# Infinite Ramsey's Theorem

## Contents

## Theorem

Let $k, n \in \N$.

For any set $S$, let $S^{\left({n}\right)}$ denote the set $\left\{ {\left\{ {s_1, \ldots, s_n}\right\}: \text{each } s_i \in S}\right\}$ of cardinality $n$ subsets of $S$.

Let $X$ be an infinite set.

Then:

- for every partition $P$ of $X^{\left({n}\right)}$ into $k$ many components
- there is an infinite subset $Y \subseteq X$

such that:

- each member of $Y^{\left({n}\right)}$ is in the same component of $P$.

## Proof

We will prove the theorem for fixed $k$ by induction on $n$.

### Basis for the Induction

The case $n = 1$ ("Infinite Pigeonhole Principle"):

In this case, $X^{\left({n}\right)}$ is the set of singletons of elements of $X$.

So to ease exposition, we will identify $X^{\left({n}\right)}$ with $X$ and just speak of partitions of $X$.

Let $P$ be a partition of $X$ into $k$ many components $X_1, \dots, X_k$.

We have:

- $X = X_1 \cup \cdots \cup X_k$

If each $X_i$ were finite, then $X$ would be a finite union of finite sets, and hence finite.

Thus, at least one of the $X_i$ must be infinite.

But taking $Y$ to be an infinite $X_i$ gives us the required subset for the theorem.

This is the basis for the induction.

### Induction Hypothesis

The induction hypothesis is that:

For every partition $P$ of $X^{\left({n - 1}\right)}$ into $k$ many components

- there is an infinite subset $Y \subseteq X$

such that:

- each member of $Y^{\left({n - 1}\right)}$ is in the same component of $P$.

It is to be proved that:

For every partition $P$ of $X^{\left({n}\right)}$ into $k$ many components

- there is an infinite subset $Y \subseteq X$

such that:

- each member of $Y^{\left({n}\right)}$ is in the same component of $P$.

### Induction Step

This is the induction step:

Suppose the theorem holds for all $i < n$.

It is to be proved that it holds for $n$.

Since $X$ is infinite, there is an injection of $\N$ into $X$.

All that is needed to be done to prove the theorem is to find an infinite subset of $X$

So, to ease exposition, $\N$ will be viewed as a subset of $X$.

The proof will proceed by finding an infinite subset of $\N$.

Write the components of $P$ as $S_1, \ldots, S_k$.

For each $a \in \N$, let $P_a$ be the partition on $\left({\N \setminus \left\{ {a}\right\} }\right)^{ \left({n - 1}\right)}$ into $k$ components:

- $S_{a, 1}, \dots, S_{a, k}$

defined by assigning:

- $\left\{ {b_1, \ldots, b_{n - 1} }\right\} \in \left({\N \setminus \left\{ {a}\right\} }\right)^{\left({n - 1}\right)}$

to $S_{a, j}$ if and only if:

- $\left\{ {a, b_1, \ldots, b_{n - 1} }\right\} \in \N^{\left({n}\right)}$ is in $S_j$

Note that the theorem holds for $P_a$ by the induction hypothesis.

Now, we inductively define $a_0 < a_1 < \dots$ and infinite sets $X_0 \supseteq X_1 \supseteq \cdots$ such that $a_i$ is the smallest element of $X_i$.

Let $a_0 = 0$, and let $X_0 = \N$.

Once $X_i$ and $a_i$ are defined:

Let $X_{i + 1}$ be an infinite subset of $X_i \setminus \left\{{a_i}\right\}$ satisfying the theorem for the partition $P_{a_i}$ of $n-1$ size subsets.

This exists by the induction hypothesis.

Let $a_{i + 1}$ be the smallest element of $X_{i+1}$.

This exists by the Well-Ordering Principle.

Note that $a_{i + 1} > a_i$ since $X_{i + 1}$ is a subset of $X_i$ not containing the smallest element $a_i$ of $X_i$.

We constructed each $X_{i + 1}$ so that each element of $X_{i + 1}^{\left({n - 1}\right)}$ is in the same component of $P_{a_i}$.

So, we may consider the sequence $\left({k_i}\right)_{i \in \N}$ where each $k_i$ is such that all elements of $X_{i + 1}^{\left({n - 1}\right)}$ are in the component $S_{a_i, k_i}$ of $P_{a_i}$.

Since each $k_i$ is one of $1, \ldots, k$, but there are infinitely many $i \in \N$, this forms a partition of $\N$ to which we can apply the base case of this proof.

This means that there is some $c$ among $1, \ldots, k$, there is an infinite set $I = \left\{ {i \in \N : k_i = c}\right\}$.

Define $Y = \left\{ {a_i: i \in I}\right\}$.

$Y$ is infinite since $I$ is infinite and each $a_i$ is distinct by construction.

We now verify that each element of $Y^{\left({n}\right)}$ is in the same component $S_c$ of $P$.

Let $\left\{ {y_1, \ldots, y_n}\right\} \in Y^{\left({n}\right)}$.

We may take $y_1 < \cdots < y_n$ by relabeling if necessary.

There is some $i \in I$ for which $y_1 = a_i$.

By construction:

- $y_1 = a_i$ is the least element of $X_i$
- each $y_2, \dots, y_n$ is in $X_i$

and:

- each element of $X_i^{\left({n - 1}\right)}$ is in the component $S_{a_i, c}$ of the partition $P_{a_i}$.

But then, by definition of $P_{a_i}$, since $\left\{ {y_2, \ldots, y_n}\right\} \in S_{a_i , c}$, we have:

- $\left\{{y_1, y_2, \ldots, y_n}\right\} = \left\{ {a_i, y_2, \ldots, y_n}\right\} \in S_c$

$\blacksquare$

## Also known as

The case $n = 1$ of this theorem is often referred to as an infinite Pigeonhole Principle, since it essentially says that:

- if an infinite set is partitioned into finitely many components, there are infinitely many elements that are sent to the same component.

## Also defined as

Partition theorems like this one are commonly stated in terms of graph colorings.

Using this language, this theorem could be stated as:

- For any coloring of a graph whose vertices correspond to the size $n$ subsets of an infinite set $X$ by $k$-many colors, there is an infinite subset $Y \subseteq X$ such that each vertex corresponding to a size $n$ subset of $Y$ is colored the same.

## Source of Name

This entry was named for Frank Plumpton Ramsey.