Let $S\subseteq \mathop{\mathrm{acc}}(\omega_1)$, fix a ladder system $\langle C_\alpha \mid \alpha\in S \rangle$ such that for every $\alpha \in S$, the ladder $C_\alpha$ is cofinal in $\alpha$ with order-type $\omega$. We will abuse the notation and let $C_\alpha(n)$ for $n<\omega$ denote the unique $\beta\in C_\alpha$ such that $\mathop{\mathrm{otp}}(C_\alpha\cap \beta )=n$. Consider the the set $L_S:=\{C_\alpha \mid \alpha\in S\}$ ordered by the lexicographic order, we will call such a linear order a Baumgartner type. The set $S$ will be called the support.

We begin with the following usefull Lemma about Baumgartner types.

Lemma 1. For every stationary subset $T\subseteq \omega_1$, there are two disjoint stationary subsets $T^-,T^+\subseteq T$ such that for all $f^-\in L_{T^-}$ and $f^+\in L_{T^+}$, $f^-<_{\mathop{\mathrm{lex}}} f^+$.

Proof.
We construct by recursion on $n<\omega$ a decreasing sequence $\langle T_n \mid n<\omega \rangle$ of stationary subsets of $T$. At the base stage let $T_0:=T$. In the $n$-step, using Fodor’s Lemma, define $T_{n+1}$ to be a stationary subset of $T_n$ such that $C_{\alpha}(n)$ is constant for all $\alpha\in T_{n+1}$ with minimal value $c_{n}$. We define the following sets:

$$T_{n+1}:=\{ \alpha\in T \mid \forall m\leq n [C_\alpha(m) = c_m]\}$$ $$D_{n}:=\{\beta \in T \mid \exists m\leq n \forall k<m[C_{\beta}(k)=c_k \wedge C_\alpha(m)< c_m]\}$$ $$U_n:=\{ \alpha\in T \mid \exists m\leq n \forall k<m[C_{\beta}(k)=c_k \wedge C_\alpha(m)> c_m]] \}$$

Notice that if the set $T^+_{n+1}:=\{\alpha\in T_n \mid C_{\alpha}(n)> c_{n}\}$ is stationary, then we are done as for all $f^-\in L_{T_{n+1}}$ and $f^+\in L_{T^+_{n+1}}$, we have $f^-<_{\mathop{\mathrm{lex}}} f^+$.
By the minimality of $c_n$, we get that $T^-_{n+1}:=\{\alpha\in T_n \mid C_{\alpha}(n)< c_{n}\}$ is non-stationary. Clearly $T=D_n\cup U_n \cup T_{n+1}$, $U_n= \bigcup_{m\leq n} T^+_{m+1}$ and $D_n = \bigcup_{m\leq n} T^-_{m+1}$.

Suppose the recursion haven’t stopped along the way. Clearly $\bigcap_{n<\omega} T_{n}$ has at most one element. But this is absurd as $T=(\bigcup_{n<\omega} (D_n \cup U_n)) \cup (\bigcap_{n<\omega} T_{n})$ is non-stationary as it is a countable union of non-stationary sets. $\square$

Recall that a linear order is said to be scattered if it does not contain a copy of the rationals. A linear order is called $\sigma$-scattered if it is a countable union of scattered linear orders, similarly a linear order is called $\sigma$-well ordered if it is a countable union of well-orders.

Theorem 2 (Baumgartner, Therorem 3.7(i)). Suppose $S\subseteq \mathop{\mathrm{acc}}(\omega_1)$.
1) If $S$ is stationary, then $L_S$ is not $\sigma$-scattered.
2) If $S$ is not stationary, then $L_S$ is $\sigma$-well ordered.

It is easy to check that a well ordered set is scattered, so we get the following Corollary:

Corollary 3. The linear order $L_S$ is $\sigma$-scattered if and only if $S\subseteq \mathop{\mathrm{acc}}(\omega_1)$ is not a stationary subset.

Let us now prove the theorem.

Proof.

1) Suppose $S\subseteq \mathop{\mathrm{acc}}(\omega_1)$ is a stationary set, we will show that $L_S$ is not $\sigma$-scattered, i.e. $L_S$ cannot be represented as a countable union of scattered linear orders.
Suppose we are given a countable partition of $S$, $\langle S_i \mid i<\omega \rangle$. As $S$ is stationary, we can find some $i<\omega$ such that $S_i$ is also stationary.

We claim that it is possible to embed the rationals, i.e. $({}^{<\omega}2, \leq_{\mathop{\mathrm{lex}}})$ into $L_{S_i}$.
We construct an order-preserving function $f:{}^{<\omega}2\rightarrow L_{S_i}$ by recursion on the length of $s\in {}^{<\omega}2$. At each step of the recursion we will find for a given stationary set $T^s$ two stationary sub-sets $T^{s^\frown 0}, T^{s^\frown 1}$.

$\blacktriangleright$ Base stage $s=\emptyset$: As $S_i$ is stationary, using the Lemma we can find a stationary subsets $R\subset S_i$ and $\alpha\in S\setminus R$ such that $C_\alpha <_{\mathop{\mathrm{lex}}} C_\beta$ for all $\beta\in R$.
Let $f(s)=C_\alpha$ and $T^{s}:= R$.

$\blacktriangleright$ Stage $n$: Suppose $s\in {}^{n}2$. As $T^s$ is stationary, we may use the Lemma twice to find two disjoint stationary subsets $T^{s^\frown 0},T^{s^\frown 1}\subseteq T^s$ and $\alpha_0, \alpha_1 \in T^s\setminus (T^{s^\frown i}\cup T^{s^\frown 1})$ such that for all $f^-\in L_{T^{s^\frown 0}}$ and $f^+\in L_{T^{s^\frown 1}}$:

$$C_{\alpha_0}<_{\mathop{\mathrm{lex}}} f^-<_{\mathop{\mathrm{lex}}} C_{\alpha_1} <_{\mathop{\mathrm{lex}}} f^+$$

For each $i<2$ let $f(s^\frown i)=C_{\alpha_i}$. Clearly $f$ has the required properties.

2) Suppose $S$ is non-stationary, we will show that $L_S$ is $\sigma$-well ordered. Fix a club $C\subseteq \omega_1$ such that $S\cap C =\emptyset$. We fix an enumeration $\langle c_\beta \mid \beta<\omega_1 \rangle$ of $C$, for convenience we assume $c_0=0$.

For $\alpha\in S$, we let $K(\alpha):=\{ \beta<\omega_1 \mid \exists \gamma\in C_\alpha [ c_\beta\leq \gamma < c_{\beta+1}] \}$, clearly $K(\alpha)$ is finite.

For each $\beta<\omega_1$, consider the set $S_\beta:=\{ C_\alpha \cap \beta \mid \alpha \in S \}$. Notice that $S_\beta\subseteq [\beta]^{<\omega} \cup \{ C_\alpha \mid \alpha \in S\cap \beta\}$ which is countable and fix a witnessing bijection $f_\beta :S_\beta\rightarrow \omega$.

We define a function $G:S\rightarrow {}^{<\omega} \omega$ as follows: Fix $\alpha\in S$. Suppose $n= |K(\alpha)|$ and $K(\alpha)=\{\beta_1,\dots, \beta_n\}$. For each $1\leq i\leq n$, let $m_i:=f_{\beta_{i+1}}(C_\alpha\cap \beta_{i+1})$. Define $G(\alpha):=(n,m_1,\dots, m_n)$.

As ${}^{<\omega} \omega$ is countable, the following claim will give us that $L_S$ is $\sigma$-well ordered.

Claim. For every $t\in {}^{<\omega} \omega$, the set $\{C_\alpha \mid G(\alpha)=t\}$ is well-ordered.

Subproof.
Fix $t=(n,m_1,\dots, m_n)$ and a $<_{\mathop{\mathrm{lex}}}$-decreasing sequence $\{C_{\alpha_k}\mid k<\omega \}$ such that for every $k<\omega$ we have $f(\alpha_k )=t$. We will show that the sequence must become constant from some point.

For each $k<\omega$, let $K(\alpha_k) :=(\beta^k_1,\dots,\beta^k_n)$. We use finite induction on $1\leq i\leq n$ to show that we can find some $k_i<\omega$ large enough such that for all $m>k_i$, we have $\beta^m_i = \beta^{k_i}_i$ and moreover $C_{\alpha_m} \cap \beta^{k_i}_i = C_{\alpha_{k_i}} \cap \beta^{k_i}_i$.

$\blacktriangleright$ Base stage: Let $k<\omega$, notice that $C_{\alpha_{k+1}} <_{\mathop{\mathrm{lex}}} C_{\alpha_{k}}$ imply that $\beta^{k+1}_1 \leq \beta^{k}_1$. As the ordinals are well-ordered, we can find some $k_1<\omega$ such that $\beta^m_1$ is constant for all $m\geq k_1$.
By $f_{\beta_1}$ definition and the fact that $G(\alpha_m)$ is constant, we get that $C_{\alpha_m} \cap \beta^{k_1}_1 = C_{\alpha_{k_1}} \cap \beta^{k_1}_1$.

$\blacktriangleright$ $i+1$-stage: We follow the same strategy as in the base stage. Notice that for $m\geq k_i$, we know that $C_{\alpha_{m+1}} <_{\mathop{\mathrm{lex}}} C_{\alpha_{m}}$ and $C_{\alpha_m} \cap \beta^{k_i}_i = C_{\alpha_{k_i}} \cap \beta^{k_i}_i$, which imply that $\beta^{m+1}_{i+1}\leq \beta^{m}_{i+1}$.
Again as the ordinals are well-ordered, we can find some $k_{i+1}<\omega$ such that $\beta^m_{i+1}$ is constant for all $m\geq k_{i+1}$. By $f_{\beta_i}$ definition and the fact that $G(\alpha_m)$ is constant, we get that $C_{\alpha_m} \cap \beta^{k_{i+1}}_{i+1} = C_{\alpha_{k_{i+1}}} \cap \beta^{k_{i+1}}_{i+1}$.

Finally, we got that $C_{\alpha_m} = C_{\alpha_{k_n}}$ for every $m>k_n$, hence $\{C_\alpha \mid G(\alpha)=t\}$ is well-ordered by the lexicographic order as wanted. $\blacksquare$ ◻

It turns out that the support of a Baumgartner type is a useful invariant. The following Proposition is used in the proof of [Lamei Ramandi, Proposition 2.8].

Proposition 4. Suppose $X,Y\subseteq \mathop{\mathrm{acc}}(\omega_1)$ are two stationary subsets, $X\setminus Y$ is stationary, $L_X =\{ x_\alpha \mid \alpha\in X \}$ and $L_Y =\{ y_\alpha \mid \alpha\in Y \}$ are Baumgartner type, then there is no $<_{\mathop{\mathrm{lex}}}$-order preserving embedding from $L_X$ to $L_Y$.

Proof. Suppose $f:L_X\rightarrow L_Y$ is a $<_{\mathop{\mathrm{lex}}}$-order preserving function. Notice that the stationary set $Z:=X\setminus Y$ is the union of the two sets:

$$S_0:=\{\alpha \in Z \mid \exists \beta <\alpha[\beta\in Y \wedge f(x_\alpha)=y_\beta] \}$$ $$S_1:=\{\alpha \in Z \mid \exists \beta >\alpha[\beta\in Y \wedge f(x_\alpha)=y_\beta ]\}.$$

Clearly one of them is stationary, we will show that this leads to a contradiction.

$\blacktriangleright$ Suppose $S_0$ is stationary, then by Fodor’s Lemma, there exists some $\beta\in Y$ and a stationary subset $Z_0\subseteq S_0$ such that $f(x_\alpha) = \beta$ for every $\alpha \in Z_0$. But this is absurd, as $f$ is an order-preserving embedding.

$\blacktriangleright$ Suppose $S_1$ is stationary. By Corollary 2 we get that $\{x_\alpha \mid\alpha \in S_1\}$ in not $\sigma$-scattered. Hence, the isomorphic linear order $\{f(x_\alpha) \mid \alpha \in S_1 \}$ is also not $\sigma$-scattered. By Corollary 2, we get that the support $A=\{\beta\in Y \mid \exists \alpha \in S_1[ f(x_\alpha)=y_\beta]\}$ is stationary. Notice that for every $\beta \in A$, we have that for some $\alpha<\beta$ we have $f(x_\alpha)=y_\beta$. Then by Fodor’s Lemma, there exists some $\alpha\in Z$ and a stationary subset $B\subseteq A$ such that $f(x_\alpha) = \beta$ for every $\beta \in B$. But this is absurd, as $f$ is an order-preserving embedding. ◻

Corollary 5. There are $2^{\aleph_1}$ many non-isomorphic Baumgartner types of linear orders.

Proof. Fix by Ulam theorem a partition of $\mathop{\mathrm{acc}}(\omega_1)$ to $\omega_1$ many disjoint stationary subsets, $\langle S_\alpha \mid \alpha<\omega_1 \rangle$. Also fix a $C$-sequence, i.e. a ladder system $\langle C_\alpha \mid \alpha\in \mathop{\mathrm{acc}}(\omega_1) \rangle$ such that for every $\alpha \in S$, the ladder $C_\alpha$ is cofinal in $\alpha$ with order-type $\omega$.
For every $A\subseteq \omega_1$, we consider the linear order $L^A:= \{ C_\alpha \mid \alpha \in \bigcup_{\alpha\in A} S_\alpha\}$ ordered by the lexicographic order $<_{\mathop{\mathrm{lex}}}$. Notice that by the above Proposition for every two different $A,B\subseteq \omega_1$, we have that $L^A$ is not isomorphic to $L^B$. ◻

In pcf theory we consider the following two orders:

Definition 1. Let $\langle \kappa_n \mid n<\omega \rangle$ be an increasing sequence of uncountable regular cardinals. Their supremum, $\mu$, is a singular cardinal of countable cofinality.
Notice that usually the set of cardinals $A$ that we work with is such that $|A|<\min(A)$. Let $\prod_{n<\omega} \kappa_n := \{ f\mid f:\omega\rightarrow \mu\}$.
For $f,g\in \prod_{n<\omega} \kappa_n$ and $m<\omega$ write $f <_m g$ iff $ f(n) < g(n)$ for all $n>m$. Let $f <^{*} g$ iff for some $f <_m g$ for some $m<\omega$. Similarly we define $f=_m g$ and $f=^{*}g$.
We also write $f < g$ if $f(n) < g(n)$ for all $n<\omega$. Notice that $(\prod_{n<\omega } \kappa_n, <^{*})$ is a quasi-ordered set.

Let $\triangleleft \in \{<, <^{*}\}$.
A sequence $\bar f = \langle f_\alpha \mid \alpha<\lambda\rangle \subseteq \prod_{n<\omega}\kappa_n$ is called $\triangleleft$-increasing if for all $\alpha<\beta<\lambda$ it follows that $f_\alpha \triangleleft f_\beta$.
A sequence $\bar f = \langle f_\alpha \mid \alpha<\lambda\rangle \subseteq \prod_{n<\omega}\kappa_n$ is called $\triangleleft$-cofinal in $\prod_{n<\omega}\kappa_n$ if for all $f\in \prod_{n<\omega}\kappa_n$ there exists $\alpha<\lambda$ so that $f\triangleleft f_\alpha$.

Definition 2. Let $\mu$ be a singular of countable cofinality and let $\mu<\lambda$ be a regular cardinal. A $(\mu,\lambda)$-scale is a pair $(\bar \kappa, \bar f)$ where $\bar \kappa = \langle \kappa_n \mid n<\omega \rangle$ is a strictly increasing sequence of uncountable regular cardinals with $\sup\{\kappa_n \mid n<\omega\}= \mu$ and $\bar f = \langle f_\alpha \mid \alpha<\lambda \rangle \subseteq \prod_{n<\omega} \kappa_n$ is $<^{*}$-increasing and $<^{*}$-cofinal in $\prod_{n<\omega} \kappa_n$.

A function $g:\omega \rightarrow \mathop{\mathrm{On}}$ is an exact upper bound (eub) of a $<^{*}$-increasing sequence $\langle f_\alpha \mid \alpha<\theta \rangle \subseteq {}^\omega \theta$, where $\theta$ is a limit ordinal, if $f_\alpha <^{*} g$ for all $\alpha<\theta$ and for all $h <^{*} g$ there is some $\alpha<\theta$ so that $ h <^{*} f_\alpha$.

The following is a useful Fact.

Fact 3. Suppose $m<\omega$, $\langle \kappa_n\mid n\leq m \rangle$ is a finite sequence of regular cardinals, $\theta$ is a regular cardinal and $\bar t = \langle t_\alpha \mid \alpha<\theta \rangle$ is a sequence in $\prod_{n\leq m}(\kappa_n+1)$.
Then there exists $t\in \prod_{n\leq m} (\kappa_n +1)$ and a cofinal sub-seqence of $\bar t$ that coverges to $t$.

Proof. Successively thin out $\bar t$ using the Pigeon-hole principle so that for each $n\leq m$ the sequence $\langle t_\alpha(n )\mid \alpha<\theta\rangle$ is either constant or strictly increasing of order-type $\theta$. ◻

Lemma 4. Suppose $\bar f =\langle f_\alpha \mid \alpha<\delta \rangle \subseteq {}^{\omega}\mathop{\mathrm{On}}$ is a $ <^{*} $-increasing sequence, $\delta$ is a limit ordinal and $\mathop{\mathrm{cf}}(\delta)=\theta>\aleph_0$. Then the following conditions are equivalent:

1. There exists an eub $g$ of $\bar f$ so that $\mathop{\mathrm{cf}}(g(n))=\theta$ for all $n<\omega$;

2. There exists a $<$-increasing sequence $\bar h = \langle h_i \mid i<\theta \rangle \subseteq {}^{\omega}\theta$ so that for every $i<\theta$ there is $\alpha<\delta$ with $ h_i <^{ * } f_\alpha $ and for every $\alpha<\delta$ there is $i<\theta$ with $f_\alpha <^{*} h_i$;

3. For every unbounded set $C\subseteq \delta$ there is some $m_0<\omega$ and an unbounded set $A\subseteq C$ with $\mathop{\mathrm{otp}}(A) =\theta$ so that $f_\alpha<_{m_0} f_\beta$ for all $\alpha<\beta$ in $A$.

Proof. $(1)\Rightarrow (2):$ Suppose $\bar f$ and $\theta$ are as above and $g$ is an eub of $\bar f$ with $\mathop{\mathrm{cf}}(g(n))=\theta$ for all $n<\omega$.
For each $n<\omega$ we fix a strictly increasing sequence $\langle \beta^n_i \mid i<\theta\rangle$ with supremum $g(n)$.
For $i<\theta$ define a function $h_i:\omega\rightarrow \mathop{\mathrm{On}}$ by $h_i(n):=\beta^n_i$.
Thus, $i< j <\theta$ implies $h_i< h_j$, and $\sup\{ h_i \mid i<\theta\}= g$.
So the sequence $\bar h = \langle h_i \mid i<\theta \rangle$ is $<$-increasing.

Fix $i<\theta$, as $g$ is an eub and $h_i< g$, there is an $\alpha<\delta$ such that $h_i<^{*} f_\alpha$. Fix $\alpha<\delta$, we know that $f_\alpha<^{*}g$.
Hence, there exists some $m<\omega$ such that $f_\alpha<_m g$. For every $m< n <\omega$, there exists some $i_n<\theta$ such that $f_\alpha(n)< h_{i_n}(n)$.
As $\theta$ is regular, there exists some $\sup\{i_n \mid n<\omega\}< i<\theta$.
As $\bar h$ is $<$-increasing, we have that $f_\alpha<^{*} h_i$ as sought.

$(2)\Rightarrow (3):$ Suppose $\bar h = \langle h_i \mid i<\theta \rangle \subseteq {}^{\omega}\theta$ is an $<$-increasing sequence and $C\subseteq \delta$ is an unbounded set.
For each $i<\theta$, let $\alpha(i)\in C$ be chosen such that $h_i<^{*} f_{\alpha(i)}$, $i< j\implies \alpha(i)<\alpha(j)$ and $A$ is unbounded in $C$. Let $A=\{\alpha(i)\mid i<\theta\}$. So $\mathop{\mathrm{otp}}(A)=\theta$.
By recursion over $i<\theta$ we may thin out $\bar h$ such that $f_{\alpha(i)}<^{*} h_{i+1}$. Let $m(i)$ be such that $h_i<_{m(i)}< f_{\alpha(i)} <_{m(i)} h_{i+1}$. Since $\theta>\aleph_0$ is regular, we may assume, by shrinking $A$ and re-enumerating it increasingly, that there is some fixed $m_0<\omega$ for which $m(i)=m_0$ for all $i<\theta$.
Suppose that $\alpha(i)<\alpha(j)$ are in $A$. Then $f_{\alpha(i)}<_{m_0} h_{i+1} \leq h_j <_{m_0} f_{\alpha(j)}$ as required.

$(3)\Rightarrow (1):$ Suppose $A\subseteq \delta$ is unbounded of order-type $\theta$ and $m_0<\omega$ such that for every $\alpha<\beta$ in $A$ implies $f_\alpha <_{m_0} f_\beta$. We define the function $g$ as follows:

$$g(n):=\begin{cases} \sup\{f_{\alpha}(n)\mid \alpha \in A\} & \text{if }m_0<n<\omega, \\ \theta & \text{if }n\leq m_0. \end{cases}$$

Now $g$ is clearly an eub of $\bar f$ and $\mathop{\mathrm{cf}}(g(n))=\theta$ for all $n>m_0$. ◻

A $<^{*}$-increasing $\bar f = \langle f_\alpha \mid \alpha<\delta \rangle$ with $\mathop{\mathrm{cf}}(\delta)=\theta>\aleph_0$ that satisfies one (equivalently, all) of the conditions in the previous Lemma is called flat.

Suppose $(\bar \kappa, \bar f)$ is a $(\mu,\lambda)$-scale for a singular $\mu$ of countable cofinality and a regular $\lambda>\mu$.

An ordinal $\alpha<\lambda$ is called a flat point of $f$ if $\mathop{\mathrm{cf}}(\alpha)>\aleph_0$ and $\bar f\restriction \alpha$ is flat.

A $(\mu,\lambda)$-scale is called good if for all $\alpha<\lambda$ with $\aleph_0<\mathop{\mathrm{cf}}(\alpha)<\mu$, $\alpha$ is a flat pint in $\bar f$ and $f_\alpha$ is an eub of $\bar f \restriction \alpha$.

Theorem 5.

• For every $\mu$ singular cardinal of countable cofinality there exists a $(\mu,\mu^+)$-scale in $\mathop{\mathrm{ZFC}}$.

• The existence of a supercompact cardinal implies it is consistent there are no good $(\aleph_\omega,\aleph_{\omega+1})$-scales.

• $\square_\mu$, and even the weaker prinicple $\mu^+\in I[\mu^+]$, implies the existence of a good $(\mu,\mu^+)$-scale.

We overview some results of Todorčević with a goal to unravel the following, mysterious at first sight, definitions.
Recall that in the previous post we gave some definitions about trees.
Let $T$ be a tree and $t\in T$, suppose $\alpha\leq \mathop{\mathrm{ht}}(t)$ then we denote by $t\mathop{\mathrm{\restriction}}\alpha$ the $s\leq t$ in $T$ such that $\mathop{\mathrm{ht}}(s)=\alpha$.

A partial map $g$ from a tree $S$ into a tree $T$ is called Lipschitz is $g$ is level-preserving and $\Delta(g(x),g(y))\geq \Delta(x,y)$ for all $x,y\in \mathop{\mathrm{dom}}(g)$.

Definition. A Lipschitz tree is any Aronszajn tree $T$ with the property that every level-preserving map from an uncountable subset of $T$ into $T$ is Lipschitz on an uncountable subset of its domain.

The following are two useful Lemmas about Lipschitz trees that are worth mentioning although we will not need them for the rest of the post.
Lemma 1. Suppose $T$ is a Lipschitz tree, $0<n<\omega$ and that $A$ is an uncountable subset of the $n$-th power of $T$. Then there existsan uncountable $B\subseteq A$ such that $\Delta(a_i,b_i)= \Delta(a_j,b_j)$ for all $a\neq b$ in $B$ and $i,j<n$. It follows that the $n$-th power of $T$ is a Lipschitz tree as well.

Lemma 2. Every uncountable subset of a Lipschitz tree $T$ contains an uncountable antichain. More generally, every family $A$ if pairwise disjoint finite subset of $T$ contains an uncountable sub-family $B$ such that $\bigcup B$ is an antichain of $T$.

Theorem (Todorčević): (Theorem 4.2.4). Every special coherent tree $T$ is Lipschitz.

Proof. Let $(T,\subseteq)$ be a special coherent tree $T\subseteq {}^{<\omega_1}\omega$ where the nodes are functions from countable ordinals to $\omega$ and the tree is closed with respect to restrictions of nodes to countable ordinals.
By our assumption, the tree $T$ can be decomposed into countably many anti chains. So let $c:T\rightarrow \omega$ be fixed map such that $c(s)\neq c(t)$, whenever $s<_T t$ and such that $a$ is one-to-one on the levels of $T$.
For every $t,s\in T$ we have $\Delta(s,t)= \min\{\xi < \min(\mathop{\mathrm{ht}}(s),\mathop{\mathrm{ht}}(t)) \mid t(\xi)\neq s(\xi)\}$
Consider a partial level-preserving map $f$ from an uncountable subset $X$ of $T$ into $T$.
As the tree $T$ is special, we may refine and assume both the domain and the image of $f$ are ncountable anti-chains.
Moreover, we may assume that $X$ contains no two different nodes of the same height. For $t\in X$, set $D_t:=\{\xi <\mathop{\mathrm{ht}}(t)\mid t(\xi)\neq f(t)(\xi)\}$.
Let $p_t:D_t\rightarrow \omega$ and $q_t:D_t\rightarrow \omega$ be the functions defined by $p_t(\xi) = c(t\mathop{\mathrm{\restriction}}\xi) \text{ and }q_t(\xi)=c(f(t)\mathop{\mathrm{\restriction}}\xi).$
Applying the $\Delta$-system, pigeon-hole principle and shrinking $X$, we assume that $\langle D_t \mid t\in X \rangle$ forms a $\Delta$-system with root $R$ such that:

• $D_s\setminus R < D_t \setminus R$ whenever $s,t\in X$ are such
  that $\mathop{\mathrm{ht}}(s)<\mathop{\mathrm{ht}}(t)$.

And for every $s,t\in X$ we have,
$s\mathop{\mathrm{\restriction}}D_s \cong t\mathop{\mathrm{\restriction}}D_t$, $f(s)\mathop{\mathrm{\restriction}}D_s \cong f(t)\mathop{\mathrm{\restriction}}D_t$, $p_s\mathop{\mathrm{\restriction}}D_s \cong p_t\mathop{\mathrm{\restriction}}D_t,$ and $q_s\mathop{\mathrm{\restriction}}D_s \cong q_t\mathop{\mathrm{\restriction}}D_t.$
Fix $s$ and $t$ in $X$, let us show that $\Delta(s,t)=\Delta(f(s),f(t))$.
Let us first establish the inequality $\Delta(f(s),f(t))\geq \Delta(s,t).$

Let $\alpha=\Delta(s,t)$. Then $\alpha<\mathop{\mathrm{ht}}(t),\mathop{\mathrm{ht}}(s)$, since $s$ and $t$ are incomparable.
Let us show that $D_s\cap \alpha=D_t\cap \alpha$. Suppose $(D_s \cap \alpha) \setminus D_t$ is non-empty and contains $\beta$. Then by the properties of our parameters $s$ and $t$, $s\mathop{\mathrm{\restriction}}\beta = t\mathop{\mathrm{\restriction}}\beta$
and $c(s\mathop{\mathrm{\restriction}}\beta) = c(t\mathop{\mathrm{\restriction}}\gamma)$ for some $\gamma\neq \beta$ in $D_t$.
Notice that$s\mathop{\mathrm{\restriction}}\beta$ and $t\mathop{\mathrm{\restriction}}\gamma$ are comparable as they both restrictions of $t$ which is a contradiction to the assumption each homogeneous set of the coloring $c$ is an anti-chain. Similarly we can show that $(D_t \cap \alpha) \setminus D_s = \emptyset$.
By the properties of our parameters $s$ and $t$, we have that $D_s\cap \alpha=D_t\cap \alpha$ and that $f(s)$ and $f(t)$ agree on this set.
From the definition of $D_s$ and $D_t$, we conclude that $f(s)$ and $f(t)$ must agree below $\alpha$, and so $\Delta(f(s),f(t))\geq\alpha$.
A completely symmetric argument will give us the other inequality $\Delta(s,t)\geq \Delta(f(s),f(t)).$ ◻

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 $B\subseteq A_\xi$. ◻