Bases for Aronszajn Trees

We overview some results by Baumgartner.
Recall that a poset $\mathbf T=(T,{<_T})$ is a $\kappa$-tree iff all of the following hold:

• $|T|=\kappa$;
• for every $x\in T$, $(x_\downarrow,<_T)$ is well-ordered.

For every $x\in T$, denote $\mathop{\mathrm{ht}}(x):=\mathop{\mathrm{otp}}(x_\downarrow,<_T)$.
For every $A\subseteq\kappa$, let $T\restriction A:=\{ x\in T\mid \mathop{\mathrm{ht}}(x)\in A\}$.
Note that, for every $\alpha<\kappa$, $T_\alpha:=\{ x\in T\mid \mathop{\mathrm{ht}}(x)=\alpha\}$ and $T\restriction\alpha$ have size ${<}\kappa$.
A maximal linearly ordered subset of $T$ is called a branch.

A $\kappa$-Aronszajn tree is a $\kappa$-tree with no chains of size $\kappa$.

A $\kappa$-Souslin tree is a $\kappa$-tree with no chains or anti-chains of size $\kappa$.

A $\kappa$-Kurepa tree is a $\kappa$-tree with at least $\kappa^+$ branches.

An $\aleph_1$-tree $T$ is special if there is a function $f:T\rightarrow \omega$ such that whenever $s<_T t$ we have $f(t)\neq f(s)$.

The next well-known lemma shows that Souslin trees are similar to Luzin spaces in the sense that every large subset of a Souslin tree is somewhere dense.

Lemma (folklore). Suppose $\mathbf T=(T,{<_T})$ is a $\kappa$-Souslin tree and $B\subseteq T$ is a subset with $|B|=\kappa$. Then there exists $w\in T$ such that $w^\uparrow\cap B$ is cofinal in $w^\uparrow$.

Definition. Let $(T,<_T)$ be a normal $\omega_1$-tree, a sub-tree of $T$ is a subset $S$ which is a normal $\omega_1$-tree and $S$ is downward closed under $<_T$ (i.e., if $t\in S$ and $s<_T t$ then $s\in S$).

A base for $T$ is a set $\mathcal B$ of sub-trees such that for every sub-tree $S$ of $T$ there is $A\in \mathcal B$ with $A\subseteq S$.

Remark. Let us observe that any $\kappa$-Souslin tree $T$ has a basis of size $\kappa$. Define for every $t\in T$ the set $T_t:=\{s\in T \mid s\leq_T t\text{ or }t\leq_T s\}$, then by the Lemma the set $\{ T_t \mid t\in T \}$ is a base for $T$.

Theorem (Todorčević): Suppose there is an $\aleph_1$-Kurepa tree with at least $\kappa$-branches. Then there is a special $\aleph_1$-Aronszajn tree for which every base has cardinality greater equal to $\kappa$.

Proof. Let $(K,\leq_K)$ be an $\aleph_1$-Kurepa tree with $\kappa$ branches and let $T$ be a special $\aleph_1$-Aronszajn tree. Let $S=K\otimes T :=\{ (k,t) \mid k\in K, t\in T\text{ and }\mathop{\mathrm{ht}}(k)=\mathop{\mathrm{ht}}(t) \}$ with the coordinatewwise ordering, i.e. $(k,t)< (a,b) \iff k<_K a \wedge t<_T b$. Then $S$ is clearly an $\aleph_1$-Aronszajn tree, for $S_\alpha = K_\alpha\times T_\alpha$ for all $\alpha<\omega_1$ and if $B\subseteq S$ is an uncountable branch then $\{t\in T \mid \exists k[(k,t)\in B]\}$ would be an uncountable branch through $T$, which is impossible.

If $f:T\rightarrow \omega$ is a witnesses that $T$ is special then we define the function $g:S\rightarrow \omega$ by $g(k,t)=f(t)$, this clearly witnesses that $S$ is also special.

Finally, let $\langle b_\xi \mid \xi<\kappa \rangle$ be a sequence of disjoint uncountable branches through $K$. Let $A_\xi := \{ (k,t)\in S \mid k\in b_\xi\}$ then it is easily to see that
for every $\xi\neq \eta$ in $\kappa$, we have $b_\xi\cap b_\eta$ is countable, hence $A_\xi \cap A_\eta$ is also countable, and it is clear that $A_\xi$ is a subtree of $S$.

Suppose we have a basis $\mathcal B$ of cardinality less than $\kappa$, but this is impossible since for every $\xi<\kappa$ there must exists some $B\in \mathcal B$ such that $B\subseteq A_\xi$. ◻