Infinite Ramsey theorem for fixed finite arity and colors

For arity one, the color fibers partition $\omega$ into $r$ sets. If all were finite, their finite union would be finite, whereas $\omega$ is infinite. Hence one fiber is infinite and homogeneous by F1. More generally, the same conclusion holds for any finite coloring of an infinite subset of $\omega$.

Assume the assertion at arity $n\ge 1$ and let $c:[\omega {]}^{n+1}\to r$. The induction assertion applies to every infinite subset $S\subseteq \omega$: enumerate it increasingly and pull back the coloring to $[\omega {]}^n$, then push forward an infinite homogeneous set. Fix a well-order of $\mathcal{P}(\omega )$ by F4 and A1, so whenever the induction assertion supplies homogeneous infinite subsets we can take the first one in this well-order. This is the explicit choice use in the construction.

Set $S_0=\omega$. Given infinite $S_i$, set $m_i=\min S_i$ and consider on $[S_i\setminus \{m_i\}{]}^n$ the coloring $u\mapsto c(\{m_i\}\cup u)$. Its argument has size $n+1$ because $m_i\mathrm{\notin }u$. By step 2.1 choose the first infinite homogeneous $S_{i+1}\subseteq S_i\setminus \{m_i\}$, and let $j_i<r$ be its color. The color is unique, since an infinite set has an $n$-element subset. Store $S_i$ and the stage as a state to apply F2. All subsequent $m_k$ for $k>i$ belong to $S_{i+1}$, and $m_{i+1}>m_i$ because $m_i$ was its predecessor set's minimum.

By step 1.1 some color $j<r$ has infinitely many indices $K=\{i:j_i=j\}$. Put $H=\{m_i:i\in K\}$. It is infinite since the $m_i$ increase strictly. Given any $n+1$ members, order their indices $i_0<\cdots <i_n$. The last $n$ nodes lie in $S_{i_0+1}$, so step 3.1 gives $c(\{m_{i_0},\dots ,m_{i_n}\})=j_{i_0}=j$. Thus $H$ is homogeneous. This proves the successor assertion; with step 1.1, F3 proves the theorem for every positive finite arity.