Let $C$ and $D$ be plane cubic curves over the real numbers. Suppose that the real loci in the projective plane of $C$ and $D$ consist of two connected components. Denote $C\_0$ and $D\_0$ the bounded component (which is often referred as "oval") of $C$ and $D$, and denote $C\_1$ and $D\_1$ the unbounded component of $C$ and $D$ respectively. 


I have following question: 
What is the maximal number of real intersections of $C\_1$ and $D\_1$ in the projective plane?


Of course a trivial bound is 9 given by Bezout's theorem, but I conjecture that the bound is rather 5. I think this should be well known, but I couldn't find any reference. 



Real intersections of plane cubic curves 
The answer is $9$. Choose a curve $C$ with the required topology. Choose $8$ points $x\_1$, $x\_2$, ..., $x\_8$ on $C\_1$. The conditions of passing through the $x\_i$ impose $8$ linear conditions on the $10$ dimensional space of cubics, so we can find a second cubic $E$ passing through the $x\_i$ and not proportional to $C$. Let $x\_9$ be the ninth intersection of $C$ and $E$.


I claim that $x\_9$ is also on $C\_1$. Proof: Consider $C\_1$ as a loop in $\mathbb{RP}^2$, and let $\tilde{C}\_1$ be the preimage in the universal cover $S^2$. The loop $C\_1$ is not contractible so, if we travel all the way around $C\_1$, we will go from $(x,y,z)$ to $(-x,-y,-z)$ in $S^2$. Since $E$ is an odd degree polynomial, $E(x,y,z) = - E(-x,-y,-z)$. So $E$ must change sign an odd number of times along $C\_1$. We already know that it changes sign at $x\_1$, ..., $x\_8$; the only place to get an additional sign change is at $x\_9$. (Note that this argument shows that $C\_1 \cap D$ is always odd, counting with multiplicity. I think that a careful count will show that $C\_1 \cap D\_1$ is odd, and the other three possibilities are always even.)


Now, $E$ may not have the right topology. But, for $\epsilon$ sufficiently small, the curve $D= C+\epsilon E$ also passes through $x\_1$, ..., $x\_9$, and it does have the right topology.


