Existence of trees with height $\omega$, size $\aleph_1$ and $\aleph_2$ maximal branches

Question

Definition A tree means a set-theoretic tree, that is a poset $(T,<)$ so that for each $x\in T$, the set $\{y\in T\mid y<x\}$ is well-ordered.
Question: I would like to know if it is consistent with ZFC, the existence of a tree with height $\omega$ and each level of it has at most $\aleph_1$ elements, with maximal branches at least $\aleph_2$ but less than $2^{\aleph_0}$ (We assume the continuum hypothesis fails).
We know that if each level of the tree has at most countable elements the maximal branches are countable or $2^{\aleph_0}$, but I haven't seen any results if the levels are $\aleph_1$.

Answer 1

Let $X$ be the poset of all maximal branches in the tree $T$ (for simplicity, assume each element in $X$ belongs to a maximal branch). Then we can endow $X$ with a metric $d$ where if $x,y$ are distinct branches and $n$ is the smallest level where the branches $x,y$ differ, then $d(x,y)=1/n$. The metric space $(X,d)$ is a complete metric space.

Let $\text{dc}(X)$ denote the smallest cardinality of a dense subset of $X$. Then Theorem 8.3 from the Handbook of Set-Theoretic Topology states that $|X|=\text{dc}(X)$ or $|X|=\text{dc}(X)^{\aleph_0}$ whenever $X$ is a complete metric space. Since $\text{dc}(X)=\aleph_1$, we can conclude that $|X|=\aleph_1$ or $|X|=\aleph_1^{\aleph_0}=2^{\aleph_0}.$

Answer 2

Suppose $T$ is a tree of height $\omega$ with $<2^\omega$-many branches. Then it must be the case that for all $t \in T$, there is $s \geq t$ such that all $x \geq s$ are comparable. Otherwise we could split into $2^\omega$-many branches.

Next take the Cantor-Bendixson derivative. Let $T_0 = T$, and for an ordinal $\alpha$, let $T_{\alpha+1} = T_\alpha$ minus the set of nodes above which $T_\alpha$ does not split. At limit $\lambda$, let $T_\lambda = \bigcap_{\alpha<\lambda} T_\alpha$. By the above paragraph, $T_{\alpha+1}$ is strictly smaller whenever $T_\alpha$ is nonempty. Thus the process converges to $T_\alpha = \emptyset$ at some $\alpha^*$, and this $\alpha^*$ must be $<|T|^+$, so $<\omega_2$ under your assumptions.

For every branch $b$ through $T$, let $\alpha_b \leq \alpha^*$ be the least ordinal $\beta$ such that $b$ is not a branch through $T_\beta$. Each $\alpha_b$ must be a successor ordinal. This means that for some node $s \in b$, there is no splitting above $s$ in $T_{\alpha_b-1}$, and $s \notin T_{\alpha_b}$. Thus $b$ is the unique branch through $T_{\alpha_b-1}$ extending $s$. If $s_b$ is the shortest node with this property, then $b \mapsto s_b$ is an injection. Thus the set of all branches has cardinality at most the size of $T$.