n-2k}={n-k-2\choose k-2}.$$

If $k=4$ then $u=4$ and $\gamma_{l+1}={n-4\choose k-3}={n-4\choose 1}$. We get that

$$|{\mathcal F}|\le {n-1\choose k-1}-{n-4\choose k-1}+{n-4\choose k-3}={n-1\choose k-1}-{n-5\choose k-1}+{n-5\choose k-4}-\epsilon,$$

where

$$\epsilon = {n-5\choose k-2}-{n-5\choose k-3}={n-5\choose 2}-{n-5\choose 1}={n-6\choose 2}-1.$$

The second part of the corollary is proved in Section \[sec33\].

Our next goal is to state the “conceptual” version of Theorem \[thmfull1\]. It requires certain preparations. We will use the framework and some of the ideas from [@FK1], as well as from [@KZ]. First of all, we switch to the cross-intersecting setting. Given an intersecting family ${\mathcal F}$ with $\Delta({\mathcal F})=\delta_1({\mathcal F})$, consider the families $$\begin{aligned}
{\mathcal F}(1):=&\{F\setminus\{1\}\ :\ 1\in F\in{\mathcal F}\}\ \ \ \ \ \text{and}\\
{\mathcal F}(\bar 1):=&\{F\ :\ 1\notin F\in{\mathcal F}\}.\end{aligned}$$ Remark that $\gamma({\mathcal F})=|{\mathcal F}(\bar 1)|$. Applying Theorem \[thmHil\], from now on and until the end of Section \[sec3\] we assume that ${\mathcal F}(1)=\mathcal L([2,n],|{\mathcal F}(1)|,k-1)$ and ${\mathcal F}(\bar 1) = \mathcal L([2,n],|{\mathcal F}(\bar 1)|, k)$. Note that ${\mathcal F}(1),{\mathcal F}(\bar 1)\subset 2^{[2,n]}$. For shorthand, we denote ${{\mathcal A}}:={\mathcal F}(1), {{\mathcal B}}:={\mathcal F}(\bar 1)$. While proving Theorem \[thmfull1\], we will work with the ground set $[2,n]$, in order not to confuse the reader and to keep clear the relationship between the diversity of intersecting 