TITLE: Noncombinatorial proofs of Ramsey's Theorem?
QUESTION [13 upvotes]: I know of 2(.5) proofs of Ramsey's theorem, which states (in its simplest form) that for all $k, l\in \mathbb{N}$ there exists an integer $R(k, l)$ with the following property: for any $n>R(k, l)$, any $2$-coloring of the edges of $K_n$ contains either a red $K_k$ or a blue $K_l$.
Both the finite and the infinite versions (the latter being--a 2-coloring of the edges of $K_\mathbb{N}$ contains an infinite monochrome $K_\mathbb{N}$) are proven on Wikipedia, and one may deduce the finite version from the infinite one by compactness, or equivalently using Konig's lemma.  The infinitary proof does not give effective bounds on $R(k, l)$, but can be converted to one that does as follows (this is the .5 proof): 
Consider a $2$-coloring of the edges of a complete graph on $N=2^{k+l}$ vertices, $v_1, ..., v_N$.  Let $V_0$ be the set of all vertices, and let $V_i$ be the largest subset of $V_{i-1}$ connected to $v_i$ by edges of a single color, $c_i$.  After $k+l$ steps, at least $k$ of the $c_i$ are read or $l$ of the $c_i$ are blue by pigeonhole; let the set of indices for which this happens be denoted $S$.  Then $(v_i)_{i\in S}$ is the desired subgraph.
My question is:

Does anyone have a fundamentally different proof of this theorem?  In particular, I am curious to know if there are any of a less combinatorial flavor.

REPLY [14 votes]: One can prove Ramsey's theorem by using a minimal ultrafilter on the infinite permutation group to make a graph monochromatic, see
http://terrytao.wordpress.com/2008/01/21/254a-lecture-5-other-topological-recurrence-results/
(This proof was first discovered by Neil Hindman.)  But this proof is somewhat idiosyncratic and might not be to most people's taste.  (Also, on some level it is equivalent to the usual iterated pigeonhole proof, though heavily disguised through several applications of the axiom of infinity and axiom of choice.)  It is also closely related to the proof that Pietro mentioned.