Theorem6.27Sperner's Theorem
For each \(t\ge1\text{,}\) the width of the subset lattice \(\bftwo^t\) is the maximum size of a rank, i.e., \begin{equation*} \width(\bftwo^t)=\binom{t}{\lfloor\frac{t}{2}\rfloor} \end{equation*}
When \(X\) is a finite set, the family of all subsets of \(X\text{,}\) partially ordered by inclusion, forms a subset lattice 1 . We illustrate this in Figure 6.26 where we show the lattice of all subsets of \(\{1,2,3,4\}\text{.}\) In this figure, note that we are representing sets by bit strings, and we have further abbreviated the notation by writing strings without commas and parentheses.
For a positive integer \(t\text{,}\) we let \(\bftwo^t\) denote the subset lattice consisting of all subsets of \(\{1,2,\dots,t\}\) ordered by inclusion. Some elementary properties of this poset are:
The height is \(t+1\) and all maximal chains have exactly \(t+1\) points.
The size of the poset \(\bftwo^t\) is \(2^t\) and the elements are partitioned into ranks (antichains) \(A_0, A_1,\dots, A_t\) with \(|A_i|=\binom{t}{i}\) for each \(i=0,1,\dots,t\text{.}\)
The maximum size of a rank in the subset lattice occurs in the middle, i.e. if \(s=\lfloor t/2\rfloor\text{,}\) then the largest binomial coefficient in the sequence \(\binom{t}{0}, \binom{t}{1},\binom{t}{2},\dots,\binom{t}{t}\) is \(\binom{t}{s}\text{.}\) Note that when \(t\) is odd, there are two ranks of maximum size, but when \(t\) is even, there is only one.
For the width of the subset lattice, we have the following classic result of Sperner.
For each \(t\ge1\text{,}\) the width of the subset lattice \(\bftwo^t\) is the maximum size of a rank, i.e., \begin{equation*} \width(\bftwo^t)=\binom{t}{\lfloor\frac{t}{2}\rfloor} \end{equation*}