\((m -1)\)-Regular Antichains on [m]

Let \( \mathcal{B} \subseteq 2^{[m]} \) be an antichain of size \( |\mathcal{B}| =: n \). \( 2^{[m]} \) is ordered by inclusion. An antichain \( \mathcal{B} \) is called \( k \)-regular (\( k \in \mathbb{N} \)), if for each \( i \in [m] \) there are exactly \( k \) sets \( B_1, B_2, \ldots, B_k \in \mathcal{B} \) containing \( i \). In this case, we say that \( \mathcal{B} \) is a \( (k, m, n) \)-antichain.

Let \( m \geq 2 \) be an arbitrary natural number. In this note, we show that an \( (m-1, m, n) \)-antichain exists if and only if \( n \in [m+2, \binom{m}{2} – 2] \cup \{m, \binom{m}{2}\} \).