We prove the <Hilbert cube theorem> directly by <mathematical induction> on its dimension. Dimension zero is immediate. Suppose every finite coloring contains a monochromatic Hilbert $n$-cube, and let $c:\mathbb N\to[k]$ be a $k$-coloring. Refine it to the finite coloring
$$
C(t)=\bigl(c(t),c(t+1),\ldots,c(t+k)\bigr).
$$
By the induction hypothesis, some Hilbert $n$-cube
$$
H=\left\{a+\sum_{i\in I}x_i:I\subseteq[n]\right\}
$$
is monochromatic for $C$. Consequently, for every $j\in\{0,\ldots,k\}$ the translate $H+j$ is monochromatic for $c$; denote its color by $\gamma_j$. By the <pigeonhole principle>, two of the $k+1$ colors $\gamma_0,\ldots,\gamma_k$ agree, say $\gamma_s=\gamma_t$ with $s<t$. Then
$$
(H+s)\cup(H+t)
=
\left\{a+s+\sum_{i\in I}x_i+\epsilon(t-s):I\subseteq[n],\ \epsilon\in\{0,1\}\right\}
$$
is a monochromatic Hilbert $(n+1)$-cube. This proves the result without using the <Finite sums theorem> or the <Van der Waerden theorem>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 130 / 1 / iii / Solution

{c}

For every dimension $m$, every <finite coloring> of the <positive integer>[positive integers] contains $x_1,\ldots,x_m$ for which all nonempty sums of distinct $x_i$ have one color. This is the case $p=c=1$ of the <monochromatic m-p-c set theorem>.

= Folkman theorem
{c}
{synonym}

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 Incoming links: Finite sums theorem