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\begin{proof}[Proof 2]
  Choose, uniformly at random, a maximal chain
  $\mathcal{C}$ (i.e. $C_0 \subset C_1 \subset C_2
  \subset \cdots \subset C_n$, with $|C_r| = r$
  for all $r$).
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/b75c6878885a4e32.png}
  \end{center}
  For any \gls{rset} $A$, $\Pbb(A \in \mathcal{C}) =
  \frac{1}{{n \choose r}}$ (all \glspl{rset} are
  equally likely).
  So $\Pbb( \mathcal{C} \text{ meets } \mathcal{A}_r)
  = \frac{| \mathcal{A}_r|}{{n \choose r}}$ (as
  events are disjoint) and hence
  \[
    1 \ge \Pbb( \mathcal{C} \text{ meets }
    \mathcal{A}) = \sum_{r = 0}^n \frac{|
    \mathcal{A}_r|}{{n \choose r}}
  . \qedhere \]
\end{proof}

Equivalently: (if you want to lose the intuition
about how this works) then: $\text{\#maximal
chains} = n!$, and $\text{\#through any fixed
\gls{rset}} = r!(n - r)!$, hence
\[
  \sum_r | \mathcal{A}_r| r!(n - r)! \le n!
.\]

\subsection{Shadows}

For $\mathcal{A} \subset X\rsubs$, know $|\shadow
\mathcal{A}| \frac{r}{n - r + 1}$. Equality is
rare -- only for $\mathcal{A} = \emptyset$ or
$X\rsubs$. What happens in between?
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/2092a36e55894d5d.png}
\end{center}
In other words, given $|\mathcal{A}|$, how should
we choose $\mathcal{A} \subset X\rsubs$ to minimise $|\shadow
\mathcal{A}|$?

Believable that if $|\mathcal{A}| = {k \choose r}$
then we sholud take $\mathcal{A} = [k]\rsubs$.

What if ${k \choose r} < |\mathcal{A}| < {k + 1
\choose r}$?

Believable that should take $[k]\rsubs$ plus some
\glspl{rset} in $[k + 1]\rsubs$. For example, for
$\mathcal{A} \subset X\rsubs$ with $|\mathcal{A}|
= {8 \choose 3} + {4 \choose 2}$, take
$\mathcal{A} = [8]\rsubs[3] \cup \{9 \cup B : B
\in [4]\rsubs[2]\}$.

\subsection{Two total orders on $X^{(r)}$}

Let $A$ and $B$ be distinct \glspl{rset}: say $A =
a_1, \ldots, a_r$, $B = b_1, \ldots, b_r$ where
$a_1 < \cdots < a_r$ and $b_1 < \cdots < b_r$.

\glsnoundefn{lex}{lexicographic}{NA}%
Say that $A < B$ in the \emph{lexicographic} (or
\emph{lex}) ordering if for some $j$ we have $a_i
= b_i$ for $i < j$ and $a_j < b_j$.

Slogan: ``Use small elements'' (``dictionary
order'').

\begin{example*}
  \Gls{lex} on $[4]\rsubs[2]$: $12, 13, 14, 23,
  24, 34$.

  \Gls{lex} on $[6]\rsubs[3]$: $123, 124, 125,
  126, 134, 135, 136, 145, 146, 156, 234, 235,
  236, 245, 256, 345$, $346, 356, 456$.
\end{example*}

\glsnoundefn{colex}{colexicographic}{NA}%
Say that $A < B$ in the \emph{colexicographic} (or
colex) ordering if for some $j$ we have $a_i =
b_i$ for all $i > j$ and $a_j < b_j$.

Slogan: ``Avoid large elements'' (note that this
is not quite the same as ``use small elements'',
which is what we had before).

\begin{example*}
  \Gls{colex} on $[4]\rsubs[2]$: $12, 13, 23, 14,
  24, 34$.

  \Gls{colex} on $[6]\rsubs[3]$: $123, 124, 134,
  234, 125, 135, 235, 145, 245, 345, 126, 136,
  236, 146, 246, 346$, $156, 256, 356, 456$.
\end{example*}

Note that, in \gls{colex}, $[n - 1]\rsubs[r]$ is
an initial segment (first $t$ elements, for some
$t$) of $[n]\rsubs[r]$.

This is \emph{false} for lex.

So we could view \gls{colex} as an enumeration of
$\Nbb\rsubs[r]$.

\begin{remark*}
  $A < B$ in \gls{colex} if and only if $A^c <
  B^c$ in ``\gls{lex} with ground set order
  reversed''.
\end{remark*}

\textbf{Aim:} \gls{colex} initial segments are
best for $\shadow$, i.e. if $\mathcal{A} \subset
X\rsubs$ and $\mathcal{C} \subset X\rsubs$ is the
initial segment of \gls{colex} with $|\mathcal{C}|
= |\mathcal{A}|$, then $|\shadow \mathcal{C}| \le
|\shadow \mathcal{A}|$.

In particular, $|\mathcal{A}| = {k \choose r}
\implies |\shadow \mathcal{A}| \ge {k \choose r -
1}$.

\subsection{Compressions}

\textbf{Idea:} try to transform $\mathcal{A}
\subset X\rsubs$ into some $A' \subset X\rsubs$
such that:
\begin{enumerate}[(i)]
  \item $|\mathcal{A}'| = |\mathcal{A}|$.
  \item $|\shadow \mathcal{A}'| \le |\shadow
    \mathcal{A}|$.
  \item $\mathcal{A}'$ looks more like'
    $\mathcal{C}$ than $\mathcal{A}$ did.
\end{enumerate}
Ideally, we'd like a family of such
`compressions': $\mathcal{A} \to \mathcal{A}' \to
\mathcal{A}'' \to \mathcal{A}''' \to \cdots \to
\mathcal{B}$ such that either $\mathcal{B} =
\mathcal{C}$ or $\mathcal{B}$ is so similar to
$\mathcal{C}$ that we can directly check that
$|\shadow \mathcal{B}| \ge |\shadow \mathcal{C}|$.
