# Number of connected components of complement to a reducible real algebraic hypersurface.[EDITED]

Let $X_1,\ldots X_k$ be irreducible(may be singular) affine real algebraic hypersurfaces in $R^n$ with $x_1,\ldots, x_k$ connected components, respectively.

Let $G_1,\ldots, G_l$ be their intersections of dimension $n-2$, with $g_1,\ldots, g_l$ connected components, respectively.

How we can estimate an upper bound for the number of connected components of $R^n\setminus (X_1\cup\ldots\cup X_k)$?

For example, such an estimate for $n=2$, and that(together with Harnack and Bezout theorems) leads us to an exact upper bound for a number of connected components of a complement to a real plane algebraic curve.

Of course, due to unsolvedness of Harnack problem in general case we cannot think of an exact upper bound in that case and the result of Bihan(Asymptotic behaviour of Betti numbers of real algebraic surfaces, Comm.Math.Helv. 78 (2003), 227-244) contradicts it's obvious generalization. (He gave an asymptotics for number of connected components of an irreducible variety in the case of surfaces as $dq^3$, where $q$ is the degree of surface and $d\in [\frac{13}{36},\frac{5}{12}]$, while obvious generalization give us $\frac{q^3+5q+6}{6}$ components of complement to a reducible -- that is a number of parts generated by planes in general position). But, of course, some upper bounds could be obtained using Comessati-Petrovsky-Oleinik inequality.

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Isn't Biahn's result in the projective case though? – Thierry Zell Oct 19 '11 at 16:28
Yes, of course in projective, but the maximal number of components in affine case is obviously greater than in projective. – Grey Violet Oct 19 '11 at 16:43

Let $p_1,\ldots, p_m$ be real polynomials in $n$ variables, each of degree $d$ or less. Let $N(p_i)$ be a set of zeros of $p_i$.
The number of topological components of the set $R^n\setminus\cup_iN(p_i)$ does not exceed $\sum_{k=0}^n2(2d)^n2^kC_{m,k}$, where $C_{m,k}$ is the usual binomial coefficient, except that $C_{m,k}=0$ for $m< k$