Proof. First we show that no anti-chain is larger than these two. We attempt to partition $\ds 2^{[n]}$ into $k={n\choose \lfloor n/2\rfloor}$ chains, that is, to find chains $$\eqalign{ &A_{1,0}\subseteq A_{1,1}\subseteq A_{1,2}\subseteq\cdots\subseteq A_{1,m_1}\cr &A_{2,0}\subseteq A_{2,1}\subseteq A_{2,2}\subseteq\cdots\subseteq A_{2,m_2}\cr &\vdots\cr &A_{k,0}\subseteq A_{k,1}\subseteq A_{k,2}\subseteq\cdots\subseteq A_{k,m_k}\cr }$$ so that every subset of $[n]$ appears exactly once as one of the $\ds A_{i,j}$. If we can find such a partition, then since no two elements of an anti-chain can be in the same chain, no anti-chain can have more than $k$ elements.

For small values of $n$ this can be done by hand; for $n=3$ we have $$\eqalign{ &\emptyset\subseteq\{1\}\subseteq\{1,2\}\subseteq\{1,2,3\}\cr &\{2\}\subseteq\{2,3\}\cr &\{3\}\subseteq\{1,3\}\cr }$$ These small cases form the base of an induction. We will prove that any $\ds 2^{[n]}$ can be partitioned into such chains with two additional properties:

1. Each set in a chain contains exactly one element more than the next smallest set in the chain.

2. The sum of the sizes of the smallest and largest element in the chain is $n$.

Note that the chains for the case $n=3$ have both of these properties. The two properties taken together imply that every chain "crosses the middle'', that is, every chain contains an element of $\sbs{n}{n/2}$ if $n$ is even, and an element of both $\sbs{n}{\lfloor n/2\rfloor}$ and $\sbs{n}{\lceil n/2\rceil}$ if $n$ is odd. Thus, if we succeed in showing that such chain partitions exist, there will be exactly $n\choose \lfloor n/2\rfloor$ chains.

For the induction step, we assume that we have partitioned $\ds 2^{[n-1]}$ into such chains, and construct chains for $\ds 2^{[n]}$.

First, for each chain $A_{i,0}\subseteq A_{i,1}\subseteq\cdots\subseteq A_{i,m_i}$ we form a new chain $A_{i,0}\subseteq A_{i,1}\subseteq\cdots\subseteq A_{i,m_i}\subseteq A_{i,m_i}\cup \{n\}$. Since $|A_{i,0}|+|A_{i,m_i}|=n-1$, $|A_{i,0}|+|A_{i,m_i}\cup \{n\}|=n$, so this new chain satisfies properties (1) and (2).

In addition, if $m_i>0$, we form a new chain $A_{i,0}\cup \{n\}\subseteq A_{i,1}\cup \{n\}\subseteq\cdots\subseteq A_{i,m_i-1}\cup \{n\}$. Now $$\eqalign{ |A_{i,0}\cup \{n\}|+|A_{i,m_i-1}\cup \{n\}|&= |A_{i,0}|+1+|A_{i,m_i-1}|+1\cr &=|A_{i,0}|+1+|A_{i,m_i}|-1 +1\cr &=n-1+1=n\cr }$$ so again properties (1) and (2) are satisfied.

Because of the first type of chain, all subsets of $[n-1]$ are contained exactly once in the new set of chains. Also, we have added the element $n$ exactly once to every subset of $[n-1]$, so we have included every subset of $[n]$ containing $n$ exactly once. Thus we have produced the desired partition of $\ds 2^{[n]}$.

Now we need to show that the only largest anti-chains are $\sbs{n}{\lfloor n/2\rfloor}$ and $\sbs{n}{\lceil n/2\rceil}$.

Suppose that $A_1,A_2,…,A_m$ is an anti-chain; then $A_1^c,A_2^c,…,A_m^c$ is also an anti-chain, where $A^c$ denotes the complement of $A$. Thus, if there is an anti-chain that contains some $A$ with $|A|>\lceil n/2\rceil$, there is also one containing $A^c$, and $|A^c|< \lfloor n/2\rfloor$. Suppose that some anti-chain contains a set $A$ with $|A|< \lfloor n/2\rfloor$. We next prove that this anti-chain cannot be of maximum size.

Partition $\ds 2^{[n]}$ as in the first part of the proof. Suppose that $A$ is a subset of the elements of a one or two element chain $C$, that is, a chain consisting solely of a set $S_1$ of size $n/2$, if $n$ is even, or of sets $S_1$ and $S_2$ of sizes $\lfloor n/2\rfloor$ and $\lceil n/2\rceil$, with $A\subseteq S_1\subseteq S_2$, if $n$ is odd. Then no member of $C$ is in the anti-chain. Thus, the largest possible size for an anti-chain containing $A$ is ${n\choose \lfloor n/2\rfloor}-1$.

If $A$ is not a subset of the elements of such a short chain, we now prove that there is another chain partition of $\ds 2^{[n]}$ that does have this property. Note that in the original chain partition there must be a chain of length 1 or 2, $C_1$, consisting of $S_1$ and possibly $S_2$; if not, every chain would contain a set of size $\lfloor n/2\rfloor -1$, but there are not enough such sets to go around. Suppose then that $A=\{x_1,\ldots,x_k\}$ and the set $S_1$ in $C_1$ is $S_1=\{x_1,\ldots,x_q,y_{q+1},\ldots y_l\}$, where $0\le q< k$ and $l>k$.

Let $\sigma$ be the permutation of $[n]$ such that $\sigma(x_{q+i})=y_{q+i}$ and $\sigma(y_{q+i})=x_{q+i}$, for $1\le i\le k-q$, and $\sigma$ fixes all other elements. Now for $U\subseteq [n]$, let $\overline U=\sigma(U)$, and note that $U\subseteq V$ if and only if $\overline U\subseteq\overline V$. Thus every chain in the original chain partition maps to a chain. Since $\sigma$ is a bijection, these new chains also form a partition of $\ds 2^{[n]}$, with the additional properties (1) and (2). By the definition of $\sigma$, $A\subseteq\overline S_1$, and $\{\overline S_1,\overline S_2\}$ is a chain, say $\overline C_1$. Thus, this new chain partition has the desired property: $A$ is a subset of every element of the 1 or 2 element chain $\overline C_1$, so $A$ is not in an anti-chain of maximum size.

Finally, we need to show that if $n$ is odd, no anti-chain of maximum size contains sets in both $\sbs{n}{\lfloor n/2\rfloor}$ and $\sbs{n}{\lceil n/2\rceil}$. Suppose there is such an anti-chain, consisting of sets $A_{k+1},\ldots,A_l$ in $\sbs{n}{\lceil n/2\rceil}$, where $l={n\choose \lceil n/2\rceil}$, and $B_1,\ldots,B_k$ in $\sbs{n}{\lfloor n/2\rfloor}$. The remaining sets in $\sbs{n}{\lceil n/2\rceil}$ are $A_1,\ldots,A_k$, and the remaining sets in $\sbs{n}{\lfloor n/2\rfloor}$ are $B_{k+1},\ldots,B_l$.

Each set $B_i$, $1\le i\le k$, is contained in exactly $\lceil n/2\rceil$ sets in $\sbs{n}{\lceil n/2\rceil}$, and all must be among $A_1,\ldots,A_k$. On average, then, each $A_i$, $1\le i\le k$, contains $\lceil n/2\rceil$ sets among $B_1,\ldots,B_k$. But each set $A_i$, $1\le i\le k$, contains exactly $\lceil n/2\rceil$ sets in $\sbs{n}{\lfloor n/2\rfloor}$, and so each must contain exactly $\lceil n/2\rceil$ of the sets $B_1,\ldots,B_k$ and none of the sets $B_{k+1},\ldots,B_l$.

Let $A_1=A_{j_1}=\{x_1,\ldots,x_r\}$ and $B_{k+1}=\{x_1,\ldots,x_s,y_{s+1},\ldots,y_{r-1}\}$. Let $B_{i_m}=A_{j_m}\backslash\{x_{s+m}\}$ and $A_{j_{m+1}}=B_{i_m}\cup\{y_{s+m}\}$, for $1\le m\le r-s-1$. Note that by the preceding discussion, $1\le i_m\le k$ and $1\le j_m\le k$. Then $A_{j_{r-s}}=\{x_1,\ldots,x_s,y_{s+1},\ldots,y_{r-1},x_r\}$, so $A_{j_{r-s}}\supseteq B_{k+1}$, a contradiction. Hence there is no such anti-chain. $\qed$