%! TEX root = EMC.tex
% vim: tw=50
% 18/02/2025 09AM

  \vspace{-2em}
  where $\h(p) = p \log \frac{1}{p} + (1 - p) \log
  \frac{1}{1 - p}$ and $s = \cent{X_{k + 1}}{X_1,
  \ldots, X_k}$.

  It turns out that this is maximised when $p =
  \frac{2^s}{2^s + 1}$. Then we get
  \[
    \frac{2^s}{2^s + 1} (\log(2^s + 1) - \log 2^s)
    + \frac{\log(2^s + 1)}{2^s + 1} +
    \frac{s2^s}{2^s + 1}
    = \log(2^s + 1)
  .\]
  This proves the claim.

  Let $r = 2^{\cent{X_d}{X_1, \ldots, X_{d -
  1}}}$. Then
  \begin{align*}
    \ent X
    &= \ent{X_1} + \cdots + \cent{X_d}{X_1,
    \ldots, X_{d - 1}} \\
    &\ge \log r + \log(r + 1) + \cdots + \log(r +
    d - 1) \\
    &= \log \left( \frac{(r + d - 1)!}{(r - 1)!}
    \right) \\
    &= \log \left( d! {r + d - 1 \choose d} \right)
  \end{align*}
  Since $\ent X = \log \left( d! {t \choose d}
  \right)$, it follows that
  \[
    r + d - 1 \le t, \qquad r \le t + 1 - d
  .\]
  It follows that
  \begin{align*}
    \ent{X_1, \ldots, X_{d - 1}}
    &= \log \left( d! {t \choose d} \right) - \log
    r \\
    &\ge \log \left( d! \frac{t!}{d!(t - d)!(t + 1
    - d)} \right) \\
    &= \log \left( (d - 1)! {t \choose d - 1} \right)
    \qedhere
  \end{align*}
\end{proof}

\newpage
\section{The union-closed conjecture}

\begin{fcdefnstar}[Union-closed]
  \glsadjdefn{uclosed}{union-closed}{family}%
  Let $\mathcal{A}$ be a (finite) family of sets.
  Say that $\mathcal{A}$ is \emph{union closed} if
  for any $A, B \in \mathcal{A}$, we have $A \cup B
  \in \mathcal{A}$.
\end{fcdefnstar}

\begin{fcconjstar}[]
  If $\mathcal{A}$ is a non-empty \gls{uclosed}
  family, then there exists $x$ that belongs to at
  least $\half |\mathcal{A}|$ sets in
  $\mathcal{A}$.
\end{fcconjstar}

\glsnoundefn{gilmer}{Justin Gilmer's Theorem}{}%
\begin{fcthmstar}[Justin Gilmer]
  % \label{thm:5.1}
  There exists $c > 0$ such that if $\mathcal{A}$
  is a \gls{uclosed} family, then there exists $x$
  that belongs to at least $c|\mathcal{A}|$ of the
  sets in $\mathcal{A}$.
\end{fcthmstar}

Justin Gilmer's constant was about
$\frac{1}{100}$.

His method has a ``natural barrier'' of $\frac{3 -
\sqrt{5}}{2}$.

We will briefly and ``informally'' discuss this.

A reason for this is that if we weaken the
property \gls{uclosed} to ``almost union-closed''
(if we pick two elements randomly, then with high
probability the union is in the family), then
$\frac{3 - \sqrt{5}}{2}$ is the right bound.

Let $\mathcal{A} = [n]^{(pn)} \cup [n]^{(\ge (2p -
p^2 - o(1))n)}$. With high probability, if $A, B$
are random elements of $[n]^{(pn)}$, then $|A \cup
B| \ge (2 p -p^2 - o(1))n$.

If $1 - (2p - p^2 - o(1)) = p$ then almost all of
$\mathcal{A}$ is $[n]^{(pn)}$.
\begin{center}
  \includegraphics[width=0.3\linewidth]{images/3dc18982fa724acf.png}
\end{center}
One of the roots of the quadratic $1 - 3p + p^2 =
0$ is $p = \frac{3 - \sqrt{5}}{2}$.

If we want to prove \gls{gilmer}, it is natural
to let $A, B$ be independent uniformly random
elements of $\mathcal{A}$ and to consider $\ent{A
\cup B}$. Since $\mathcal{A}$ is \gls{uclosed}, $A
\cup B \in \mathcal{A}$, so $\ent{A \cup B} \le
\log|\mathcal{A}|$. Now we would like to get a
lower bound for $\ent{A \cup B}$ assuming that no
$x$ belongs to more than $p|\mathcal{A}|$ sets in
$\mathcal{A}$.
\[
  \h(xy)
  \ge c(x\h(y) + y\h(x)),
  \qquad
  \h(x^2)
  \ge 2cx\h(x)
.\]
