By Coste M.

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Bd for the chosen ordering. 3 The curve selection lemma The triangulation theorem allows one to give a short proof of the following. 3. 13 (Curve selection lemma) Let S ⊂ Rn be a semialgebraic set. Let x ∈ clos(S), x ∈ S. Then there exists a continuous semialgebraic mapping γ : [0, 1] → Rn such that γ(0) = x and γ((0, 1]) ⊂ S. Proof. Replacing S with its intersection with a ball with center x and radius 1, we can assume S bounded. Then clos(S) is a compact semialgebraic set. By the triangulation theorem, there is a ﬁnite simplicial complex K and a semialgebraic homeomorphism h : |K| → clos(S), such that x = h(a) for a vertex a of K and S is the union of some open simplices of K.

By the triangulation theorem, there is a ﬁnite simplicial complex K and a semialgebraic homeomorphism h : |K| → clos(S), such that x = h(a) for a vertex a of K and S is the union of some open simplices of K. In particular, since x is in the closure of S and not in S, there is a simplex σ of K whose a ◦ is a vertex, and such that h(σ) ⊂ S. Taking a linear parametrization of the segment joining a to the barycenter of σ, we obtain δ : [0, 1] → σ such that ◦ δ(0) = a and δ((0, 1]) ⊂σ. Then γ = h◦δ satisﬁes the property of the theorem.

D. algorithm described in Chapter 2 does not suﬃce to eliminate quantiﬁers. d. are adjacent to others, except for the cells in a cylinder. We do not know, in general, what happens when we pass from a cylinder to another. d. adapted to the sphere, it is not diﬃcult to determine the topology from the cell decomposition. The two functions on the disk x2 + y 2 < 1, whose graphs are the two open hemispheres, have an obvious extension by continuity on the closed disk. We show an example where this is not so.