I am reading from the paper here and have a doubt in the proof of Proposition 3.8. The author says that the extensive property is immediate but I do not understand how.
The extensive property to be established is $J\subset F_K[J]$. If $j\in J$ and $j\in K$ then a $(j,K)$-PBD containing a flat of order $j'$ for some $j'\in J$ is the $(j,K)$-PBD consisting of a single block. If $j\in J$ and $j\not\in K$ then how do we exhibit a $(j,K)$-PBD containing a flat of order $j'$ for some $j'\in J$?
The paper defines $B[K]$ to be the set of positive integers $v$ for which there exists a $(v,K,1)$-PBD. It defines $F_K[u] \subseteq B[K]$ to be the set of positive integers $v$ for which there exists $(v,K,1)$-PBD with a flat of order $u$.
A flat of a block design is a subset $F$ of its points such that for every pair of points in $F$, $F$ also contains the block containing that pair. (Thinking of blocks as lines in some geometry, flats are the affine subspaces, like planes and such.) In particular, the set of all points is a flat, so any $(u,K,1)$-PBD has a flat of order $u$. Therefore, for any $u \in B[K]$ (that is, for any $u$ for which there exists any $(u,K,1)$-PBD to begin with) we also have $u \in F_K[u]$.
The operation $F_K[J]$ is defined as $$ F_K[J] = \bigcup_{u \in J} F_K[u]. $$ When $J \subseteq B[K]$ (which is assumed in the proposition) then each $F_K[u]$ will contain $u$, and therefore the union of all the $F_K[u]$'s will contain all of $J$.