Independence numbers of disjoint families and complete joins

Let $S$ be independent in $D$. Its intersection with the $j$th tagged part corresponds to an independent set of $J_j$, so it has at most $\alpha (J_j)$ vertices. The intersections partition $S$; the finite sum rule gives $\mathrm{\mid }S\mathrm{\mid }\le {\sum }_{j=1}^t\alpha (J_j)$. Conversely, choose an independent set of size $\alpha (J_j)$ in each $J_j$. Their tagged copies have no edges between them, so their union is independent in $D$ and has size ${\sum }_{j=1}^t\alpha (J_j)$. Therefore $\alpha (D)$ equals that sum.

By the complete-connection definition, successive joins retain all internal edges and make every two distinct tagged parts completely adjacent. Thus an independent set of $Q$ meets at most one part, and its size is at most ${\max }_j\alpha (J_j)$; this also holds for the empty set. Choose an index attaining that finite maximum and a maximum independent set in its part. It remains independent in $Q$, so $\alpha (Q)={\max }_j\alpha (J_j)$.

If $t\ge 2$, choose an index $r$ attaining the maximum. Every $J_j$ is nonempty, so every $\alpha (J_j)\ge 1$; in particular the sum over indices other than $r$ is positive. Splitting the sum at $r$ gives ${\sum }_j\alpha (J_j)>\alpha (J_r)={\max }_j\alpha (J_j)$. Together with the preceding steps this proves all assertions.