Consider a set $X$ with $|X| = n\geq 1$ elements. A family $\mathcal{F}$ of distinct subsets of $X$ is said to have property $\mathcal{P}$ if there exist $A,B \in \mathcal{F}$ so that $A\subset B$ and $|B\setminus A| = 1$.

i) Determine the least value $m$, so that any family $\mathcal{F}$ with $|\mathcal{F}| > m$ has property $\mathcal{P}$.
ii) Describe all families $\mathcal{F}$ with $|\mathcal{F}| = m$, and not having property $\mathcal{P}$.

([i]Dan Schwarz[/i])
### Part (i)

1. **Claim**: The least value \( m \) such that any family \(\mathcal{F}\) with \(|\mathcal{F}| > m\) has property \(\mathcal{P}\) is \( \boxed{2^{n-1}} \).

2. **Proof**:
   - Consider any element \( x \in X \).
   - Define the power set of \( X \) excluding \( x \) as \(\mathcal{P}(X - \{x\})\).
   - For each subset \( S \in \mathcal{P}(X - \{x\}) \), consider the pairs \(\{S, S \cup \{x\}\}\).
   - The power set \(\mathcal{P}(X)\) can be expressed as:
     \[
     \mathcal{P}(X) = \bigcup_{S \in \mathcal{P}(X - \{x\})} \{S, S \cup \{x\}\}.
     \]
   - By the pigeonhole principle, if \(|\mathcal{F}| > 2^{n-1}\), then there must exist a pair \(\{S_0, S_0 \cup \{x\}\} \in \mathcal{F}\) such that \(S_0 \subset S_0 \cup \{x\}\) and \(|(S_0 \cup \{x\}) \setminus S_0| = 1\). This satisfies property \(\mathcal{P}\).

### Part (ii)

1. **Families with \(|\mathcal{F}| = 2^{n-1}\) and not having property \(\mathcal{P}\)**:
   - Consider the following two families:
     \[
     \mathcal{F}_1 = \{S \in \mathcal{P}(X) : |S| \equiv 0 \pmod{2}\},
     \]
     \[
     \mathcal{F}_2 = \{S \in \mathcal{P}(X) : |S| \equiv 1 \pmod{2}\}.
     \]
   - Both \(\mathcal{F}_1\) and \(\mathcal{F}_2\) have \(2^{n-1}\) elements and do not have property \(\mathcal{P}\).

2. **Proof**:
   - **Lemma 1**: Given \(0 \leq k < n\), if \(\{S \in \mathcal{P}(X) : |S| = k\} \subset \mathcal{F}\), then \(\{S \in \mathcal{P}(X) : |S| = k+1\} \cap \mathcal{F} = \emptyset\).
     - **Proof**: For any \(B \in \mathcal{P}(X)\) with \(|B| = k+1\), take \(x \in B\) and \(A = B - \{x\}\). Then \(A \in \mathcal{F}\), and so \(B \not\in \mathcal{F}\), since otherwise \(\mathcal{F}\) would have property \(\mathcal{P}\). \(\blacksquare\)
   
   - **Lemma 2**: If \(\{S \in \mathcal{P}(X) : |S| = k\} \cap \mathcal{F} = \emptyset\), then \(\{S \in \mathcal{P}(X) : |S| = k+1\} \subset \mathcal{F}\).
     - **Proof**: For any \(B \in \mathcal{P}(X)\) with \(|B| = k+1\), take \(x \in B\) and \(A = B - \{x\}\). Then \(A \not\in \mathcal{F}\), and since \(\mathcal{F}\) must contain exactly one element of each pair in the partition, we must have \(B \in \mathcal{F}\). \(\blacksquare\)