Red-blue clique partitions and (1-1)-transversals

Motivated by the problem of Gallai on $(1-1)$-transversals of $2$-intervals, it was proved by the authors in 1969 that if the edges of a complete graph $K$ are colored with red and blue (both colors can appear on an edge) so that there is no monochromatic induced $C_4$ and $C_5$ then the vertices of $K$ can be partitioned into a red and a blue clique. Aharoni, Berger, Chudnovsky and Ziani recently strengthened this by showing that it is enough to assume that there is no induced monochromatic $C_4$ and there is no induced $C_5$ in {\em one of the colors}. Here this is strengthened further, it is enough to assume that there is no monochromatic induced $C_4$ and there is no $K_5$ on which both color classes induce a $C_5$. We also answer a question of Kaiser and Rabinovich, giving an example of six $2$-convex sets in the plane such that any three intersect but there is no $(1-1)$-transversal for them

An Erd\H{o}s--Hajnal analogue for permutation classes

Let $\mathcal{C}$ be a permutation class that does not contain all layered permutations or all colayered permutations. We prove that there is a constant $c$ such that every permutation in $\mathcal{C}$ of length $n$ contains a monotone subsequence of length $cn$

Bounds on piercing and line-piercing numbers in families of convex sets in the plane

A family of sets has the $(p, q)$ property if among any $p$ members of it some $q$ intersect. It is shown that if a finite family of closed convex sets in $\mathbb{R}^2$ has the $(p+1,2)$ property then it is pierced by $\lfloor \frac{p}{2} \rfloor +1$ lines. As a result, the following is proved: Let $\mathcal{F}$ be a finite family of closed convex sets in the plane with no isolated sets, and let $\mathcal{F}'$ be the family of its pairwise intersections. If $\mathcal{F}$ has the $(p+1,2)$ property and $\mathcal{F}'$ has the $(r+1,2)$ property, then $\mathcal{F}$ is pierced by $(\lfloor \frac{r}{2} \rfloor ^2 +\lfloor\frac{r}{2} \rfloor)p$ points when $r\ge 2$, and by $p$ points otherwise. The proof uses the topological KKM theorem