Partition Calculus and Ramsey theory question

Question

These topics are outside of my area of research, so I am not quite sure where in the literature to find the answers.

In what follows, if $X$ is partially ordered and $n$ is a natural number, let $[[X]]^n$ denote the set of $S\subset X$ such that $|S|=n$ and $S$ is linearly ordered. A tree will be a partially ordered set $X$ such that for each $x\in X$, $$X|x:=\{y\in X: y\leqslant x\}$$ is a well-ordered set. The height of the tree $X$ is the supremum of the order types of $X|x$ as $x$ ranges through $X$. We define $d(X)$ to be the tree obtained by removing from $X$ all of its maximal members. We then define $$d^0(X)=X,$$ $$d^{\xi+1}(X)=d(d^\xi(X)),$$ and if $\xi$ is a limit ordinal, $$d^\xi(X)=\bigcap_{\zeta<\xi}d^\zeta(X).$$ We then define the rank of $X$ to be the minimum $\xi$ (assuming one exists) such that $d^\xi(X)=\emptyset$.

For the following questions, I suspect the answers are known, and if so, I would a reference. If the answers are not known, I would like to know what is the best partial result in this direction and some of the most relevant related results in the literature.

Question $1$: Given an ordinal $\xi$ and a tree $X$ with height $\omega^\xi$, is it true that either $(a)$ there exists a linearly ordered subset $B$ of $X$ such that the height of $B$ is $\omega^\xi$, or $(b)$ there exists a collection $(X_i)_{i\in I}$ of incomparable subsets of $X$ such that $\sup_{i\in I}\text{height}(X_i)=\omega^\xi$ EDIT and each $X_i$ is linearly ordered.

Question $2$: For a fixed natural number $n$, for which ordinals $\xi$ is the following true? If $X$ is a tree with height $\xi$ and if $[[X]]^n$ is colored with finitely many colors, there exists a subtree $Y$ of $X$ with height $\xi$ such that $[[Y]]^n$ is monochromatic.

Question $3$: Same as Question $2$, but with height replaced by rank.

Answer 1

Question 1 is true for countable $\omega^\xi$ but it is at least consistent that this is false at $\omega_1.$ There might be a ZFC counterexample but the first thing that popped to mind is a Suslin tree. It is consistent with ZFC that there is a Suslin tree but it is also consistent that there aren't any.

To see that a Suslin tree $X$ is a counterexample, recall that a Suslin tree is a tree of height $\omega_1$ where every linearly ordered set is countable and every antichain is countable. Suppose we have a collection $(X_i)_{i \in I}$ of incomparable nonempty linearly ordered subsets of $X$. Let $x_i$ be the minimal element of $X_i$. This set $\{x_i \mid i \in I\}$ is an antichain, which must therefore be countable. Since each $X_i$ is countable, the supremum of the heights of the $X_i$ must be countable as well since $\omega_1$ is a regular cardinal.

Suslin trees generalize to higher levels and provide similar counterexamples higher up. It's harder to get rid of higher Suslin trees than it is to get rid of those with height $\omega_1$ but these are still conditional counterexamples since we can't prove the existence of such trees in ZFC.

For Question 2, if $n \geq 2$, it is necessary that $\xi$ is either $\omega$ or a weakly compact cardinal otherwise the result is false when $X$ is linearly ordered. The result is true for $\xi=\omega$ because of Ramsey's Theorem. There are two cases:

First, the easy case, if $X$ has a branch of height $\omega$ then apply the infinite form of Ramsey's Theorem to that branch.

Otherwise, by Kőnig's Lemma, the tree must have an infinite level. For simplicity, let's chop the base of the tree so that we can assume that the tree (or rather forest) has infinitely many roots. This gives a partition of $X$ into infinitely many rooted trees $(X_i)_{i \in I}$. Since $X$ has height $\omega$, the heights of these trees are unbounded and they are all finite. From a selection of these trees pick branches $Y_1,Y_2,\ldots$ so that the finite form of Ramsey's Theorem ensures that if $[Y_m]^n$ is $k$-colored then there is a homogeneous set $Z_m$ of size $m$ (where $n$ and $k$ are given by your setup). Apply Ramsey's Theorem to each $Y_m$ using your coloring and assemble the subchains $Z_m$ into a homogeneous tree of height $\omega$.

The same trick applies when $\xi$ is weakly compact. The basic idea is that a weakly compact cardinal satisfies the analogue of the infinite form of Ramsey's Theorem and it also has an analogue of Kőnig's Lemma. As a replacement for the finite form of Ramsey's Theorem, one needs to use the Erdős-Rado Theorem along with the observation that weakly compact cardinals are inaccessible.