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

 * $\forall l ,n, r \in \N: \exists m \in \N: m \rightarrow (l)_r^n$

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 $m\in\N$ 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 $V=\{v_i:i\in\N\}$ and let the hyperedges of $G<$ be enumerated by $E=\{E_i:E_i\subset\N, |E_i|=n\}$. 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 König's Tree 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's Theorem. This contradiction proves the theorem.