# Ramsey theorem and ultrafilters (II)

This is a continuation of the post Has Ramsey theorem anything to do with the ultrafilter monad?  In this post I want to write the shortest proof of Ramsey theorem (in finite or infinite version) I can make for the moment. This is probably well-known, please tell me if so, with references, and I shall update the post.

The motivation is the fact that I don’t get why one has to reason by compactness and contradiction, so I am looking for a direct proof (which is of course still non constructive).

Notation: for any $k \in \mathbb{N}$ with $k > 0$,  we denote by $[k]$ the set $\left\{ 1, ... , k \right\}$ and by $\lceil k \rceil$ the set $\left\{0,1, ... , k\right\}$.

Definition: A function $c: \mathbb{N} \times \mathbb{N} \times \mathbb{N} \rightarrow \lceil k \rceil$ is a $k$color if it has the following distance-like properties

• it is symmetric, i.e. for any permutation $(abc)$ of the set $[3]$ and for any $x, y, z \in \mathbb{N}$ we have $c(x_{a}, x_{b}, x_{c}) = c(x_{1}, x_{2}, x_{3})$,
• it is reflexive, i.e. for any $x \leq y \leq z$ in $\mathbb{N}$$c(x,y,z) = 0$ if and only if $x=y$.

Explanation: a sequence of coloured complete graphs $K_{n}$ is a function $c(x,y,n) = c_{n}(x,y)$, defined for any $x (in order to colour the non-oriented edge from $x$ to $y$) and for any $y \leq n$ (because is a colouring of the complete graph with $n$ vertices). We complete the colouring by $c(x,x,n) = 0$.  In case the color function does not depend on $n$ then it’s just a colouring of $K_{\infty}$.

Fact. Any color function admits an unique continuous extension to $U\mathbb{N}^{3}$ , which is identified with $\left(U\mathbb{N}\right)^{3}$.

Proof of Ramsey theorem: for any non-principal ultrafilter $\beta \in U\mathbb{N} \setminus \mathbb{N}$, there exists and is  unique a colour $j \in [k]$ such that $c(\beta, \beta, \beta) = j$. By continuity,

$\lim_{x \rightarrow \beta} \lim_{y \rightarrow \beta} \lim_{n \rightarrow \beta} c(x,y,n) = j$.

Can you see why this is Ramsey theorem? [added: curry and lazy evaluate, so to say, the three limits.]

1. Well, it means the set of $(x,y,n)$ with the equation $c(x,y,n)=j$ is in the ultrafilter $(\beta,\beta,\beta)$, which is nonprincipal for some fairly-obvious reason.