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% 24/10/2024 09AM

\begin{fclemma}[]
  \label{lemma_1_7}
  % Lemma 1 7
  Assuming:
  - $U, V \subset X$
  - $|U| = |V|$
  - $U \cap V = \emptyset$
  - $\max U < \max V$
  - $\mathcal{A} \subset X\rsubs$
  - \phantom{} \\[-2.1\baselineskip] \[
    \forall u \in U ~\exists v \in V \text{ such that
    $\mathcal{A}$ is \glsref[uvc]{$(U-u,
    V-v)$-compressed}}
    \tag{$*$}
    \label{comp_cond_eq}
  \]
  Then:
  $|\shadow \CUV(\mathcal{A})| \le |\shadow
  \mathcal{A}|$.
\end{fclemma}

\begin{proof}
  Let $\mathcal{A}' = \CUV(\mathcal{A})$. For $B
  \in \shadow \mathcal{A}' - \shadow \mathcal{A}$,
  we'll show $U \subset B$, $V \cap B =
  \emptyset$, and $B \cup V - U \in \shadow
  \mathcal{A} - \shadow \mathcal{A}'$. (Then
  done).
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/7368fcd23c50404c.png}
  \end{center}
  Have $B \cup x \in \mathcal{A}'$ for some $x$,
  and $B \cup x \notin \mathcal{A}$, so $U \subset
  B \cup x$, $V \cap (B \cup x) = \emptyset$, and
  $(B \cup x) \cup V - U \in \mathcal{A}$ (by
  definition of $\CUV$).

  \underline{If $x \in U$:} there exists $y \in U$
  such that $\mathcal{A}$ is \glsref[uvc]{$(U - x,
  V - y)$-compressed}, so from $(B \cup x) \cup V
  - U \in \mathcal{A}$ we have $B \cup y \in
  \mathcal{A}$ -- contradicting $B \notin \shadow
  \mathcal{A}$, contradiction.

  Thus $x \notin U$, and so $U \subset B$, $V
  \cap B = \emptyset$. Certainly $B \cup V - U \in
  \shadow \mathcal{A}$ (because $(B \cup x) \cup V
  - U \in \mathcal{A}$), so just need to show that
  $B \cup V - U \notin \shadow \mathcal{A}'$.

  Suppose $B \cup V - U \in \shadow \mathcal{A}'$:
  so $(B \cup V - U) \cup w \in \mathcal{A}'$, for
  some $w$. Also have $(B \cup V - U) \cup w \in
  \mathcal{A}$ (for example, as $V$ contained in
  it).

  \underline{If $w \in U$:} know $\mathcal{A}$ is
  \glsref[uvc]{$(U - w, V - z)$-compressed} for
  some $z \in V$, so $B \cup z \in \mathcal{A}$ --
  contradicting $B \notin \shadow$.

  \underline{If $w \notin U$:} have $V \subset (B
  \cup V - U) \cup w$, $U \cap ((B \cup V - U)
  \cup w) = \emptyset$, so by definition of $\CUV$
  we must have that \emph{both} $(B \cup V - U)
  \cup w$ and $B \cup w \in \mathcal{A}$ --
  contradicting $B \notin \shadow \mathcal{A}$, a
  contradiction.
\end{proof}

\begin{fcthm}[Kruskal-Katona]
  \label{KK}
  % Theorem 1 8
  Assuming:
  - $\mathcal{A} \subset X\rsubs$, $1 \le r \le n$
  - $\mathcal{C}$ is the initial segment of
  \gls{colex} on $X\rsubs$ with $|\mathcal{C}| =
  |\mathcal{A}|$
  Then:
  $|\shadow \mathcal{C}| \le |\shadow
  \mathcal{A}|$.

  In particular: if $|\mathcal{A}| = {k \choose
  r}$, then $|\shadow \mathcal{A}| \ge {k
  \choose r - 1}$.
\end{fcthm}

\begin{proof}
  Let $\Gamma = \{(U, V) : |U| = |V| > 0, U \cap V
  = \emptyset, \max U < \max V\} \cup
  \{(\emptyset, \emptyset)\}$. Define a sequence
  $\mathcal{A}_0, \mathcal{A}_1, \ldots$ of set
  systems in $X\rsubs$ as follows:
  \begin{itemize}
    \item Set $\mathcal{A}_0 = \mathcal{A}$.
    \item Having chosen $\mathcal{A}_0, \ldots,
      \mathcal{A}_k$, if $\mathcal{A}_k$ is
      \gls{uvced} for all $(U, V) \in \Gamma$ then
      stop. Otherwise, choose $U, V \in \Gamma$
      with $|U| = |V| > 0$ minimal such that
      $\mathcal{A}_k$ is not \gls{uvced}.

      Note that $\forall u \in U ~ \exists v \in V$
      such that $(U - u, V - v) \in \Gamma$ (namely,
      use $v = \min V$). So \eqref{comp_cond_eq} is
      satisfied.

      So \cref{lemma_1_7} tells us that $|\shadow
      \CUV(\mathcal{A}_k)| \le |\shadow
      \mathcal{A}_k|$.

      Set $\mathcal{A}_{k + 1} = \CUV(\mathcal{A}_k)$,
      and continue.
  \end{itemize}
  Must terminate, as $\sum_{A \in \mathcal{A}_k}
  \sum_{i \in A} 2^i$ is strictly decreasing. The
  final term $\mathcal{B} = \mathcal{A}_k$
  satisfies $|\mathcal{B}| = |\mathcal{A}|$,
  $|\shadow \mathcal{B}| \le |\shadow
  \mathcal{A}|$ and is \gls{uvced} for all $(U, V)
  \in \Gamma$.

  So $\mathcal{B} = \mathcal{C}$ by
  \cref{lemma_1_6}.
\end{proof}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item Equivalently: if
      \[
        |\mathcal{A}| = {k_r \choose r} + {k_{r - 1}
        \choose r - 1} + \cdots + {k_s \choose s}
      ,\]
      where $k_r > k_{r - 1} > \cdots > k_s$, and
      $s \ge 1$, then
      \[
        |\shadow \mathcal{A}| \ge {k_r \choose r -
        1} + {k_{r - 1} \choose r - 2} + \choose +
        {k_s \choose s - 1}
      .\]
    \item Equality in \nameref{KK}? Can check that
      if $|\mathcal{A}| = {k \choose r}$ and
      $|\shadow| = {k \choose r - 1}$ (i.e.
      equality in a step of the proof of
      \nameref{KK}), then $A = Y\rsubs$, for some
      $Y \subset X$ with $|Y| = k$.
    \item However, not true in general that
      $|\shadow \mathcal{A}| = |\shadow
      \mathcal{C}|$ implies that $\mathcal{A}$ is
      isomorphic to $\mathcal{C}$. (uset systems
      $\mathcal{A}, \mathcal{B}$ are
      \emph{isomorphic} if there exists
      permutation of the ground set $X$ sending
      $\mathcal{A}$ to $\mathcal{B}$).
  \end{enumerate}
\end{remark*}

\glssymboldefn{ushadow}%
For $A \subset X\rsubs$, $0 \le r \le n$, the
\emph{upper shadow} of $\mathcal{A}$ is
\[
  \partial^+ \mathcal{A} = \{A \cup x : A \in
  \mathcal{A}, x \notin A\} \subset X\rsubs[r + 1]
.\]

\begin{corollary}
  % Corollary 1 9
  Let $\mathcal{A} \subset X\rsubs$, where $0 \le
  r \le n$, and let $\mathcal{C}$ be the initial
  segment of \gls{lex} on $X\rsubs$ with
  $|\mathcal{C}| = |\mathcal{A}|$. Then $|\ushadow
  \mathcal{A}| \ge |\ushadow \mathcal{C}|$.
\end{corollary}

\begin{proof}
  From \nameref{KK}, since $A < B$ in \gls{colex}
  if and only if $A^c < B^c$ in \gls{lex} with
  ground-set order reversed.
\end{proof}

Note that the \gls{shad} of an initial segment of
\gls{colex} on $X\rsubs$ is an initial segment of
\gls{colex} on $X\rsubs[r - 1]$ -- as if
$\mathcal{C} = \{A \in X\rsubs : A \le a_1 \ldots
a_r \text{ in \gls{colex}}\}$ then $\shadow
\mathcal{C} = \{B \in X\rsubs[r - 1] : B \le a_2
\ldots a_r \text{ in \gls{colex}}\}$.
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/d90ec4416a3c4d0a.png}
\end{center}
This fact gives:

\begin{corollary}
  % Corollary 1 10
  Let $\mathcal{A} \subset X\rsubs$, and
  $\mathcal{C}$ is the initial segment of
  \gls{colex} on $X\rsubs$ with $|\mathcal{C}| =
  |\mathcal{A}|$. Then $|\shadow^t \mathcal{C}|
  \le |\shadow^t \mathcal{A}|$ for all $1 \le t \le
  r$.
\end{corollary}

\begin{proof}
  If $|\shadow^t \mathcal{C} \le |\shadow^t
  \mathcal{A}|$, then $|\shadow^{t + 1}
  \mathcal{C}| \le |\shadow^{t + 1} \mathcal{A}|$,
  because $\shadow^t \mathcal{C}$ is an initial
  segment of \gls{colex}. Done by induction.
\end{proof}

\begin{note*}
  If $|\mathcal{A}| = {k \choose r}$, then
  $|\shadow^t \mathcal{A}| \ge {k \choose r - t}$.
\end{note*}