Looking for a counterexample for a conjecture about union-closed families of sets

Question

(I have posted a corrected version of this question. The limit $\lceil n/2 \rceil$ must be replaced with $\lceil (n+1)/2 \rceil$.)

I have already asked basically the same question here, but now I have found a way to rephrase it simply, so this new formulation might be more interesting.

Consider a union-closed family $\mathcal{F}$ of $n$ finite sets with $\mathcal{F} \not = \{ \emptyset \}$.

Let $\mathcal{H} \subseteq \mathcal{F}$ be the family of all sets in $\mathcal{F}$ which are (not necessarily proper) supersets of at least $\lceil n/2 \rceil$ of the sets in $\mathcal{F}$.

I conjecture that there always exists a non-empty set in $\mathcal{F}$ which is a subset of at least $| \mathcal{H} | - 1$ of the sets in $\mathcal{H}$.

Can we say something or find a counterexample for this conjecture?

I have tried many examples but couldn't find a counterexample.

Proving the conjecture should be difficult, because I believe it implies the union-closed sets conjecture, however finding a counterexample might be easier and could provide a "difficult" example for the union-closed sets conjecture.

If someone wants to experiment, I have written a python program: given an input family on the standard input (use an empty line for the empty set), it removes duplicates and adds all missing unions of some of its sets, in order to obtain a closed family, then verifies the conjecture:

Here it is run over this example, and here over an example similar to this one where the conjecture is satisfied for $| \mathcal{H} | - 1$ sets in $\mathcal{H}$ and not for all of them.

Answer 1

Counterexample with $|\mathcal F|=n=20$. First partition the set $[18]=\{1,2,3,\dots,18\}$ into the disjoint sets
$A=\{1,2,3,4\}$, $B=\{5,6,7,8\}$, $C=\{9,10,11,12\}$,
$X=\{13,14\}$, $Y=\{15,16\}$, $Z=\{17,18\}$. Now define
$\mathcal F=\{\varnothing,A_0,A_1,A_2,A_3,B_0,B_1,B_2,B_3,C_0,C_1,C_2,C_3,X_0,X_1,Y_0,Y_1,Z_0,Z_1,T\}$ where:
$A_0=B\cup C\cup Z$, $A_1=A_0\cup\{1\}$, $A_2=A_1\cup\{2\}$, $A_3=A_2\cup\{3\}$,
$B_0=A\cup C\cup Y$, $B_1=B_0\cup\{5\}$, $B_2=B_1\cup\{6\}$, $B_3=B_2\cup\{7\}$,
$C_0=A\cup B\cup X$, $C_1=C_0\cup\{9\}$, $C_2=C_1\cup\{10\}$, $C_3=C_2\cup\{11\}$,
$X_0=A\cup B\cup C\cup Y\cup Z$, $X_1=X_0\cup\{13\}$,
$Y_0=A\cup B\cup C\cup X\cup Z$, $Y_1=Y_0\cup\{15\}$,
$Z_0=A\cup B\cup C\cup X\cup Y$, $Z_1=Z_0\cup\{17\}$,
$T=A\cup B\cup C\cup X\cup Y\cup Z=[18]$.

To see that $\mathcal F$ is union-closed and $|\mathcal F|=20$, observe that:
$\varnothing\subsetneqq A_0\subsetneqq A_1\subsetneqq A_3$,
$\varnothing\subsetneqq B_0\subsetneqq B_1\subsetneqq B_3$,
$\varnothing\subsetneqq C_0\subsetneqq C_1\subsetneqq C_3$,
$A_0\cup B_0=X_0\subsetneqq X_1\subsetneqq T$,
$A_0\cup C_0=Y_0\subsetneqq Y_1\subsetneqq T$,
$B_0\cup C_0=Z_0\subsetneqq Z_1\subsetneqq T$,
$A_0\cup B_0\cup C_0=T$.

Finally, $\mathcal H=\{X_0,X_1,Y_0,Y_1,Z_0,Z_1,T\}$, so $|\mathcal H|=7$, while each of the minimal nonempty elements of $\mathcal F$, namely $A_0$, $B_0$, and $C_0$, is contained in just $5$ elements of $\mathcal H$.

Answer 2

The following non-counterexample (it had a mistake) can serve to show that the factor $1/2$ can’t be lowered.

Consider for instance, for $r\in \mathbb N$, the family $\mathcal F:=\{F\subset[2r]: |F|\ge r\}$, so that $n:=|\mathcal F|=\sum_{i\ge r}{2r\choose i}=2^{2r-1}+\frac12{2r\choose r}$. Then in $ \mathcal H$ there are all $H\in\mathcal F$ of cardinality $|H|= 2r-1$, because every such $H$ includes $\sum_{i\ge r}{2r-1\choose i}=2^{2r-2}\sim\frac n2 $ sets $F\in\mathcal F$. But every $F\in\mathcal F$ fails to be a subset of at least $r$ elements $H$ of $\mathcal H$, namely $H_i:=[2r]\setminus \{i\}$ for $i\in F$.