# How much can we "shrink" intersecting families

Motivation. An intersecting family is a collection of subsets $${\cal S}\subseteq {\cal P}(X)$$ of a set $$X\neq \emptyset$$ such that $$A\cap B\neq \emptyset$$ for all $$A,B\in {\cal S}$$. The intersections of members of $${\cal S}$$ can be quite large. I was wondering whether it is always possible to "shrink" $${\cal S}$$ such that a lot of intersections between $$2$$ sets become singletons.

Shrinkings. If $${\cal S}\subseteq {\cal P}(X)$$ is intersecting, we say that $${\cal S}^*\subseteq {\cal P}(X)$$ is a shrinking of $${\cal S}$$ if $${\cal S}^*$$ is intersecting and there is a bijection $$\varphi:{\cal S} \to {\cal S^*}$$ such that for all $$A \in {\cal S}$$ we have that $$\varphi(A)\subseteq A$$.

For any set $$Z$$, let $$[Z]^2=\big\{\{a,b\}:a\neq b\in Z\big\}$$. Moreover, we set $$[{\cal S}]^2_1:=\big\{\{A,B\} \in [{\cal S}]^2: |A\cap B|=1\big\}$$. We say $${\cal S}$$ is intersection-efficient if $$\big|[{\cal S}]^2_1\big| \geq \big|[{\cal S}]^2 \setminus [{\cal S}]^2_1\big|.$$

Question. If $${\cal S}\subseteq {\cal P}(X)$$ is intersecting, is there a shrinking $${\cal S}^*\subseteq {\cal P}(X)$$ of $${\cal S}$$ such that $${\cal S}^*$$ is intersection-efficient?

Counterexample. Let $$X=\{1,2,3,\dots,n\}$$ where $$n\ge8$$, and let $$\mathcal S=\{\{1,2\},\ \{1,3\},\ \{2,3\}\}\cup\{\{2,3,x\}:3\lt x\le n\}\subset\mathcal P(X).$$ Then $$\mathcal S$$ is an intersecting family with no proper shrinking, and $$\mathcal S$$ is not intersection-efficient, since $$|[\mathcal S]^2_1|=2n-3\lt\binom{n-2}2=|[\mathcal S]^2\setminus[\mathcal S]^2_1|.$$
Another counterexample. Let $$X=\{0,1,2,\dots,n\}$$ where $$n\ge3$$, and let $$\mathcal S=\{A\subseteq X:0\in A\}\subset\mathcal P(X).$$ Then $$\mathcal S$$ is an intersecting family with no proper shrinking, and $$\mathcal S$$ is not intersection-efficient, since $$|[\mathcal S]^2_1|=\frac{3^n-1}2\lt\binom{2^n}2-\frac{3^n-1}2=|[\mathcal S]^2\setminus[\mathcal S]^2_1|.$$