# Infinite Ramsey's Theorem implies Finite Ramsey's Theorem

## Theorem

- $\forall l, n, r \in \N: \exists m \in \N: m \to \left({l}\right)_r^n$

where $\alpha \to \left({\beta}\right)^n_r$ means that:

- for any assignment of $r$-colors to the $n$-subsets of $\alpha$
- there is a particular color $\gamma$ and a subset $X$ of $\alpha$ of size $\beta$ such that all $n$-subsets of $X$ are $\gamma$.

## Proof

Aiming for a contradiction, assume that there is a $l$ such that:

- $\forall m \in \N: m \nrightarrow \left({l}\right)_r^n$

Let $\hat{K_i}$ denote a hypergraph on $i$ vertices where all possible $n$-subsets of the vertices are the hyperedges.

Let $G$ be a hypergraph with vertices $V = \left\{ {v_i: i \in \N}\right\}$.

Let the hyperedges of $G <$ be enumerated by:

- $E = \left\{ {E_i: E_i \subset \N, \left\vert{E_i}\right\vert = n}\right\}$

We construct a (rooted) tree $T$ as follows:

$(1): \quad$ First introduce a root node $r t$.

$(2): \quad$ Each node is allowed to have at most $r<$ children which correspond to the $r$-colors, subject to it satisfying the criteria:

- A child is always labeled by one among the $r$-colors.

- Let the colors be denoted $c_1, c_2, \ldots, c_r$.

$(3): \quad$ 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, if the hyperedges $E_1, E_2, \ldots, E_k$ are colored by the colors used in the path in the same order, then the corresponding subgraph of $G$ does not contain a monochromatic $\hat{K_l}$.

- For example: if the introduction of a child $c_i$ creates the $k$ length path $r t, c_a, c_b, \ldots, c_i$ and the hyperedges $E_1, E_2, \ldots, E_k$ when colored $c_a, c_b, \ldots, c_i$ do not 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.

Thus 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.

So each level is finite.

By König's Tree Lemma there will be an infinite path $P$ in $T$.

$P$ provides a $r$-coloring of $G$ that contains no monochromatic $\hat{K_i}$.

Hence $P$ contains no monochromatic infinite hypergraph.

This contradicts the Infinite Ramsey's Theorem.

The result follows by Proof by Contradiction.

$\blacksquare$