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\begin{fcthm}[Sperner's Lemma]
  \label{sperner}
  % Theorem 1
  Assuming:
  - $\mathcal{A} \subset \powset X$ is an
  \gls{antichain}
  Then:
  $|\mathcal{A}| \le {n \choose \left\lfloor
  \frac{n}{2} \right\rfloor }$.
\end{fcthm}

\textbf{Idea:} Motivated by ``a \gls{chain} meets each layer in
$\le 1$ point, because a layer is an
\gls{antichain}'', we will try to decompose the
cube into \glspl{chain}.

\begin{center}
  \includegraphics[width=0.2\linewidth]{images/21baed414ea34d70.png}
\end{center}

\begin{proof}
  We'll decompose $\powset X$ into ${n \choose
  \half n}$ \glspl{chain} -- then done. To achieve
  this, it is sufficient to find:
  \begin{enumerate}[(i)]
    \item For each $r < \frac{n}{2}$, a matching
      from $X\rsubs$ to $X\rsubs[r + 1]$ (recall
      that a matching here means a set of disjoint
      edges, one for each point in $X\rsubs$).
    \item For each $r \ge \frac{n}{2}$, a matching
      from $X\rsubs$ to $\rsubs[r - 1]$.
  \end{enumerate}
  We then put these together to form our
  \glspl{chain}, each passing through
  $X\rsubs[\halfn]$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/732344e403874cf8.png}
  \end{center}
  By taking complements, it is enough to prove
  (i).

  Let $G$ be the (bipartite) subgraph of $\Q_n$
  spanned by $X\rsubs \cup X\rsubs[r + 1]$: we
  seek a matching from $X\rsubs$ to $X\rsubs[r +
  1]$. For any $S \subset X\rsubs$, the number of
  $S-\Gamma(S)$ edges in $G$ is $|S|(n - r)$
  (counting from below) and $\le |\Gamma(S)| (r +
  1)$ (counting from above).
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/3dfaeab3a06f48cb.png}
  \end{center}
  Hence, as $r < \frac{n}{2}$,
  \[
    |\Gamma(S)| \ge \frac{|S|(n - r)}{r + 1} \ge
    |S|
  .\]
  Thus by Hall's Marriage theorem, there exists a
  matching.
\end{proof}

Equality in \nameref{sperner}? Proof above tells us nothing.

\textbf{Aim:} If $\mathcal{A}$ is an
\gls{antichain} then
\[
  \sum_{r = 0}^{n}  \frac{| \mathcal{A} \cap
  X\rsubs|}{{n \choose r}} \le 1
.\]
\begin{center}
  \includegraphics[width=0.2\linewidth]{images/395cae44c8e24ed3.png}
\end{center}
``The percentages of each layer occupied add up to
$\le 1$.''

Trivially implies \nameref{sperner} (think about it).

\begin{fcdefn}[Shadow]
  \glsnoundefn{shad}{shadow}{shadows}%
  \glssymboldefn{shadow}%
  For $\mathcal{A} \subset X\rsubs$ ($1 \le r \le
  n$), the
  \emph{shadow} of $\mathcal{A}$ is $\partial \mathcal{A} =
  \partial^- \mathcal{A} \subset X\rsubs[r - 1]$
  defined by, $\partial \mathcal{A} = \{B \in
  X\rsubs[r - 1] : \exists A \in \mathcal{A}, B
  \subset A\}$.
\end{fcdefn}

\begin{example*}
  If $\mathcal{A} = \{123, 124, 134, 137\} \subset
  X\rsubs[3]$, then $\shadow A = \{12, 13, 23, 14,
  24, 34, 17, 37\} \subset X\rsubs[2]$.
\end{example*}

\begin{fcprop}[Local LYM]
  \label{local_lym}
  Assuming:
  - $\mathcal{A} \subset X\rsubs$
  - $1 \le r \le n$
  Then:
  \[
    \frac{|\shadow \mathcal{A}|}{{n \choose r - 1}} \ge
    \frac{|\mathcal{A}|}{{n \choose r}}
  .\]
\end{fcprop}

``The fraction of the level occupied by $\shadow
\mathcal{A}$ is $\ge$ the fraction for
$\mathcal{A}$''.

\begin{remark*}
  LYM = Lubell, Meshalkin, Yamamoto.
\end{remark*}

\begin{proof}
  The number of $\mathcal{A}-\shadow \mathcal{A}$
  edges in $\Q_n$ is $| \mathcal{A}| r$ (counting
  from above) and is $\le |\shadow \mathcal{A}|(n
  - r + 1)$ (counting from above). So
  \[
    \frac{|\shadow \mathcal{A}|}{| \mathcal{A}|}
    \ge \frac{r}{n - r + 1}
  .\]
  But $\frac{{n \choose r - 1}}{{n \choose r}} =
  \frac{r}{n - r + 1}$, so done.
\end{proof}

Equality in \nameref{local_lym}? Must have that
$\forall A \in \mathcal{A}$, $\forall i \in A$,
$\forall j \notin A$ have $A - \{i\} \cup \{j\}
\in A$. So $A = \emptyset$ or $X\rsubs$.

\begin{fcthm}[LYM Inequality]
  \label{lym_ineq}
  Assuming:
  - $\mathcal{A} \subset \powset X$ is an
  \gls{antichain}
  Then:
  \[
    \sum_{r = 0}^{n} \frac{| \mathcal{A} \cap
    X\rsubs|}{{n \choose r}} \le 1
  .\]
\end{fcthm}

\begin{notation*}
  \glssymboldefn{intr}%
  We will now start writing $\mathcal{A}_r$ for
  $\mathcal{A} \cap X\rsubs$.
\end{notation*}

\begin{proof}[Proof 1]
  ``Bubble down with \nameref{local_lym}''.

  Have $\frac{| \mathcal{A}_n|}{{n \choose n}} \le
  1$. Now, $\shadow \mathcal{A}\intr[n]$ and
  $\mathcal{A}\intr[n - 1]$ disjoint (as
  $\mathcal{A}$ is an \gls{antichain}), so
  \[
    \frac{|\shadow \mathcal{A}\intr[n]|}{{n \choose
    n - 1}} + \frac{| \mathcal{A}\intr[n - 1]|}{{n
    \choose n - 1}} = \frac{|\shadow
    \mathcal{A}\intr[n] \cup \mathcal{A}\intr[n -
    1]}{{n \choose n - 1}} \le 1
  ,\]
  whence
  \[
    \frac{| \mathcal{A}\intr[n]|}{{n \choose n}} +
    \frac{| \mathcal{A}\intr[n - 1]|}{{n \choose n
    - 1}} \le 1
  \]
  by \nameref{local_lym}.

  Now, note $\shadow(\shadow \mathcal{A}\intr[n]
  \cup \mathcal{A}\intr[n - 1])$ is disjoint from
  $\mathcal{A}\intr[n - 2]$ (since $\mathcal{A}$
  is an \gls{antichain}), so
  \[
    \frac{|\shadow (\shadow \mathcal{A}\intr[n]
    \cup \mathcal{A}\intr[n - 1])|}{{n \choose n -
    2}} + \frac{| \mathcal{A}\intr[n - 2]|}{{n
    \choose n - 2}} \le 1
  ,\]
  whence
  \[
    \frac{|\shadow \mathcal{A}\intr[n] \cup
    \mathcal{A} \intr[n - 1]|}{{n \choose n - 1}}
    + \frac{| \mathcal{A} \intr[n - 2]|}{{n
    \choose n - 2}}
  .\]
  (\nameref{local_lym}) so
  \[
    \frac{| \mathcal{A}\intr[n]|}{{n \choose n}} +
    \frac{| \mathcal{A}\intr[n - 1]|}{{n \choose n
    - 1}} + \frac{| \mathcal{A}\intr[n - 2]|}{{n
    \choose n - 2}} \le 1
  .\]
  Continue inductively.
\end{proof}

Equality in \nameref{lym_ineq}? Must have had
equality in each use of \nameref{local_lym}. Hence
equality in \nameref{lym_ineq} needs: max $r$ with
$\mathcal{A}\intr \neq \emptyset$ has $\mathcal{A}\intr =
X\rsubs$.

So: equality in \nameref{local_lym} $\iff$
$\mathcal{A} = X\rsubs$ for some $r$.

Hence: equality in \nameref{sperner} if and only
if $\mathcal{A} = X\rsubs{\frac{n}{2}}$ (if $n$
even), and $\mathcal{A} = X\rsubs[\halfn]$ or
$\mathcal{A} = X\rsubs[\left\lceil \frac{n}{2}
\right\rceil]$.
