Order-theoretic tree with infinitely repeating colouring

Question

This seems likely to be an instance of general combinatorial theorem. But which one?

Definition 1:

1. Define an order-theoretic tree $T$ to be a poset such that for any node, the set of elements less than that node is well-ordered.
2. The size of $T$ is the cardinality of the set of nodes in $T$.
3. For two nodes $n_1\neq n_2\in T$, if $n_1< n_2$ and $n_1$ is greatest strictly less than $n_2$ (and equivalently, $n_2$ is least greater than $n_1$), then call $n_1$ the parent of $n_2$ (uniqueness is by condition 1 above) and $n_2$ a child of $n_1$.

All trees are order-theoretic henceforth. If we write "order-theoretic tree", this is just for emphasis.

Definition 2: Suppose $\beta$ is a cardinal. Then say a tree $T$ is $\beta$-branching when every node in $T$ has strictly fewer than $\beta$ children.
(For example, an $\omega$-branching tree is what would normally be called finitely branching.)

Definition 3: Suppose $C$ is a set. A $C$-colouring on a tree $T$ is an assignment of an element $c\in C$ to each node $n$ of $T$.

Fix the following data:

• A pair of cardinals $\alpha<\beta$ where $\alpha$ is uncountable. (Note that we are free to take $\beta$ much larger than $\alpha$.)
• A set of colours $C$ of cardinality $\alpha$.
• A $\beta$-branching order-theoretic tree $T$ of size $\beta$.

So intuitively, writing "large" for $\beta$, "small" for "less than $\beta$", and "very small" for $\alpha$: $T$ is "large", local branches are "small", and available colours are "very small".

Lemma 1 (required to prove): There exists some colour $c\in C$ and some node $n\in T$ such that any path $p$ in $T$ starting from $n$ is either shorter than $\beta$ or has length $\beta$ and encounters $\beta$ many nodes of colour $c$.

To sum up intuitively:

If a large order-theoretic tree has small branches and a very small number of colours, then there exists a large subtree and a colour $c$ such that any large path in the subtree has a large number of $c$-coloured nodes.

Example: Consider a large tree such that one path from the root is coloured red, and the rest of the tree is coloured blue. Then take $n$ to be any blue node in $T$.

I wonder if Lemma 1 is a standard result, or is easily proved from standard results?

Answer 1

Lemma 1 is false. Take $\beta$ to be any limit ordinal and colours to be $C=\{0,1\}$. Let nodes of the tree be functions $n:\alpha\to\{0,1\}$ for any $\alpha<\beta$ ($\alpha$ in this answer has nothing to do with the $\alpha$ in the question).

Define the colouring $f$ by: $$ \begin{array}{r@{\ }l@{\qquad}l} f(\alpha)=&c & \text{if }\exists c{\in}C.\exists \alpha'{<}\alpha.\forall \alpha'\leq\alpha''{<}\alpha.n(\alpha'')=c \\ f(\alpha)=&0 & \text{otherwise} \end{array} $$

Unpacking this in plain English:

• The root node $n:0\to\{0,1\}$ is coloured with $0$.
• A non-limit node is coloured with its last element --- so if $n:\alpha{+}1\to\{0,1\}$ then $f(n)=n(\alpha)$.
• A limit node tended to by ones, is coloured with $1$.
• A limit node not tended to by ones, is coloured with $0$.

Then any node sees a branch coloured all with $0$s, and a branch coloured all with $1$s.