\documentclass[11pt]{scrartcl}
\usepackage{evan}

\begin{document}
\title{Twitch 113.2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 113}
\maketitle

\section*{Problem}
Let $f \colon [0, 1] \to [0, 1]$ be monotonically increasing.
Show that there exists $\xi \in [0,1]$ with $f(\xi)=\xi$.

\section*{Video}
\href{https://www.youtube.com/watch?v=mUDShxX8XNk&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/mUDShxX8XNk}}



\newpage

\section*{Solution}
Assume not. Then
\begin{align*}
  A &\coloneqq \left\{ f(x) < x \right\} \ni 1 \\
  B &\coloneqq \left\{ f(x) > x \right\} \ni 0
\end{align*}
This gives a partition $A \sqcup B = [0,1]$.

Let $\xi = \sup B$. We consider two cases:
\begin{itemize}
    \ii If $\xi \in B$, then $f(a) < a$ for all $a > \xi$.
    Hence \[ f(\xi) < f(a) < a \qquad \forall a > \xi \]
    so $f(\xi) \le \xi$, contradiction to $\xi \in B$.

    \ii Otherwise, for all $\eps > 0$,
    the set $(\xi-\eps, \xi) \cap B$ is nonempty.
    So \[ f(\xi) > f\bigg(\text{something in $(\xi-\eps,\xi) \cap B$}\bigg)
      > \xi-\eps \]
    for all $\eps > 0$, ergo $f(\xi) \le \xi$.
    This violates the assumption $\xi \in A$.
\end{itemize}
Since both cases give a contradiction, we're done.

\end{document}
