball seems to work. In 3D we can easily imagine such an arrangement by using the stereographic projection of the sphere to the plane, putting the points along the parabola $y=x^2$ accumulating to $0$ and choosing the neighborhoods so small compared to the distance to the origin that no line or circle will be able to produce that order of tangency. In 4D and higher my vision blurs a bit but I see no principal obstacles. $\endgroup$ – fedja Dec 8 '17 at 0:02
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    $\begingroup$ In fact the latter is not a problem. Michel Coste kindly told me that it follows from the Zariski closure of the boundary being of wrong dimension (as otherwise it will contain infinitely many planes), whereas the Zariski closure of the boundary of a semialgebraic set must be algebraic of the same dimension, cf. Thm 3.20 in perso.univ-rennes1.fr/michel.coste/polyens/SAG.pdf $\endgroup$ – Dima Pasechnik Dec 10 '17 at 18:36
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Here I record an answer based on very kind comments above (and offline). The construction starts from a closed ball $B\subset\mathbb{R}^3$ and an infinite sequence $\mathcal{L}:=\{L^+_k\}$ of open half-spaces, satisfying the condition $$B\cap L^+_i\cap L^+_j=\emptyset\quad\text{ if and only if $i\neq j$.}\qquad\qquad (*)$$ Denote $L^-_k:=\mathbb{R}^3\setminus L_k^+$, and set $$B^-:=B\cap L^-_1\cap L^-_2\cap\dots\cap L^-_k\cap\dots$$ Zariski closure of the boundary $\partial B^-$ of $B^-$ contains infinitely many planes $\partial L_k^-$, and thus it cannot be a surface. Theorem 3.20 in lecture notes by Michel Coste says that the Zariski closure of the boundary $\partial S$ of a semialgebraic set $S$ is algebraic of the same dimension as $\partial S$. Applying it to the $B^-$, we obtain that $\partial B^-$ is not semialgebraic, threfore $B^-$ is not semialgebraic, as well.

It remains to construct an appropriate $\mathcal{L}$, so that any plane $P$ intersects only finitely many $B\cap L_k$. Each $B\cap L_k$ is a spherical cap of radius $r_k$ and centre $C_k\in\partial B$. Let $\pi$ be the stereographic projection of $\partial B$ from the north pole onto the tangent plane $\Pi$ at the south pole $O\in\Pi$. Let $M\gg 0$, $c_k:=((k+M)^{-1},(k+M)^{-2})\in\Pi$, for $k\geq 1$; set $C_k=\pi^{-1}(c_k)$ and $r_k:=e^{-k-M}$, thus specifying a particular $\mathcal{L}$. Then $(*)$ holds. As $O$ is the limit point of $\{C_k\}$, any plane $\Omega$, to have a chance to intersect infinitely many $B\cap L_k$, and thus provide non-semialgebraic $B^-\cap\Omega$, must pass through $O$.

To establish that such an $\Omega$ does not exist, it suffices to show that any line $\ell$ in $\Pi$ intersects only finitely many disks $D_k\subset\Pi$ with centre at $c_k$ of radius $r_k$. As the points $c_k$ lie on a parabola $y=x^2$, $\ell$ won't hit inifinitely many $D_k$, unless it is tangent to it at $O$. So it remains to consider the case of $\ell$ being the line $y=0$. As $r_k$ decreases exponentially, but the $y$-coordinate $(k+M)^{-2}$ of $c_k$ only quadratically, there will be $K$ so that for any $k>K$ the intersection of $\ell$ and $D_k$ is empty. Thus any plane intersects $B^-$ in a semialgebraic set, and we are done.

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  • $\begingroup$ So likewise in higher dimensions $n$, using a rational normal curve $(t,t^2,t^3,\ldots,t^n)$ to generalize a parabola? $\endgroup$ – Noam D. Elkies Dec 11 '17 at 0:56
  • $\begingroup$ Yes, and probably we can even have examples of non-semialgebraic convex sets with not only all linear, but also all non-linear semialgebraic sections (provided such sections do not contain rational normal curves). $\endgroup$ – Dima Pasechnik Dec 11 '17 at 11:13
  • $\begingroup$ Maybe that proviso can be avoided too by using a transcendental curve in place of the rational normal one. $\endgroup$ – Noam D. Elkies Dec 11 '17 at 15:42
  • $\begingroup$ this is probably much harder: indeed, "non-wild" transcendental curves are approximated well by algebraic ones, and we only care about approximate behaviour in this proof. Perhaps one can go different way, placing smaller and caps, as we approach a singular point, on an oscillating curve... $\endgroup$ – Dima Pasechnik Dec 11 '17 at 19:50

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