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I'm working on the following questions but with no luck so I was hoping maybe someone can come up with help.

Let $T$ be a tournament on $n$ vertices, say $\left\{v_{1},\ldots,v_{n}\right\}$, and let $B(T)$ be the graph on the same vertices $v_{1},\ldots,v_{n}$ so that for $i<j$, $v_{i}v_{j}$ is an edge of $B(T)$ if $v_{j} \rightarrow v_{i}$ is an edge of $T$. In other words, $B(T)$ is the graph of backward edges of $T$. A transitive subset of $T$ is a subtournament that has no directed cycle. Furthermore, let $u$ be the size of the largest transitive set of $T$ and $y$ be the size of the largest clique or stable set of $G$.

  1. If $x$ is bounded (independent of $n$) does it follow that $y$ is bounded?
  2. If $y$ is bounded does it follow that $x$ is bounded?

Considering the literature, I feel like the answer for 1 should be yes and for 2 no, but I really don't see it. Maybe someone can help me. I think the problem is very interesting. The analogy between transitive subsets of tournaments and cliques and stable sets comes basically from the dualization of the Erdos-Hajnal conjecture for tournament. Erdos proved some bounds for x but I really don't think we need them here.

Note. I also posted this on MathOverflow because I'm not really sure how difficult it can be... it's either very easy or hard.

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up vote 2 down vote accepted

If I haven't misunderstood your question (I assume $u$ means $x$), I think it does fall under "very easy". Note that $x \ge y$ for any $T$: any clique in $B(T)$ corresponds to an "all-backwards" subtournament of $T$ which is necessarily transitive. So the answer to 1 is yes.

The answer to 2 is no in a strong sense: if $y$ is bounded then $x$ is necessarily unbounded as a function of $n$. This follows readily from Ramsey's theorem: since $B(T)$ has no clique of size $y+1$, it must contain an independent set of size $x$ once $n$ is larger than $R(x,y+1)$. This independent set corresponds to an "all-forward" subtournament of $T$ which is again transitive.

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Darn, you are right. Everything seems correct. I guess I just confused myself too much with these diagrams... – TJIF Feb 12 '13 at 6:47

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