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We will prove only the 2 dimensional version of
this theorem, but a more general version works in
$\RR^n$.

\begin{flashcard}[sperners-lemma]
\begin{lemma*}[Sperner's Lemma]
  \label{sperners_lemma}
  \cloze{Take an equilateral triangle formed from
  a collection of smaller triangles by using $n$
  equally spaced lines parallel to each of the
  sides of the big triangle. Now colour all the
  vertices red, green or blue subject to the rule
  that if $XYZ$ are the vertices of the big
  triangle, then all the vertices on $XY$ except
  $X$ are red, all vertices on $YZ$ except $Z$ are
  blue and all vertices on $ZX$ except $X$ are
  green.

  Then at least one small triangle has vertices of
  all $3$ colours.}
\end{lemma*}

\begin{proof}
  \cloze{Let $\Gamma$ be the set of small vertices. Once
  the colouring has been chosen, we will assign each $T
  \in \Gamma$ an integer $\zeta(T)$ according to a
  rule. For vertices of $\alpha, \beta$ of $T$,
  define $\zeta(\alpha\beta)$ to be:
  \[ \zeta(\alpha \beta) = \begin{cases}
    1 & \text{RG, BR, GB} \\
    0 & \text{RR, BB, GG} \\
    -1 & \text{GR, RB, BG}
  \end{cases} \]
  Now define
  \[ \zeta(T) = \zeta(\alpha\beta) +
  \zeta(\beta\gamma) + \zeta(\gamma\alpha) \]
  where $\alpha, \beta, \gamma$ are the vertices
  listed counter clockwise.

  Observe $\zeta(T) = 3$ or $\zeta(T) = -3$ if $T$
  is $3$-coloured, and $\zeta(T) = 0$ otherwise
  (to see this, just consider the cases where we
  have exactly 2 of the same colour, or having 3
  of the same colour). Now look at $\sum_T
  \zeta(T)$. Note that all inner edges cancel, so
  \[ \sum_T \zeta(T) = \sum_{xy} \zeta(xy) \]
  where $xy$ ranges over all sides of the big
  triangle, pointing counter clockwise. This sum
  on the right equals $3$ (exercise). So there
  must be a $3$ coloured small triangle.}
\end{proof}
\end{flashcard}

Now we wish to revisit our cousin of
\nameref{no_retrac_thm}. Suppose we have $A, B, C$
as stated. Cut up the triangle as in
\nameref{sperners_lemma}. Label each vertex
$\bf{x}$ to be R, G or B according to whether it
lies in $A$, $B$ or $C$. If it lies in multiple,
we pick arbitrarily, except for on the boundary,
where we make sure to choose such that the
boundaries are coloured in the way that
\nameref{sperners_lemma} requires.

Now \nameref{sperners_lemma} tells us that there
exist vertices $\bf{a}_n \in A$, $\bf{b}_n \in B$,
$\bf{c}_n \in C$ which form a small triangle. If
we cut up small enough, we can make the pairwise
distances between $\bf{a}_n, \bf{b}_n, \bf{c}_n$
be at most $\frac{1}{n}$. By compactness, there
exists $\bf{a} \in \ol{\triangle}$ and $n(j) \to
\infty$ such that $\bf{a}_{n(j)} \to \bf{a}$.
Automatically, $\bf{b}_{n(j)}, \bf{c}_{n(j)} \to
\bf{a}$. Apply closure to deduce $\bf{a} \in A
\cap B \cap C$.

Now, we can simply follow our chain of
equivalences backwards to deduce
\nameref{brouwer}.

\subsubsection*{* Non-examinable material}

Brouwer came to disbelieve his theorem. Look at
our $3$ colour theorem (the cousin of
\nameref{no_retrac_thm}). It says that there is a
point $\bf{x} \in A \cap B \cap C$, but it gives
no recipe for finding such an $\bf{x}$.

** This is the end of the non-examinable comments.

\begin{flashcard}[mc-eigenval-lemma]
\begin{lemma*}
  If $A = (a_{ij})_{\substack{1 \le i \le 3\\1 \le
  j \le 3}}$ (a $3 \times 3$ matrix). Suppose
  $a_{ij} \ge 0$ and $\sum_{i = 1}^3 a_{ij} = 1$
  for $i = 1, 2, 3$. Then there exists $x_1, x_2,
  x_3 \ge 0$ not all zero such that $\sum_i a_{ij}
  x_j = x_i$ (i.e. $\bf{x}$ is an eigenvector with
  eigenvalue $1$).
\end{lemma*}
  
\begin{proof}
  \cloze{Consider $\ol{\triangle} = \{\bf{x} \in \RR^3
  \st x_1 + x_2 + x_3 = 1, x_j \ge 0\}$. If
  $T\bf{x} = A\bf{x}$, then $y_i = \sum_j a_{ij}
  x_j$. Then $y_i \ge 0$ and
  \[ \sum_i y_i = \sum_i \sum_j a_{ij} x_j =
  \sum_j \sum_i a_{ij} x_i = \sum_j x_j = 1 \]
  So $\bf{y} \in \ol{\triangle}$, and $T$ is a
  continuous map $\ol{\triangle}$ to
  $\ol{\triangle}$, so has a fixed point.}
\end{proof}
\end{flashcard}

\subsubsection*{Nash on economics}

Nash did two things. First is study of non-zero
sum games.

$A$ choose strategy $1$ with
probability $p$, and chooses strategy $2$ with
probability $1 - p$. $B$ chooses strategy $1'$
with probability $q$, $2'$ with probability $1 -
q$. Expected value to $A$ is
\begin{align*}
  A(p, q)
  &= \sum_{ij} a_{ij} p_i q_j \\
  B(p, q)
  &= \sum_{ij} b_{ij} p_i q_j
\end{align*}