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% Example Class: Friday, 1:30PM, MR3
% Hand in Q4, Q8 by 5pm Wednesday
% Also hand in last part of last question -- "And
% what if $\mathcal{A} \subset \Nbb^{(r)}$?"

\begin{remark*}
  Very few isoperimetric inequalities are known
  (even approximately).

  For example, ``isoperimetric in a layer'' -- in
  the graph $X\rsubs$, with $x, y$ joined if $|x
  \cap y| = r - 1$ (i.e. $d(x, y) = 2$ in $\Q_n$).
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/91ee5b97fec1440b.png}
  \end{center}
  This is open. Nicest special
  case is $r = \frac{n}{2}$, where it is
  conjectured that balls are best -- i.e. sets of
  the form $\{x \in [r]\rsubs : |x \cap [r]| \ge
  t\}$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/e9660efb73684aa6.png}
  \end{center}
\end{remark*}

\newpage
\section{Intersecting Families}

\subsection{$t$-intersecting families}

\glsadjdefn{tint}{$t$-intersecting}{subset}%
$A \subset \powset X$ is called
\emph{$t$-intersecting} if $|x \cap y| \ge t$ for
all $x, y \in A$.

How large can a $t$-intersecting family be?

\begin{example*}
  $t = 2$. Could take $\{x : 1, 2 \in x\}$ -- has
  size $\quarter 2^n$. Or $\{x : |x| \ge
  \frac{n}{2}  + 1\}$ -- has size $\sim \half
  2^n$.
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/c8d2064ea95a4bdf.png}
  \end{center}
\end{example*}

\begin{fcthm}[Katona's $t$-intersecting Theorem]
  \label{thm_3_1}
  % Theorem 3 1
  Assuming:
  - $A \subset \powset X$ is \gls{tint}
  - $n + t$ even (to make the proof simpler --
  same proof works for odd)
  Then:
  $|A| \le |X^{\left( \ge \frac{n + t}{2}
  \right)}|$.
\end{fcthm}

\begin{proof}
  For any $x, y \in A$: have $|x \cap y| \ge t$,
  so $d(x, y^c) \ge t$. So, writing $\ol{A}$ for
  $\{y^c : y \in A\}$, have $d(A, \ol{A}) \ge t$
  -- i.e. $A_{\ineigh{t - 1}}$ disjoint from
  $\ol{A}$. Suppose that $|A| > \left|X^{\left( \ge
  \frac{n + t}{2} \right)}\right|$.

  Then, by \nameref{harper}, we have
  \[
    |A_{\ineigh{t - 1}}| \ge \left|X^{\left( \ge
    \frac{n + t}{2} - (t - 1) \right)}\right|
    = \left|X^{\left( \ge \frac{n - t}{2} + 1
    \right)}\right|
  .\]
  But $A_{\ineigh{t - 1}}$ disjoint from $\ol{A}$,
  which has size $> \left|X^{\left( \le \frac{n - t}{2}
  \right)}\right|$ contradicting $|A_{\ineigh{t - 1}}| +
  |\ol{A}| \le 2^n$.
\end{proof}

What about \gls{tint} $A \subset
X\rsubs$?

Might guess: best is $A_0 = \{x \in X\rsubs : [t]
\subset x\}$.

Could also try $A_\alpha = \{x \in X\rsubs : |x
\cap [t + 2\alpha]| \ge t + \alpha\}$, for $\alpha
= 1, 2, \ldots, r - t$.

\begin{example*}
  For \glsref[tint]{$2$-intersecting} in:
  \begin{itemize}
    \item $[7]\rsubs[4]$: $|A_0| = {5 \choose 2} =
      10$, $|A_1| = 1 + {4 \choose 3} {3 \choose
      1} = 13$, $|A_2| = {6 \choose 4} = 15$.
    \item $[8]\rsubs[4]$: $|A_0| = {6 \choose 2} =
      15$, $|A_1| = 1 + {4 \choose 3} {4 \choose
      1} = 17$, $|A_2| = {6 \choose 4} = 15$.
    \item $[9]\rsubs[4]$: $|A_0| = {7 \choose 2} =
      21$, $|A_1| = 1 + {4 \choose 3} {5 \choose
      1} = 21$, $|A_2| = {6 \choose 4} = 15$.
  \end{itemize}
  Note that $|A_0|$ grows quadratically, $|A_1|$
  linearly, and $|A_2|$ constant -- so $|A_0|$
  largest of these for $n$ large.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/5f6d2464e1604ae4.png}
  \end{center}
\end{example*}

\begin{fcthm}[]
  \label{thm_3_2}
  % Theorem 3 2
  Assuming:
  - $A \subset X\rsubs$ is \gls{tint}
  Then:
  for $n$ sufficiently large, we have $|A| \le
  |A_0| = {n - t \choose r - t}$.
\end{fcthm}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item Bound we get on $n$ would be $(16r)^r$
      (crude) or $2tr^3$ (careful).
    \item Often called the ``second Erd\H{o}s-Ko-Rado
      Theorem''.
  \end{enumerate}
\end{remark*}

\textbf{Idea of proof:} ``$A_0$ has $r - t$
degrees of freedom''.

\begin{proof}
  Extending $A$ to a \emph{maximal} \gls{tint}
  family, we must have some $x, y \in A$ with $|x
  \cap y| = t$ (if not, then by maximality have
  that $\forall x \in A$, $\forall i \in x$,
  $\forall j \notin x$, have $x \cup j - i \in A$
  -- whence $A = X\rsubs$, contradiction).

  May assume that there exists $z \in A$ with $x
  \cap y \not\subset z$ -- otherwise all $z \in A$
  have $x \cap y \subset z$. Whence $|A| \le {n -
  t \choose r - t} = |A_0|$.
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/9db16e0778bc4e0f.png}
  \end{center}
  So each $w \in A$ must meet $x \cup y \cup z$ in
  $\ge t + 1$ points. Thus
  \[
    |A| \le \ub{2^{3r}}_{\text{\scriptsize $w$ on
    $x \cup y \cup z$}} \left( \ub{{n \choose r - t -
    1} + {n \choose r - t - 2} + \cdots + {n
    \choose 0}}_{\text{$w$ off $x \cup y \cup z$}} \right)
  .\]
  Note that the right hand side is a polynomial of
  degree $r - t - 1$ -- so eventually beaten by
  $|A_0|$.
\end{proof}

\subsection{Modular Intersections}

For intersecting families, we ban $|x \cap y| =
0$.

What if we banned $|x \cap y| \equiv 0
\pmod{\text{something}}$?

\begin{example*}
  Want $A \subset X\rsubs$ with $|x \cap y|$ odd
  for all distinct $x, y \in A$?

  Try $r$ odd: can achieve $|A| = {\left\lfloor
  \frac{n - 1}{2} \right\rfloor \choose \frac{r -
  1}{2}}$, by picture.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/d4658c850a094e87.png}
  \end{center}
  What if, still for $r$ odd, had $|x \cap y|$
  even for all distinct $x, y \in A$? Can achieve
  $n - r + 1$, by picture.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/5acfcb9bdabd447e.png}
  \end{center}
  This is only linear in $n$. Can we improve this?

  Similarly if $r$ even:
  For $|x \cap y|$ even for all $x, y \in A$, can
  achieve $|A| = {\left\lfloor \frac{n}{2}
  \right\rfloor \choose \frac{r}{2}}$ -- picture
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/66dc860d00344298.png}
  \end{center}
  But for $|x \cap y|$ odd for all $x, y \in A$
  (distinct): can achieve $n - r + 1$ (as above).
  Can we improve this?

  Seems to be that banning $|x \cap y| = r \pmod
  2$ forces the family to be \emph{very} small
  (polynomial in $n$, in fact a linear
  polynomial).
\end{example*}
