A finite star class can require one query per alternative

Before the first query, $V=\{h_{\ast },h_1,\dots ,h_m\}$. Suppose before index $i$ the current version space is $\{h_{\ast },h_i,h_{i+1},\dots ,h_m\}$. At $x_i$, $h_i(x_i)=0$ while $h_{\ast }(x_i)=1$ and $h_j(x_i)=1$ for every $j>i$, so the point is a disagreement. Since $i\le m\le b$, the procedure queries it; [F1] returns $1$, and [F2] removes exactly $h_i$. Thus after this query the version space is $\{h_{\ast },h_{i+1},\dots ,h_m\}$. Induction proves this description at every index.

Applying step 1.1 through $i=m$ shows that all $m$ queries receive the positive target label, each removes one distinct alternative, and the final version space is $\{h_{\ast }\}$. Hence $q=m=\mathrm{\mid }\mathcal{H}\mathrm{\mid }-1$.

Set $m=2^k-1$ for integers $k\ge 2$. Then $\mathrm{\mid }\mathcal{H}\mathrm{\mid }=2^k$ and $q=2^k-1$. Repeated application of [F4]'s product law gives $\log \mathrm{\mid }\mathcal{H}\mathrm{\mid }=k\log 2$, and strict increase with $2>1$ gives $\log 2>\log 1=0$. The ratio $q\mathrm{/}(k\log 2)$ is at least $2^{k-1}\mathrm{/}(k\log 2)$; the latter grows without bound because its consecutive-term ratio is $2k\mathrm{/}(k+1)\ge 4\mathrm{/}3$. Therefore no fixed constant multiple of $\log \mathrm{\mid }\mathcal{H}\mathrm{\mid }$ bounds the query count over this family.