Maximal antichain with prescribed elements

In the partial order of the subsets of the set \( \{1{,}2, …, 8 \} \) ordered by inclusion:

Variant
Determine the size of the maximum antichain containing the sets \( \{1\} \) and \( \{8 \} \).
Solution
The other sets in the antichain must not contain elements 1 and 8. They therefore form an antichain in the order \( (\mathcal {P} \{2,…, 7 \}, \subseteq) \).

According to Sperner’s theorem, there can be at most \( \binom63 = 20 \).
Answer
The maximum antichain has 22 elements.
Variant
Find the best estimate of the size of the maximum antichain containing the set \( \{1{,}8 \} \).

You do not have to prove the optimality of this estimate.
Solution
Upper bound: The remaining sets in the antichain can be divided into three groups: those that contain 1, those that contain 8, and those that do not contain 1 or 8. According to Sperner’s theorem, each group can have at most \( \binom63 = 20 \) elements .

Together with the set \( \{1{,}8 \} \) we can get an antichain with at most 61 elements.

Lower bound: If we remove all sets containing 1 and 8 from the longest antichain in \( (\mathcal {P} \{1,…, 8 \}, \subseteq) \), we get an antichain with \( \binom84 - \binom62 = 70 - 15 = 55 \) elements. You can also include the set \( \{1{,}8 \} \).

Another view of the same construction: in comparison with the upper bound, we take in the third group of sets ‘only’ 15 four-element sets instead of 20 three-element sets, or \( 2 \binom63 + \binom64 + 1 = 56 \).
Answer
An antichain with 56 elements can be constructed and an upper bound 61 can be derived.