Proving that some intersection of $5$ (or more) unions is non-empty for a union-closed family of sets

Question

Consider a union-closed family $\mathcal{F} = \{A_1, \dotsc ,A_n\}$ of $n$ finite sets, $n$ odd, $n \ge 3$, $A_i \neq \emptyset$, $i=1,\dotsc,n$.

Let $h=\frac{n+1}{2}$. We have that:

$$\bigcup_{1 \le i_1 \lt \ldots \lt i_h \le n} A_{i_1} \cap \ldots \cap A_{i_h} = \bigcap_{1 \le i_1 \lt \ldots \lt i_h \le n} A_{i_1} \cup \ldots \cup A_{i_h}. \tag{1}\label{1}$$

This is because the LHS contains all elements that are in at least $h$ of the $A_i$ and the RHS contains all elements that are in at least $n-h+1$ of the $A_i$ (because for each element that does not belong to the RHS we can find $h$ $A_i$ that do not contain it and thus it is in maximum $n-h$ of the $A_i$) and note that in our case $h=n-h+1$.

Let $B_k = A_{i_{k,1}} \cup \dotsb \cup A_{i_{k,h}}$, $B_k \neq U(\mathcal{F})$, $1 \le i_{k,1} \lt \dotsb \lt i_{k,h} \le n$, $1 \le k \le n$.

Let's choose $1 \le i_{1,1} \lt \dotsb \lt i_{1,h} \le n$ and $1 \le i_{2,1} \lt \dotsb \lt i_{2,h} \le n$ to form $B_1$ and $B_2$, $B_1 \neq B_2$. Since $B_k \neq U(\mathcal{F})$ then $A_i \neq U(\mathcal{F})$ and $B_1$ and $B_2$ must share at least two sets $C_1 = A_{i_{1,u_1}} = A_{i_{2,z_1}}$ and $C_2 = A_{i_{1,u_2}} = A_{i_{2,z_2}}$ in the respective unions expressions $B_1 = A_{i_{1,1}} \cup \dotsb \cup A_{i_{1,h}}$ and $B_2 = A_{i_{2,1}} \cup \dotsb \cup A_{i_{2,h}}$, because $2h-(n-1) = 2$. Therefore $B_1 \cap B_2 \neq \emptyset$.

Likewise, we choose three sets $B_1$, $B_2$, $B_3$, all different. $B_1$ and $B_2$ share at least two sets $C_1$ and $C_2$ in the respective union expressions, as in the above paragraph with only two sets $B_1$ and $B_2$. If one of these sets has an element in common with one of $A_{i_{3,1}}, \dotsc, A_{i_{3,h}}$ then $B_1 \cap B_2 \cap B_3 \neq \emptyset$. Otherwise all unions of the form $C_j \cup A_{i_{3,l}}$ are in $\mathcal{F}$, and it is easy to see that they must be all different, otherwise we would have $C_1 = C_2$ and/or $A_{i_{3,l}} = A_{i_{3,m}}$. Therefore there are $2h = n+1$ such unions, which added to $C_1, C_2, A_{i_{3,1}}, \ldots, A_{i_{3,h}}$, give $n+1+2+h \gt n$, absurd. Therefore $B_1 \cap B_2 \cap B_3 \neq \emptyset$.

Now choose four sets $B_1$, $B_2$, $B_3$, $B_4$ like above. The four unions are formed in total by $4h=2n+2$ sets selected among $n-1$, thus $A_1, \dotsc ,A_n$ appear on average $2\frac{n+1}{n-1} > 2$ times in $B_1$, $B_2$, $B_3$, $B_4$. Therefore there must be at least one set $A_t$ belonging to three of the $B_1$, $B_2$, $B_3$, $B_4$. Letting $C_1 = A_t$ and reasoning like before we have that, assuming $C_1 \not\subset B_4$ and $C_1 \cap A_{i_{4,1}} = \emptyset, \dotsc, C_1 \cap A_{i_{4,h}} = \emptyset$, the sets $C_1, A_{i_{4,1}}, \dotsc, A_{i_{4,h}}$ generate $h$ different unions $C_1 \cup A_{i_{4,1}}, \dotsc, C_1 \cup A_{i_{4,h}}$, in turn different from $C_1, A_{i_{4,1}}, \dotsc, A_{i_{4,h}}$, for a total of $h+1+h = n+2 > n$, absurd. Therefore $B_1 \cap B_2 \cap B_3 \cap B_4 \neq \emptyset$.

Now I wanted to push that a little further for $5$ sets $B_1$, $B_2$, $B_3$, $B_4$, $B_5$, but while I can make it work for specific examples, I cannot get a general proof that it must be always $B_1 \cap B_2 \cap B_3 \cap B_4 \cap B_5 \neq \emptyset$.

Can someone help with the case $5$ (or even more) sets?

Edit 2022-01-26

I have seen now that for $j \neq k$ it must be $A_{i_{k,1}} \neq B_j , \dotsc , A_{i_{k,h}} \neq B_j$, otherwise $B_j \subset B_k$ and therefore $B_k$ could be removed from the intersection that would have fewer sets. But since $A_{i_{k,1}} \neq B_j , \dotsc , A_{i_{k,h}} \neq B_j$ for $j \neq k$, it follows that $A_{i_{k,1}} , \dotsc , A_{i_{k,h}}$ can be chosen in less than $n$ sets, or more precisely in $n-1-(q-1)$ where $q$ is the number of $B_i$ sets ($q=5$ in the question case). Since there are fewer choices for the $A_i$, there will be more "overlapping" for the $B_i$ union expressions, but I am not sure if it might help.

Answer 1

Each of the $5$ unions $B_k = A_{i_{k,1}} \cup \dotsb \cup A_{i_{k,h}}$, $k=1,\dotsc,5$ cannot contain another of them i.e. $B_k \not\subseteq B_j$ for $k \neq j$ otherwise we can simplify $B_1 \cap B_2 \cap B_3 \cap B_4 \cap B_5$ to get an intersection with less than $5$ sets. Therefore $A_{i_{k,l}} \neq B_j$ for $k \neq j$, $l = 1, \dotsc, h$ and then the $A_{i_{k,l}}$ can be chosen among $n-1-4 = n-5$ sets (having removed $U(\mathcal{F})$ and all $B_j$ with $j \neq k$).

The total number of occurrences, in the $B_k$ union expressions, of $A_1,\dotsc,A_n$, $A_i \neq B_1,B_2,B_3,B_4,B_5$, $A_i \neq U(\mathcal{F})$, $i=1,\dotsc,n$, will be at least $5(\frac{n+1}{2}-1)$. The $-1$ accounts for the worst case when there always exist one $A_{i_{k,l}} = B_k$ in the union expression of $B_k$. The minimum average occurrence for the $A_i \neq B_1,B_2,B_3,B_4,B_5$, $A_i \neq U(\mathcal{F})$, $i=1,\dotsc,n$ will then be $\frac{5}{2}\frac{n-1}{n-6}$.

$\frac{5}{2}\frac{n-1}{n-6} > 3$ for $6 < n < 31$ therefore when $n < 31$ there will be at least one $A_i$ occurring in at least $4$ of the union expressions for the $B_k$. Now either $A_i$ belongs to all $B_1,B_2,B_3,B_4,B_5$, or otherwise we can apply the same reasoning for the case of four union expressions $B_1,B_2,B_3,B_4$ (see the question). In any case $B_1 \cap B_2 \cap B_3 \cap B_4 \cap B_5 \neq \emptyset$.

For $n \ge 31$, let $t$ be the number of $A_i$ occurring in $3$ of the union expressions for the $B_k$. It must be $\frac{5\frac{n-1}{2}-3t}{n-6-t} \le 2$, which gives, for $n \ge 31$, $t \ge \frac{n+19}{2} \ge 20$. The $A_i$ occurring in $3$ of the $B_k$ can occur in ${5 \choose 3} = 10$ ways (the number of ways to choose $3$ of the $B_k$). We want to find $1 \le j \lt k \lt l \le 5$ such that there are as much as possible $A_i$ in the union expressions of $B_j, B_k, B_l$. In the worst case, they will be equally distributed among the ${5 \choose 3}$ combinations of $B_j, B_k, B_l$, but since $t \ge 20$ there will be at least $2 = \frac{20}{10}$ of the $A_i$, $C_1 = A_{i_1}, C_2 = A_{i_2}$, occurring in the same triplet of union expressions $B_j, B_k, B_l$. The other two $B_p$, $B_q$ unions $1 \le p \lt q \le 5$, $p,q \neq j,k,l$ will share $s = 2h - (n-6)=7$ of the $A_i$ sets in the respective union expressions. Let's call these sets $D_1, \dotsc, D_s$. If there exist a couple $C_u, C_v$ such that $C_u \cap D_v \neq \emptyset$, we are done ($B_1 \cap B_2 \cap B_3 \cap B_4 \cap B_5 \neq \emptyset$), otherwise it is easy to see that the unions of one of $C_1, C_2$ and one of $D_1, \dotsc, D_s$ will be all different between themselves and from $C_1, C_2, D_1, \dotsc, D_s$. The number of such unions is $2s$, and they cannot be in the union expressions of $B_p$ and/or $B_q$, otherwise either $C_1$ or $C_2$ would belong to $B_p$ and/or $B_q$, therefore belonging to $4$ or $5$ of the $B_k$, which we have already shown leading to $B_1 \cap B_2 \cap B_3 \cap B_4 \cap B_5 \neq \emptyset$ (see the $n < 31$ case above). On the other hand, if neither $C_1$ nor $C_2$ belong to $B_p$ or $B_q$, we can recompute the number of $A_i$ shared by the union expressions of $B_p$ and $B_q$ as $s' = 2h - (n-6-2s) = 2s+2h-n+6 = 3s > s$ and reiterating the same reasoning, recomputing the number of sets shared by the union expressions of $B_p$ and $B_q$ again, we would have $s'' > s'$ and so on without limit, therefore there must exist a couple $C_u, C_v$ such that $C_u \cap D_v \neq \emptyset$ and then $B_1 \cap B_2 \cap B_3 \cap B_4 \cap B_5 \neq \emptyset$ always.

(Edit 2022-02-04: The above argument can probably be simplified because we just need one $C_1 = A_{i_1}$ occurring in three $B_k$ to make it work, and it exists for all $n > 6$ because $\frac{5}{2}\frac{n-1}{n-6}>2$. I think it works also with six $B_k$ unions because $\frac{6}{2}\frac{n-1}{n-7}>3$: in this case we have one $C_1 = A_{i_1}$ occurring in four $B_k$. However, for the case of seven $B_k$, we have $\frac{7}{2}\frac{n-1}{n-8}>4$ only for $n < 57$).

All the above should be rewritten in a more detailed and formal way, but I think (I hope) it is correct.

Now, extending the above for $B_1,B_2,B_3,B_4,B_5,B_6$ unions or more does not seem easy. Maybe it worked for $5$ because $3$ (the number of $B_j$, $B_k$, $B_l$) and $2$ (the number of $B_p$, $B_q$) are near to $5/2$. So it seems one should try to divide the $B_k$ into two groups of about the same number each. Intuitively, if we write the unions one below the other aligning vertically the same $A_i$, since the unions use $\frac{n+1}{2}$ sets, then the $A_i$ occur on about half of the positions horizontally, therefore they must occur about half of the times vertically too.