# Conjecture about partitions of the powerset without the empty set

I would like to have some ideas about possibilities of proving or disproving the following conjecture:

For any partition $$\mathcal{F}=\{\mathcal{A_1},\ldots,\mathcal{A_m} \}$$ of the powerset without the empty set element $$\mathcal{P}([n]) \setminus \{\emptyset\}$$, and defined $$\mathcal{F}_a = \{\mathcal{A} \in \mathcal{F} : \exists B \in \mathcal{A} : a \in B\}$$, there exists $$x \in [n]$$ such that $$\vert \mathcal{F}_x \vert \ge \big\lfloor \frac{m}{2} \big\rfloor$$.

For example, for $$n = 4$$, a possible family $$\mathcal{F}$$ is: $$\{\{\{1\}\}, \{\{2\}\}, \{\{1,3\}, \{2,3\}\}, \{\{4\}\}, \{\{1,4\}, \{2,4\}\}, \{\{1,2\}, \{3\}, \{1,2,3\}, \{1,2,4\}, \{3,4\}, \{1,3,4\}, \{2,3,4\}, \{1,2,3,4\}\}\}$$ with $$m = 6$$ and $$\vert \mathcal{F}_1 \vert = 4 \ge 3$$, $$\vert\mathcal{F}_2 \vert = 4 \ge 3$$, $$\vert\mathcal{F}_3 \vert = 2 \lt 3$$, $$\vert\mathcal{F}_4 \vert = 3 \ge 3$$.

I have tested all cases for $$n \le 4$$, however I don't see how to go further with brute force.

• The statement of the question is not clear. I suppose that you mean $\mathcal{P}([n]) \setminus \{\emptyset\}$, and that you note $\mathcal{F}_a = \{\mathcal{A} \in \mathcal{F} : \exists B \in \mathcal{A} : \{a\} \in B\}$ for all $a \in [n]$? Sep 20 at 17:40
• @ChristopheLeuridan no it is just at least one $a \in [n]$. Sep 20 at 18:24

Here is a counterexample for $$n=5$$. Partition the non-empty subsets of $$\{1, \dots, 5\}$$ into the singleton subsets and a sixth family containing all the other non-empty subsets. So, $$m=6$$ and $$|\mathcal{F}_i|=2$$ for all $$i \in [5]$$.
Partition the subsets of $$\{1,2,\dots, 100\}$$ as $$\{\mathcal{A_1},\ldots,\mathcal{A_6}\}$$ by defining $$\mathcal{A_i}=\{S\in \mathcal{P}([100]) \setminus \{\emptyset\}: \text{all elements of S are} \,\equiv i \pmod{6}\}$$ for $$1\le i\le 5$$, and letting $$\mathcal A_6$$ consist of all the remaining subsets. You can check that $$|\mathcal F_x|\le 2$$ for all $$x$$, providing a counterexample.