Another possible strengthening of the union-closed sets conjecture

Question

Consider a union-closed family $\mathcal{F}=\{A_1,…,A_n\}$ of $n$ finite sets. Let $\mathcal{A}(\mathcal{F})$ be the set of elements that belong to at least half of the sets of $\mathcal{F}$.

Another question asked for examples of families $\mathcal{F}$ such that:

1. $A_k \not= \mathcal{A}(\mathcal{F})$, $k=1,\ldots,n$
2. $A_k \not\subseteq \mathcal{A}(\mathcal{F})$, $k=1,\ldots,n$

Both of them do exist.

With this question I am asking for examples of families such that:

1. $A_j \cap A_k \not= \mathcal{A}(\mathcal{F})$, $j,k=1,\ldots,n$
2. $A_j \cap A_k \not\subseteq \mathcal{A}(\mathcal{F})$, $j,k=1,\ldots,n$

Answer 1

The family in this answer is an example for both cases 1. and 2. in the question.

It has $135$ sets and $105$ elements.

This is its Hasse diagram:

For more details see the linked answer.