  Let $T$ be a tree such that all its vertices of degree more than two lie on
one path, that is, $T$ is a caterpillar subdivision. We prove that there exists
$\epsilon>0$ such that for every graph $G$ with $|V(G)|\ge 2$ not containing
$T$ as an induced subgraph, either some vertex has at least $\epsilon|V(G)|$
neighbours, or there are two disjoint sets of vertices $A,B$, both of
cardinality at least $\epsilon|V(G)|$, where there is no edge joining $A$ and
$B$.
  A consequence is: for every caterpillar subdivision $T$, there exists $c>0$
such that for every graph $G$ containing neither of $T$ and its complement as
an induced subgraph, $G$ has a clique or stable set with at least $|V(G)|^c$
vertices. This extends a theorem of Bousquet, Lagoutte and Thomass\'e [JCTB
2015], who proved the same when $T$ is a path, and a recent theorem of
Choromanski, Falik, Liebenau, Patel and Pilipczuk [Electron. J. Combin. 2018],
who proved it when $T$ is a "hook".
