A sparse host has many leaf extensions, few smaller copies, or a long sparse blockade

Suppose all three outcomes fail. If $y^{2}n\le 1$, a singleton gives outcome 2 since $\mathrm{\mid }{H}^{\mathrm{\prime }}\mathrm{\mid }\ge 2$. If $y^{a+1}n\le 1\le y^{2}n$, a greedy stable-set algorithm gives at least $n\mathrm{/}(1+y^{2}n)\ge (2y^2{)}^{-1}\ge y^{-1}$ stable vertices: each selected vertex removes itself and at most $y^{2}n$ neighbors. Taking those vertices as singleton blocks gives outcome 3. Consequently $y^{a+1}n>1$.

Among sequences $(B_0,\dots ,B_k)$ of disjoint nonempty sets with $B_i$ $x$-sparse to every earlier block, $\mathrm{\mid }{B}_i\mathrm{\mid }\ge y^{a+1}n$ for $i<k$, and $\mathrm{\mid }{B}_k\mathrm{\mid }\ge (1-hy{)}^{k}n$, choose one with maximum $k$. Such a sequence exists with $k=0$ and $B_0=V(G)$; finiteness bounds $k$. Since outcome 3 fails, $k<y^{-1}$. For $0\le t\le 1\mathrm{/}2$, $1-t\ge 4^{-t}$ (for example, $\log (1-t)+t\log 4$ is concave and nonnegative at both endpoints). As $hy\le 1\mathrm{/}2$, this gives |B_k|\ge(1-hy)^kn\ge4^{-hyk}n>4^{-h}n\ge yn\ge xn.\tag{1}

The graph $G[B_k]$ has maximum degree at most $y^{2}n\le y\mathrm{\mid }{B}_k\mathrm{\mid }$ by (1). Failure of outcome 1 and (1) give ${\mathrm{ind}}_H(G[B_k])\le {\mathrm{ind}}_H(G)\le x^{2a+2h}{n}^h\le x^{2a+h}\mathrm{\mid }{B}_k{\mathrm{\mid }}^h\mathrm{.}$ Because $\mathrm{\mid }{B}_k\mathrm{\mid }>yn\ge y^{2}n$, failure of outcome 2 gives ${\mathrm{ind}}_{H^{\mathrm{\prime }}}(G[B_k])>y^{a-2}\mathrm{\mid }{B}_k{\mathrm{\mid }}^{h-1}$. Thus [L1] applied to $G[B_k]$ supplies disjoint $A,B\subseteq B_k$ such that $\mathrm{\mid }A\mathrm{\mid }\ge y^a\mathrm{\mid }{B}_k\mathrm{\mid }\ge y^{a+1}n$, $\mathrm{\mid }B\mathrm{\mid }\ge (1-hy)\mathrm{\mid }{B}_k\mathrm{\mid }\ge (1-hy{)}^{k+1}n$, and $B$ is $x$-sparse to $A$.

Replace the old final block $B_k$ by $A,B$. Since every vertex of $B_k$ was $x$-sparse to each earlier block, the same is true of its subsets $A$ and $B$; the new last pair is sparse by step 3.1. The new sequence satisfies all the size conditions of step 2.1 with index $k+1$, contradicting maximality. Therefore at least one stated outcome holds.