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% 07/11/2024 11AM

      Then $\sum_{i \in B_k}
      p^t C_i + \sum_{i \in B_1, \ldots, B_{k' -
      1}} x_i (d)^{-1} C_i \equiv 0 \pmod {p^{t +
      1}}$ ($*$).

      \textbf{Claim:} $\sum_{i \in B_k} C_i \in
      \Span_{i \in B_1 \cup \cdots \cup B_{k - 1}}
      \langle C_i \rangle$.

      We will show that given $U$ such that $U
      \cdot C_i = 0$ got sll $i \in B_1, \ldots,
      B_{k - 1}$. Then
      \[
        u \cdot \left( \sum_{i \in B_k} C_i
        \right) = 0
      \]
      which finishes the proof as this implies
      $\sum_{i \in B_k} C_i \in \Span_{i \in B_1
      \cup \cdots \cup B_{k - 1}} \langle C_i
      \rangle$. Take the inner product of ($*$)
      with $u$. Get
      \[
        p^t \sum_{i \in B_k} u \cdot C_i \equiv 0
        \pmod{p^{t + 1}}
      ,\]
      which is equivalent to $\sum_{i \in B_k} u
      \cdot C_i \equiv 0 \pmod p$. Since this
      happens for infinitely many $p$, we get
      $\sum_{i \in B_k} u \cdot C_i = 0$. \qedhere
  \end{itemize}
\end{proof}

A crucial notion that puts things into perspective
is:

\begin{definition*}[$(m, p, c)$-set]
  \glsnoundefn{mpcset}{$(m, p, c)$-set}{$(m, p,
  c)$-sets}%
  An $(m, p, c)$-set $S \subseteq \Nbb$ ($m$ the
  number of generators, $p$ the range of
  coefficients, $c$ the leading coefficient) with
  $x_1, x_2, \ldots, x_m$ is the set of the
  following form:
  \[
    S = \left\{ \sum_{i = 1}^{m } \lambda_i x_i :
    \exists j, \lambda_j = c \text{ and } \forall
    i < j, \lambda_i = 0, \text{ and } \forall k >
    j, \lambda_k
    \in [-p, p] \right\}
  \]
  \begin{align*}
    c x_1 + \lambda_2 x_2 + \lambda_3 x_3 + \cdots
    + \lambda_m x_m
    & \qquad \lambda_i \in [-p, p] \\
    c x_2 + \lambda_3 x_3 + \cdots
    + \lambda_m x_m
    & \qquad \lambda_i \in [-p, p] \\
    \vdots \qquad
    & \\
    c_m x_m
    & \qquad \lambda_i \in [-p, p]
  \end{align*}
  We call these the \emph{rows} of the $(m, p,
  c)$-set.
\end{definition*}

\begin{remark*}
  An \gls{mpcset} is sort of a progression of
  progressions.
\end{remark*}

\begin{example*}
  \phantom{}
  \begin{enumerate}
    \item \mpc 2p1: generators $x_1, x_2$. Have
      $x_1 - p x_2, x_1 - (p - 1) x_2, \ldots, x_1
      + p x_2$, then $x_2$. This is an \gls{ap} of length $2p
      + 1$ with its common difference.
    \item \mpc 2p3: $3 x_1 - p x_2, 3 x_1 - (p -
      1) x_2, \ldots, 3x_1 + px_2$, then $3x_2$.
      This is an \gls{ap} of length $2p + 1$ with
      $3$ times its common difference, and its
      middle term is divisible by $3$.
  \end{enumerate}
\end{example*}

\begin{fcthm}[]
  \label{thm:2.6}
  % Theorem 2.6
  Assuming:
  - $m, p, c$ in $\Nbb$
  - a finite colouring of $\Nbb$
  Then:
  there exists a monochromatic \gls{mpcset}.
\end{fcthm}

\begin{remark*}
  An \mpc Mpc with $M \ge m$ contains a
  \gls{mpcset}.
\end{remark*}

\begin{proof}
  By the above remark, it is enough to find a
  \mpc{k(m - 1) + 1}pc set such that each row is
  monochromatic.

  Let $n$ be large enough (enough in order
  to apply everything to follow). Let $A_1 = \{c,
  2c, \ldots, c \left\lfloor \frac{n}{c}
  \right\rfloor\}$. By \nameref{VW}, there exists
  a monochromatic \gls{ap} of length $2n_1 + 1$,
  with $n_1$ is large enough.
  \[
    R_1 = \{c x_1 - n_1 d_1, c x_1 - (n_1 - 1)d_1,
    \ldots, c x_1 n_1 d_1\}
  \]
  has colour $k_1$. Let $M = m(k - 1) + 1$.
  Now we restrict attention to
  \[
    B_1 = \left\{d_1, 2 d_1, \ldots, \left\lfloor
    \frac{n}{(M - 1)p} \right\rfloor d_1\right\}
  .\]
  Observe that $cx_1 + \lambda_1 b_1 + \lambda_2
  b_2 + \cdots + \lambda_{M - 1} b_{M - 1}$ where
  $b_i\in B_i$, $\lambda_i \in [-p, p]$ is in
  $R_1$ for any $\lambda_i \in [-p, p]$ and any
  $b_1, \ldots, b_{M - 1}$. Thus has colour
  $k_1$.

  Next: look inside $B_1$:
  \[
    A_2 = \left\{ cd_1, 2cd_1, \ldots,
    \left\lfloor \frac{n_1}{(M - 1)pc}
    \right\rfloor cd_1 \right\}
  .\]
  Apply \nameref{VW} to find an \gls{ap} of length
  $2 n_2 + 1$, of colour $k_2$. Let
  \[
    R_2 = \{cx_2 - n_2 d_2, cx_2 - (n_2 - 1) d_2,
    \ldots, c x_2 + n_2 d_2\}
  ,\]
  of colour $k_2$, and let
  \[
    B_2 = \left\{ d_2, 2d_2, \ldots, \left\lfloor
    \frac{n_2}{(M - 2) p_2}\right\rfloor d_2 \right\}
  .\]
  Note that for any $b_1, \ldots, b_{M - 2}$ and
  $\lambda_1, \ldots, \lambda_{M - 2} \in [-p,
  p]$. Then
  \[
    cx_2 + \lambda_1 b_1 + \cdots + \lambda_{M -
    2} b_{M - 2}
  \]
  is in $R_2$, thus has colour $k_2$.

  Keep on doing this $M$ times. Restrict to $M$
  generators (by setting some $\lambda$ to $0$).
\end{proof}

\begin{remark*}
  For the sake of exams (and also in general):

  Being ``super'' pedantic about
  $\left\lfloor \bullet \right\rfloor$ and
  bounds is not that important.

  The \emph{idea} is important.
\end{remark*}

\begin{theorem}[Finite Sums Theorem]
  \label{fst}
  % Theorem 2.7
  Let $m$ be fixed. Then whenever we finitely
  colour $\Nbb$, there exist $\{x_1, \ldots,
  x_m\}$ such that
  \[
    \left\{ \sum_{i \in X_i} : I \subset
    [m], I \neq \emptyset \right\}
  \]
  is monochromatic.
\end{theorem}

Also known as Folkman's Theorem

\begin{proof}
  The previous theorem implies this: any \mpc m11
  contains a set of the above desired form.
\end{proof}