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 with , we denote by the set and by the set .

**Definition:** A function is a –**color** if it has the following distance-like properties

- it is symmetric, i.e. for any permutation of the set and for any we have ,
- it is reflexive, i.e. for any in , if and only if .

**Explanation:** a sequence of coloured complete graphs is a function , defined for any (in order to colour the non-oriented edge from to ) and for any (because is a colouring of the complete graph with vertices). We complete the colouring by . In case the color function does not depend on then it’s just a colouring of .

**Fact.** Any color function admits an unique continuous extension to , which is identified with .

**Proof of Ramsey theorem:** for any non-principal ultrafilter , there exists and is unique a colour such that . By continuity,

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

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Well, it means the set of with the equation is in the ultrafilter , which is nonprincipal for some fairly-obvious reason.

Yes, but curry the limit into three separate limits, do it one by one.