Question on cross-cuts contained in cross-cuts

Question

Let $X$ be a finite set with $|X| =n $ and $\mathcal{A} \subset \mathcal{P}(X)$ a set system.

Call $\mathcal{A}$ a cross-cut if $\forall B \in \mathcal{P}(X), \; \exists A \in \mathcal{A} $ such that $A \subset B $ or $B \subset A$

I am asked to show that any cross-cut $\mathcal{A}$ contains a cross-cut that has size at most ${n}\choose{\lfloor \frac{n}{2}\rfloor}$.

My thoughts are that this can be achieved by taking a minimal cross-cut $\mathcal{B}$ contained in $\mathcal{A}$. Then, by minimality, for every $B \in \mathcal{B}$, we have that:

$\exists C \in \mathcal{P}(X),\; \forall A \in \mathcal{B} - \{B\}$ such that $A \not\subset C$ and $C \not\subset A$

Then taking such a $C$, we can consider $(\mathcal{B} \cup C) - \{B\}$. Thinking about graphs where two sets are adjacent if one is contained in the other, then we see $(\mathcal{B} \cup C) - \{B\}$ has one less edge than $\mathcal{B}$.

Now, I want to continue and keep switching out sets of $\mathcal{B}$ until we get $\mathcal{C}$ which is an anti-chain, then we are done by Sperner.

However, there is no guarantee that $\mathcal{B} \cup C - \{B\}$ is contained in $\mathcal{A}$, so even if it were a cross-cut, there is no guarantee it is "minimal".

I am stuck at this point and I think that perhaps I am supposed to turn this sub-cross-cut $\mathcal{B}$ into an anti-chain in some other way, but I'm not entirely sure how that might be achieved.

Answer 1

Hint 1: there are minimal cross cuts that are not antichains, so you are right that you need a modification to get an antichain.

Example for hint 1:

Take $X=\mathbb Z/5\mathbb Z$ and the ten-element set family consisting of pairs $\{i,i+1\}$ and triples $\{i,i+1,i+2\}$ for each $i\in X.$

Hint 2:

The set of inclusion-minimal elements of a set family is always an antichain, so you just need to modify the other elements.

Solution:

Assume $\mathcal A$ is a minimal cross cut. Let $\mathcal A_0$ be the set of inclusion-minimal elements of $\mathcal A.$ For each $A\in\mathcal A\setminus\mathcal A_0$ pick a set $W_A\subset A$ such that for all $B\in\mathcal A\setminus\{A\}$ we have $W_A\not\subset B$ and $B\not\subset W_A.$ Let $\mathcal W=\{W_A\mid A\in \mathcal A\setminus\mathcal A_0\}.$ Then $\mathcal A_0\cup\mathcal W$ is an antichain and has the same order as $\mathcal A.$