Consider an undirected graph $K(n,k,i)$, with the all $k$-element subsets of $\{1,\dots,n\}$ as vertices, and two vertices connected by an edge if their sets intersect in less than $i$ elements. 


This [paper](https://cdm.ucalgary.ca/cdm/index.php/cdm/article/download/557/259) claims on page 74 (Theorem 5.1) that the chromatic number of $K(n,k,i)$ is at least $n-2k+2i$. However, their is a place in the proof of the theorem which I don't understand: They say that 



> 
> Otherwise we would have two $k$-subsets of
>  color $j$ such that each of them has at least $k - i + 1$ elements in one of two
>  disjoint hemispheres, so their intersection has at most $i-1$ elements which is
>  impossible.
> 
> 
> 


Since the intersection can be of size up to $i-1$ in each hemisphere, don't we only get that it has size at most $2i-2$ in total?



Understanding proof about chromatic number
This appears to be a genuine logical error. If $k$ is at least twice $i-1$ and $n$ is large enough that we can cover the sphere in a fine dust, then almost any pair of antipodal points will correspond to hemispheres $H\_1$, $H\_2$ containing at least $k-i+1$ points from sets $S\_1$, $S\_2$ with intersection of size $2(i-1)$.


