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UPDATE: This is dead wrong. See comments below.

Can you give a combinatorial proof of the infinite Ramsey theorem? That is:

Let $X$ be an infinite well ordered set. Let $G$ be a graph with vertex set $X$. Then there is a subset $Y$ of $X$, with $|Y|=|X|$, such that either every two elements of $Y$ are joined by an edge of $G$, or no two elements of $Y$ are joined by an edge of $G$.

If you can get that, let $G$ be the graph where there is an edge $(u,v)$ if $<_1$ and $<_2$ give the same relation between $u$ and $v$. You want to prove that the first alternative in the alternate Ramsey theorem applies.

Assume, for contradiction, that there is an infinite $Y$ such that, for $u$ and $v$ in $Y$, the inequality $u <_1 v$ implies $u >_2 v$. Then $Y$ is well ordered by $<_1$. Let $z$ be the $<_1$ smallest element of $Y$ such that $\{ y \in Y, \ y <_1 z\}$ is infinite. You should be able to give a combinatorial proof that $<_1$ restricted to $\{ y \in Y, \ y <_1 z\}$ is $\omega$. Then $<_2$ restricted to $\{ y \in Y, \ y <_1 z\}$ is not well ordered, a contradiction.

1 [made Community Wiki]

Can you give a combinatorial proof of the infinite Ramsey theorem? That is:

Let $X$ be an infinite well ordered set. Let $G$ be a graph with vertex set $X$. Then there is a subset $Y$ of $X$, with $|Y|=|X|$, such that either every two elements of $Y$ are joined by an edge of $G$, or no two elements of $Y$ are joined by an edge of $G$.

If you can get that, let $G$ be the graph where there is an edge $(u,v)$ if $<_1$ and $<_2$ give the same relation between $u$ and $v$. You want to prove that the first alternative in the alternate Ramsey theorem applies.

Assume, for contradiction, that there is an infinite $Y$ such that, for $u$ and $v$ in $Y$, the inequality $u <_1 v$ implies $u >_2 v$. Then $Y$ is well ordered by $<_1$. Let $z$ be the $<_1$ smallest element of $Y$ such that $\{ y \in Y, \ y <_1 z\}$ is infinite. You should be able to give a combinatorial proof that $<_1$ restricted to $\{ y \in Y, \ y <_1 z\}$ is $\omega$. Then $<_2$ restricted to $\{ y \in Y, \ y <_1 z\}$ is not well ordered, a contradiction.