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\begin{remark*}
  Proof of \nameref{KK} used \cref{lemma_1_6} and
  \cref{lemma_1_7}, but not \cref{lemma_1_4} or
  \cref{coro_1_5}.
\end{remark*}

\subsection{Intersecting Families}

\glsadjdefn{int}{intersecting}{families}%
Say $\mathcal{A} \subset \powset X$
\emph{intersecting} if $A \cap B \neq \emptyset$
for all $A, B \in \mathcal{A}$.

How large can an \gls{int} family be? Can have
$|\mathcal{A}| = 2^{k - 1}$, by taking
$\mathcal{A} = \{A : 1 \in A\}$.

\begin{fcprop}[]
  % Proposition 1 11
  Assuming:
  - $\mathcal{A} \subset \powset{X}$ be
  \gls{int}
  Then:
  $|\mathcal{A}| \le 2^{k - 1}$.
\end{fcprop}

\begin{proof}
  For any $A \subset X$, at most one of $A, A^c$
  can belong to $\mathcal{A}$.
\end{proof}

\begin{note*}
  Many other extremal examples. For example, for
  $n$ odd take $\{A : |A| > \frac{k}{2}\}$.
\end{note*}

What if $\mathcal{A} \subset X\rsubs$?

If $r > \frac{n}{2}$, take $\mathcal{A} =
X\rsubs$.

If $r = \frac{n}{2}$: just choose one of $A, A^c$
for all $A \in X\rsubs$: gives $|\mathcal{A}| =
\half {n \choose r}$.

So interesting case is $r < \frac{n}{2}$.

Could try $\mathcal{A} = \{A \in X\rsubs : 1 \in
A\}$. Has size ${n - 1 \choose r - 1} =
\frac{r}{n} {n \choose r}$ (while this identity
can be verified by writing out factorials, a more
useful way of observing it is by noting that
$\Pbb(\text{random \gls{rset} contains $1$}) = \frac{r}{n}$).

Could also try $\mathcal{B} = \{A \in X\rsubs :
|\mathcal{A} \cap \{1, 2, 3\}| \ge 2\}$.

\begin{example*}
  $n = 8$, $r = 3$. Then $|\mathcal{A}| = {7
  \choose 2} = 21$ and
  \[
    |\mathcal{B}| = \ub{1}_{|B \cap [3]| = 3} +
    \ub{{3 \choose 2} {5 \choose 1}}_{|B \cap [3]|
    = 2} = 16 < 21
  .\]
\end{example*}

\begin{fcthm}[Erdos-Ko-Rado Theorem]
  \label{ekr}
  % [Erd\H{o}s-Ko-Rado Theorem]
  % Theorem 1 12
  Assuming:
  - $\mathcal{A} \subset X\rsubs$ be \gls{int},
  where $r < \frac{n}{2}$
  Then:
  $|\mathcal{A}| \le {n - 1 \choose r - 1}$.
\end{fcthm}

\begin{proof}[Proof 1 (``Bubble down with
  \nameref{KK}'')]
  Note that $A \cap B \neq \emptyset \iff A \not
  \subset B^c$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/4c6f69975e424654.png}
  \end{center}
  Let $\ol{\mathcal{A}} = \{A^c : A \in
  \mathcal{A}\} \subset X\rsubs[n - r]$. Have
  $\shadow^{n - 2r} \ol{\mathcal{A}}$ and
  $\mathcal{A}$ are \emph{disjoint} families of
  \glspl{rset}.

  Suppose $|\mathcal{A}| > {n - 1 \choose r - 1}$.
  Then $|\ol{\mathcal{A}}| = |\mathcal{A}| > {n -
  1 \choose r - 1} = {n - 1 \choose n - r}$.
  Whence by \nameref{KK} we have $|\shadow^{n -
  2r} \ol{\mathcal{A}}| \ge {n - 1 \choose r}$.

  So $|\mathcal{A}| + |\shadow^{n - 2r}
  \ol{\mathcal{A}}| > {n - 1 \choose r - 1} + {n -
  1 \choose r} = {n \choose r}$, a
  contradiction.
\end{proof}

\begin{remark*}
  Calculation at the end \emph{had} to give the
  right answer, as the $\shadow$ calculations
  would all be exact if $\mathcal{A} = \{A \in
  X\rsubs : 1 \in A\}$.
\end{remark*}

\begin{proof}[Proof 2]
  Pick a \emph{cyclic ordering} of $[n]$ i.e. a
  bijection $c : [n] \to \Zbb_n$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/9560e229c2aa48be.png}
  \end{center}

  How many sets in
  $\mathcal{A}$ are intervals ($r$ consecutive
  elements) in this ordering?

  Answer: $\le r$. Because say $C_1, \ldots, C_r
  \in \mathcal{A}$. Then for each $2 \le i \le 1$,
  at most one of the two intervals $C_i C_{i + 1}
  \ldots C_{i + r - 1}$ and $C_{i - r} C_{i - r +
  1} \ldots C_{i - 1}$ can belong to $\mathcal{A}$
  (subscrpits are modulo $n$).
  
  For each \gls{rset} $A$, in how many of the $n!$
  cyclic orderings is it an interval?

  Answer: $n r! (n - r)!$ ($n =$ where, $r! =$ order inside
  $A$, $(n - r)! =$ order outside $A$).

  Hence $~\mathcal{A}| nr!(n - r)! \le n!r$, i.e.
  $|\mathcal{A}| \le \frac{n!r}{n r!(n - r)!} = {n
  - 1 \choose r - 1}$.
\end{proof}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item Numbers had to work out, given that we
      get equality $\mathcal{A} = \{A \in X\rsubs
      : 1 \in A\}$.
    \item Equivalently, we are double-counting the
      edges in the bipartite graph, with vertex
      classes $\mathcal{A}$ and all cycling
      orderings, with $A$ joined to $c$ if $A$ is
      an interval in $c$.
    \item This method is caled \emph{averaging} or
      \emph{Katona's method}.
    \item Equality in \nameref{ekr}? Our example
      is actually unique -- if $\mathcal{A}
      \subset X\rsubs$ is \gls{int} and
      $|\mathcal{A}| = {n - 1 \choose r - 1}$,
      then $\mathcal{A} = \{A \in X\rsubs : i \in
      A\}$ for some $1 \le i \le n$.

      Can get this from Proof 1 (and equality in
      \nameref{KK}) or from Proof 2 (with a bit of
      care).
  \end{enumerate}
\end{remark*}
