Infinite Ramsey's Theorem implies Finite Ramsey's Theorem

$$ \alpha\rightarrow (\beta)^n_r$$ means that for any assignment of $$ r$$-colors to the $$ n$$-subsets of $$ \alpha$$, there is a particular color (say red) and a subset $$ X$$ of $$ \alpha$$ of size $$ \beta$$ such that all $$ n$$-subsets of $$ X$$ are red. Theorem: $$\forall l,n,r\in\mathbb{N}, \exists m\in\mathbb{N}$$ such that $$m\rightarrow (l)_r^n$$.

Proof: The proof is identical to the abridged case with cosmetic changes. By way of contradiction assume that there is an $$l$$ such that for all $$m\in\mathbb{N}$$ we have $$m\nrightarrow (l)_r^n$$. We will use the notation $$\hat{K_i}$$ for a hypergraph on $$i$$ vertices where all possible $$n$$-subsets of the vertices are the hyperedges. Also let $$G$$ be a hypergraph with vertices $$V=\{v_i:i\in\mathbb{N}\}$$ and let the hyperedges of $$G$$ be enumerated by $$E=\{E_i:E_i\subset\mathbb{N}, |E_i|=n\}$$. We construct a (rooted) tree $$T$$ as follows:

1. First introduce a root vertex $$rt$$.

2. Each vertex is allowed to have at most $$r$$ children which correspond to the $$r$$-colors, subject to it satisfying the criteria below. A child is always labeled by one among the $$r$$-colors. (Call the colors $$c_1,c_2\cdots c_r$$ for convenience).

3. A child $$c_i$$ is permitted if and only if its introduction creates a path of some finite length $$k$$ starting from the root, so that if the hyperedges $$E_1,E_2\cdots E_k$$ are colored by the colors used in the path in the same order, then the corresponding subgraph in $$G$$ does not contain a monochromatic $$\hat{K_l}$$. For example if the introduction of a child $$c_i$$ creates the $$k$$ length path $$rt,c_a,c_b\cdots ,c_i$$ and the hyperedges $$E_1,E_2,\cdots E_k$$ when colored $$c_a,c_b\cdots c_i$$ don't contain a monochromatic $$\hat{K_l}$$ the child $$c_i$$ is permitted to be added to $$T$$.

Note that for all $$m$$, there always exists a coloring of $$\hat{K_m}$$ such that no monochromatic $$\hat{K_l}$$ exists within. So the situation that a child cannot be added to any vertex at a given level $$k$$ cannot arise. For we can always take a coloring of $$\hat{K_{k+n}}$$ containing no monochromatic $$\hat{K_l}$$. Since any $$k$$ hyperedges in it would yield a sequence of colors already existing in $$T$$, we know which vertex to add the child to. We give the child the color corresponding to any other edge. Hence we can forever keep adding children and so $$T$$ is infinite. It is also obvious that each level $$k$$ of $$T$$ has at most $$r^k$$ vertices and so each level is finite.

Now by Konig's lemma there will be an infinite path in $$T$$. This infinite path provides a $$r$$-coloring of $$G$$ that contains no monochromatic $$\hat{K_i}$$ and hence no monochromatic infinite hypergraph which contradicts the infinite Ramsey theorem. This contradiction proves the theorem.$$\Box$$