Consider the set $[n]=\{1,2,\dots, n\}$. A Kneser $k$-colouring of $[n]$ is a map $c:\mathcal P([n])\to S$ to some set $S$ such that for sets $A_1, A_2,\dots, A_k$ we have $c(A_i)=c(A_j)$ for all $i,j$ implies $\cap_{i=1}^kA_i\ne \varnothing$. Graph theoretically, we want to colour the vertices of a $k$-uniform hypergraph with vertices corresponding to subsets of $[n]$ and $k$ vertices form a hyperedge if their intersection is empty.
A chain in this setting is a sequence of sets $A_1\subset A_2\subset\cdots\subset A_n$. Given a Kneser $k-$colouring, I want to look at how many colours a chain gets.
An easy upper bound of $\lfloor n(1-1/k)\rfloor+1$ follows by colouring the sets in the following way : If $A\subseteq [n]$ has size at most $n(1-1/k)$, then colour it with $\min(A)$; if its size is more than $n(1-1/k)$, colour it with $n+1$. Then every chain gets at most the required many colours.
My question is : Is the bound tight? That is, does every Kneser $k$-colouring have a chain with at least these many colours?
I have verified that it holds for $k=2$ (as is the case with problems like these, it required some topology), but I cannot really see how to do it for $2<k\le n$.
Any hints or solutions are appreciated! Thanks in advance.
