Is it possible, in ZFC, to find an $\omega\_1$-branching tree $(T,\leq)$ of size and height $\omega\_1$ such that whenever $T$ is partitioned into countably many sets $T=\bigcup\_{n<\omega} T\_n$ one of the sets $T\_n$ contains a subset $S$ which is again an $\omega\_1$-branching tree of size and height $\omega\_1$
(with the tree order on $S$ being just $\leq$ restricted to $S$).



Partitioning $\omega_1$-branching trees of size and height $\omega_1$
The existence of such trees is independent of ZFC. 


On one hand, if $CH$ holds then $\omega\_1^{<\omega\_1}$ (and in fact - any $\sigma$-closed $\omega\_1$-branching tree) cannot be partitioned into $\aleph\_0$ many subsets, each does not contain an $\omega\_1$-branching tree. 


*Proof*:
Assume otherwise and let $T\_n$ be such partition. 


**Claim**: For each $n<\omega$, each $t\in T$ can be extended to a node $s\in T$ such that $s$ has no extension in $T\_n$.


*Proof*: If this failed for some $n$ then $T\_n$ would be $\omega\_1$-branching of height $\omega\_1$. $\blacksquare$


Using the claim we can construct, by recursion on $n$, a strictly increasing sequence of $s\_n$'s such that $s\_n$ has no extension in $T\_n$. Then, by $\sigma$-closedness of $T$, all of them would have a single extension $s\_\omega\in T$. However, $s\_\omega$ would have no extension in any $T\_n$ contradicting the fact that $T=\bigcup T\_n$.$\square$


On the other hand, it is consistent (relative to the existence of an inaccessible) that $2^{\aleph\_0} = \aleph\_2$ and every tree of height and size $\omega\_1$ can be partitioned into countably many subsets, each not containing an $\omega\_1$-branching subtree.


**Definition:** Let $T$ be a tree of height $\omega\_1$, $|T|=\omega\_1$. $T$ is *special* if there is a function $f:T\to \omega$ such that for every $x,y,z\in T$ if $f(x)=f(y)=f(z)$ and $x\leq y,z$ then $y\leq z$ or $z\leq y$.


If $T$ is special, as witnessed by $f$, then $T\_n = f^{-1}(n)$ is the desired partition: for every $x\in T\_n$, $x$ doesn't have any two incompatible successors in $T\_n$, since if $x\leq y,z$ and $y,z\in T\_n$ then $f(y)=f(z)=f(x)=n$ and therefore $y$ and $z$ are comparable. 


**Theorem** (Todorcevic/Baumgartner): The statement "Every tree of size and height $\omega\_1$ is special" is equiconsistent with the existence of an inaccessible cardinal. 


See "Some Combinatorial Properties of Trees" by Todorcevic, and "Aronszajn trees and the independence of the transfer property" by Mitchell for more details.


