Let $X$ be a (singular, reducible) affine plane real algebraic curve of degree $d$.


How we can estimate maximal number of connected components of it's complement in $R^2$ in terms of degree?



Maximal number of connected components of complement to an affine plane real algebraic curve
We can also prove that the maximal number of components for reducible nonsingular curve is $\frac{n^2+n+2}{2}$ by the following computation.


Let $m$ be a partition of $n$, where $m\_i$ is the degree of $i$-th irreducible component of our curve.
Denote $s=\{i|m\_i=1\}$, $t=\{i|m\_i>1\}$.
Then, using Harnack inequality and Besout's theorem(for intersections with infinite line and for intersections between irreducible components) we can write that the number of connected components is lesser than
$$
\max\_{m\vdash n}(\sum\_{i}(\frac{(m\_i-1)(m\_i-2)}{2}+1)+1+n-s+\sum\_{i\neq j}m\_im\_j)=$$
$$=\frac{n^2}{2}-n+1+\max\_{m\vdash n}(2t+s))\leq $$
$$\leq \frac{n^2}{2}-n+1+ \max\_{0\leq 2x+y\leq n,0\leq x,0\leq y}(2x+y)=\frac{n^2+n+2}{2}
$$ 


Therefore the only question we have is the following:
why do Harnack bound (in terms of degree) holds for singular curves as well as for nonsingular?


The proof could be obtained using First Harnack Theorem.


