A complete-or-weakly-sparse blockade yields a complete subblockade or an anticonnected thinning

Suppose some block $D_i$ has no anticonnected component of size at least $q\mathrm{/}\mathrm{\ell }$. Partition the anticonnected components of $G[D_i]$ into a minimum number of nonempty unions $S_0,\dots ,S_r$, each of size less than $q\mathrm{/}\mathrm{\ell }$, ordered so that $\mathrm{\mid }{S}_0\mathrm{\mid }\le \cdots \le \mathrm{\mid }{S}_r\mathrm{\mid }$. Since the unions cover $D_i$ and each has size less than $q\mathrm{/}\mathrm{\ell }$, one has $r+1>\mathrm{\ell }$. Minimality implies $\mathrm{\mid }{S}_{t-1}\mathrm{\mid }+\mathrm{\mid }{S}_t\mathrm{\mid }\ge q\mathrm{/}\mathrm{\ell }$ for every $t\ge 1$, for otherwise those two unions could be merged. Hence $\mathrm{\mid }{S}_t\mathrm{\mid }\ge q\mathrm{/}(2\mathrm{\ell })\ge q\mathrm{/}{\mathrm{\ell }}^2$ for every $t\ge 1$, because $\mathrm{\ell }\ge 2$. Distinct anticonnected components are complete to one another by [L1], so distinct unions of them are also complete to one another. Therefore $S_1,\dots ,S_{\mathrm{\ell }}$ form a complete $(\mathrm{\ell },q\mathrm{/}{\mathrm{\ell }}^2)$-blockade, proving outcome 1.

We may therefore assume that every $D_i$ has an anticonnected component $C_i$ of size at least $q\mathrm{/}\mathrm{\ell }$. By [L2], the complement $\overline{G[C_i]}$ is connected. Choose a spanning tree of $\overline{G[C_i]}$, and repeatedly delete leaves until exactly $\lceil q\mathrm{/}\mathrm{\ell }\rceil$ vertices remain. The remaining tree is connected, so the induced subgraph of $\overline{G[C_i]}$ on those vertices is connected as well. Calling that vertex set $B_i$, we have $B_i\subseteq C_i$, $\mathrm{\mid }{B}_i\mathrm{\mid }=\lceil q\mathrm{/}\mathrm{\ell }\rceil$, and $G[B_i]$ anticonnected.

If $(D_i,D_j)$ is complete, then $(B_i,B_j)$ is complete because $B_i\subseteq D_i$ and $B_j\subseteq D_j$. If $(D_i,D_j)$ is mutually $\eta$-sparse, every vertex of $B_i$ has at most $\eta \mathrm{\mid }{D}_j\mathrm{\mid }=\eta q$ neighbours in $D_j$, hence at most $\eta q\le \eta \mathrm{\ell }\mathrm{\mid }{B}_j\mathrm{\mid }$ neighbours in $B_j$ because $\mathrm{\mid }{B}_j\mathrm{\mid }\ge q\mathrm{/}\mathrm{\ell }$; the same argument with $i$ and $j$ exchanged gives the reverse direction. Thus every noncomplete pair is mutually $(\eta \mathrm{\ell })$-sparse.

Step 1.1 yields outcome 1, while steps 2.1 and 3.1 yield outcome 2. Therefore one of the two stated outcomes holds.