Let $n$ be bigger than two, and let $A$ be a subset of the $n$-dimensional Euclidean space.
Suppose that the intersection of $A$ with any $(n-1)$-dimensional affine hyperplane is semialgebraic.
Can one conclude that $A$ is semialgebraic?
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1$\begingroup$ In the plane, any convex set that is not semialgebraic will be a counterexample. For example, consider the set which is an intersection of infinitely many hyperplanes, such that no finite subset of hyperplanes defines the set. This is convex, but probably not semialgebraic; I'm not sure how to formally prove the latter, though. Maybe you know? $\endgroup$– Tobias FritzSep 29, 2015 at 13:31
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1$\begingroup$ Actually already for $n=1$ there are plenty of counterexamples, since any subset of the real line has semialgebraic sections ;) $\endgroup$– Tobias FritzSep 29, 2015 at 13:32
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3$\begingroup$ Consider the curve (t,t^2,t^3) and pick some countable infinite set on this curve, this is a counterexample. A plane contains at most two of the points and the resulting set is in fact algebraic. But the original set is not semialgebraic. $\endgroup$– PatrikSep 29, 2015 at 15:40
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$\begingroup$ @Patrik: you probably mean that any plane contains at most three of the points? (Every three points are contained in some plane, so your statement can't be quite right.) $\endgroup$– Tobias FritzSep 29, 2015 at 16:18
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$\begingroup$ Yes thats what I meant :) $\endgroup$– PatrikSep 29, 2015 at 16:20
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