The special-vertex-local structural-partition criterion implies property (*)

Set $c_1=c_3=c\mathrm{/}4>0$ and $c_2=10\mathrm{/}c>0$. Fix an arbitrary $\bar{\mathcal{H}}$-free finite graph $G$ and a special-vertex $(\mathrm{\ell },w)$-comb from Property (*) for a finite graph family, with integral $\mathrm{\ell }\ge 4$ and real $w\ge 4$. Use its supplied local partition. Suppose that all three property-$(\ast )$ outcomes with these constants fail.

If $\mathrm{\mid }{Y}_i\mathrm{\mid }\ge w\mathrm{/}2$ for some $i$, then $Y_i$ is nonempty and [F1] supplies a clique or stable set of size at least $(w\mathrm{/}2{)}^c\ge w^{c\mathrm{/}2}\ge w^{c\mathrm{/}4}$, since $w\ge 4$. This contradicts the first failure. Hence every $\mathrm{\mid }{Y}_i\mathrm{\mid }<w\mathrm{/}2$, and $\mathrm{\mid }{B}_i\mathrm{\mid }\ge w$ implies $\mathrm{\mid }{X}_i\mathrm{\mid }>w\mathrm{/}2$.

If every partition has a block $D_i=A_{j_i}^i$ of size at least $w\mathrm{/}(2\mathrm{\ell })$, choose one for each of the finitely many indices $i$. Fix distinct $i,h$. Each vertex of $D_h$ is complete or anticomplete to $D_i$ by the local external-purity clause. Two vertices of $D_h$ with opposite relations would make any vertex of the nonempty $D_i$ mixed on $D_h$, contrary to the same clause with $i,h$ reversed. Thus $D_i,D_h$ are pure. The disjoint sequence $(D_1,\dots ,D_{\mathrm{\ell }})$ is consequently a pure blockade of width at least $w\mathrm{/}(2\mathrm{\ell })\ge w\mathrm{/}{\mathrm{\ell }}^2$, contradicting the third failure.

By step 2.2 there is an index $i$ such that every $\mathrm{\mid }{A}_j^i\mathrm{\mid }<w\mathrm{/}(2\mathrm{\ell })$, hence is at most this bound. Put $X=X_i$, $t=t_i$, and reorder these blocks as $A_1,\dots ,A_t$ in nonincreasing size. Reordering preserves purity and changes the pattern only by relabelling. Since $w\mathrm{/}2<\mathrm{\mid }X\mathrm{\mid }={\sum }_{j=1}^t\mathrm{\mid }{A}_j\mathrm{\mid }\le tw\mathrm{/}(2\mathrm{\ell })$, we have $t\ge \mathrm{\ell }$. The reordered pattern is still ${\mathcal{F}}_2$-free.

Form the cutoffs of [F3]. They reach $t$: for example, ${\mathrm{\ell }}^{t\mathrm{/}2}\ge 2^t\ge t$ for the positive integer $t$, the latter elementary inequality following by induction. Let $q$ be the first index with $m_q=t$. Since $m_1\le \sqrt{\mathrm{\ell }}<\mathrm{\ell }\le t$, we have $q\ge 2$. For $2\le r\le q$, the integer $m_{r-1}+1$ is at most $t$ and at most $2{\mathrm{\ell }}^{(r-1)\mathrm{/}2}\le {\mathrm{\ell }}^{r\mathrm{/}2}$, so $m_r\ge m_{r-1}+1$. Thus all layers $C_1,\dots ,C_q$ are nonempty and partition the blocks in order.

For $1\le r<q$, put $x={\mathrm{\ell }}^{r\mathrm{/}2}\ge 2$. Then $m_r=\lfloor x\rfloor \ge 2$, so $x<m_r+1\le m_r^2$, yielding $m_r\ge \sqrt{x}={\mathrm{\ell }}^{r\mathrm{/}4}$. Also $m_1\le {\mathrm{\ell }}^{1\mathrm{/}2}$ and each $C_{r+1}$ contains at most $m_{r+1}\le {\mathrm{\ell }}^{(r+1)\mathrm{/}2}$ blocks, including when $r+1=q$.

Suppose a preterminal layer $C_r$, $1\le r<q$, has every block of size at least $w\mathrm{/}{\mathrm{\ell }}^{5r\mathrm{/}2}$. The first $m_r$ blocks all have at least that size by their nonincreasing order. Their induced pattern is nonempty and ${\mathcal{F}}_2$-free, so [F1] gives a pattern clique or stable set $S$ of integral cardinality $k\ge m_r^c\ge {\mathrm{\ell }}^{cr\mathrm{/}4}$. By [F2], the blocks indexed by $S$ form a complete or anticomplete blockade of length $k$ and width at least $w\mathrm{/}{\mathrm{\ell }}^{5r\mathrm{/}2}$.

Since $k^{10\mathrm{/}c}\ge {\mathrm{\ell }}^{5r\mathrm{/}2}$ and $k\ge {\mathrm{\ell }}^{cr\mathrm{/}4}\ge {\mathrm{\ell }}^{c\mathrm{/}4}$, this blockade has width at least $w\mathrm{/}{k}^{10\mathrm{/}c}$ and satisfies the second property-$(\ast )$ outcome. That contradicts step 1.1. Therefore every preterminal $C_r$ contains a block of size strictly less than $w\mathrm{/}{\mathrm{\ell }}^{5r\mathrm{/}2}$.

The first layer contributes at most ${\mathrm{\ell }}^{1\mathrm{/}2}w\mathrm{/}(2\mathrm{\ell })=w\mathrm{/}(2\sqrt{\mathrm{\ell }})\le w\mathrm{/}4$ vertices. For $1\le r<q$, every block in $C_{r+1}$ follows the small block in $C_r$ and has size less than $w\mathrm{/}{\mathrm{\ell }}^{5r\mathrm{/}2}$. Hence $C_{r+1}$ contributes less than $w{\mathrm{\ell }}^{(r+1)\mathrm{/}2-5r\mathrm{/}2}=w{\mathrm{\ell }}^{1\mathrm{/}2-2r}$.

Because $\mathrm{\ell }\ge 4$ and $1\mathrm{/}2-2r<0$, the sum of the latter bounds is at most $w{\sum }_{r\ge 1}{4}^{1\mathrm{/}2-2r}=(w\mathrm{/}8){\sum }_{s\ge 0}{16}^{-s}=2w\mathrm{/}15$. All layers have been counted, so $\mathrm{\mid }X\mathrm{\mid }<w\mathrm{/}4+2w\mathrm{/}15=23w\mathrm{/}60<w\mathrm{/}2$, contradicting step 2.1.

Thus one of the three outcomes holds for every special-vertex comb required by Property (*) for a finite graph family, with constants independent of $G$ and the comb. This proves that $\mathcal{H}$ has property $(\ast )$.