Strange...because, if you know some proofs of Sperner's theorem, you generally know the equality cases.... Pierre.

abbas_mehrabian wrote: Let $A$ be a family of subsets of $\{1,2,\ldots,n\}$ such that no member of $A$ is contained in another. Sperner’s Theorem states that $|A|\leq{n\choose{\lfloor\frac{n}{2}\rfloor}}$. Find all the families for which the equality holds. your's solution what is Sperner's theorem ?

to pbornsztein:students are supposed to deduce when n is odd integer why there are just two families that equality occurs for them.

Here is a solution. Firstly, we shall prove Sperner's Theorem. Sperner's Theorem: Let $S$ be a set with $n$ elements, and let $\cal S$ be a set of its subsets, such that $A\nsubseteq B$ for each $A,B\in\cal S$. Then $|\cal S|\leqslant\left(\begin{array}{c}n\\{\left\lfloor\frac{n}{2}\right\rfloor}\end{array}\right)$. Proof: Let $s_{i}$ be the number of elements of $\cal S$ with $i$ elements. Obviously, $\sum_{i=0}^{n}s_{i}=|\cal S|$. Since \[\left(\begin{array}{c}n\\{\left\lfloor\frac{n}{2}\right\rfloor}\end{array}\right)\geqslant \binom{n}{i}\quad(*)\] for each $i\in\{1,2,\dots,n\}$, it follows $\frac{s_{i}}{\left(\begin{array}{c}n\\{\left\lfloor\frac{n}{2}\right\rfloor}\end{array}\right)}\leqslant\frac{s_{i}}{\binom{n}{i}}$, and therefore $\sum_{i=0}^{n}\frac{s_{i}}{\left(\begin{array}{c}n\\{\left\lfloor\frac{n}{2}\right\rfloor}\end{array}\right)}\leqslant\sum_{i=0}^{n}\frac{s_{i}}{\binom{n}{i}}$. By the following lemma, RHS is $\leqslant 1$, and the theorem follows. $\square$ Lemma (Lubell-Yamamoto-Meshalkin inequality): With the former conditions and notation, the following holds: \[\sum_{i=0}^{n}\frac{s_{i}}{\binom{n}{i}}\leqslant 1\] Proof: By chain we mean any array $(S_{i})_{i\geqslant 0}$ of subsets of $S$ such that $S_{i}\subset S_{i+1}$ for each $i$. Obviously, the maximal chain has $n+1$ elements, and there are $n!$ maximal chains. Fix an $\overline{S}\in{\cal S}$. There are $|S|!(n-|S|)!$ maximal chains that contain $\overline{S}$, because we start from empty set, add one element at the time till we get to $\overline{S}$ in $|S|!$ ways, and then add the rest $n-|S|$ elements in $(n-|S|)!$ ways. Also note that it is impossible that two elements $A,B$from $\cal S$ are in the same chain, because that would mean $A\subset B$ or $B\subset A$. Therefore: \[\sum_{i=0}^{n}s_{i}i!(n-i!)=\sum_{\overline{S}\in\cal S}|S|!(n-|S|)!\leqslant n!\quad(**)\] and the lemma follows by dividing with $n!$ (note that $\binom{n}{i}=\frac{n!}{i!(n-i)!}$). $\square$ Now we may return to the main problem. Note that we have the equality iff there is equality in (*) and (**). For even $n$ it is easy: we note from (*) that the equality is attained iff all sets are of the size $\frac{n}{2}$. For odd $n$ it holds $\left(\begin{array}{c}n\\{\left\lfloor\frac{n}{2}\right\rfloor}\end{array}\right)=\left(\begin{array}{c}n\\{\left\lceil\frac{n}{2}\right\rceil}\end{array}\right)$ and we shall prove that the equality holds iff all the elements of $\cal S$ are simultaneously either of cardinality $\left\lfloor\frac{n}{2}\right\rfloor$ or $\left\lceil\frac{n}{2}\right\rceil$. Proof: Let the equality in Sperner's Theorem is obtained. Therefore, each element of $\cal S$ corresponds to one and only one chain, as described in the proof of Lemma. (1): If a $\left\lfloor\frac{n}{2}\right\rfloor$-elements subset of $S$ is not in $\cal S$, then when we add any of the remaining elements, we shall always get a subset which is in $\cal S$, because otherwise no element of $\cal S$ will correspond to the chain which begins with the elements of the "missing" set. (2): More, if a $\left\lceil\frac{n}{2}\right\rceil$-elements subset of $S$ is in $\cal S$, it is obvious that there is no $\left\lfloor\frac{n}{2}\right\rfloor$-elements subsets of that set in $\cal S$. Suppose that there is at least one $\left\lceil\frac{n}{2}\right\rceil$-elements set in $\cal S$, and that there is at least one $\left\lceil\frac{n}{2}\right\rceil$-elements set which is not in $\cal S$. Call them $P_{1}$ and $P'$, respectively. Look at the following "train": \[P_{1}\supset Q_{1}\subset P_{2}\supset Q_{2}\subset P_{3}\supset Q_{3}\subset\cdots\] $P_{i}$'s are pairwise different $\left\lfloor\frac{n}{2}\right\rfloor$-elements subsets of $S$ in $\cal S$, and $Q_{i}$'s are pairwise different $\left\lceil\frac{n}{2}\right\rceil$-elements subset of $S$ not in $\cal S$. In order to form $P_{i+1}$ from $Q_{i}$, we shall choose an element from $P'\setminus\left(\bigcup_{j=1}^{i}P_{j}\right)$. This is possible by (1) and (2). However, it is obvious that, by this procedure, the train will "arrive" at $P'$. Contradiction, what completes the proof. $\blacksquare$

There is a different proof for this

So can you post your proof?

Not sure if this is the same as the above, but once we get that all subsets must be of size $\left \lfloor \frac{n}{2} \right \rfloor$ or $\left \lceil \frac{n}{2} \right \rceil$, we can finish by analyzing Hall's condition. Indeed, we can clearly discard the case when $n$ is even as trivial, as let $n = 2k+1.$ Now consider the matching between subsets of $[n]$ of size $k$ and subsets of $[n]$ of size $k+1$, where we match $A$ ($|A| = k$) with $B$ ($|B| = k+1$) if $A \subset B.$ This matching is regular of degree $k+1$ and so Hall's condition is satisfied. Let $S$ be the set of subsets which are of size $k$. If $|S| = 0$ or $\binom{2k+1}{k}$, we're done. Else, let $N(S)$ be the set of neighbors of vertices of $S$ with size $k+1$, as in Hall's Theorem. It's clear that there are at most $\binom{2k+1}{k} - |N(S)|$ subsets chosen with size $k+1$, and so we just need to show that $|N(S)| > |S|.$ To do this, notice that the sets in $S$ have $|S| (k+1)$ edges coming from them, so for $|N(S)| \le |S|$ we require $|N(S)| = |S|$ and no other sets of size $k$ are adjacent to anything in $N(S).$ However, we may consider a set $C$ of size $k$ such that it differs from a set in $S$ in only one element (do this by starting with something in $S$ and switching out one element at a time until we exit $S$), and notice that $C$ is not in $S$ yet is adjacent to something in $N(S).$ This implies that $|N(S)| > |S|$ and we're done. So the only possibilities for $n = 2k+1$ are the set of $k-$element subsets and the set of $(k+1)-$element subsets. $\square$