# Abstract It is proved that if $\aleph_{\alpha}^{\aleph_{\xi}}=\aleph_\alpha$, for all $\xi<\alpha$, then there are $2^{\aleph_{\alpha+1}}$ non-isomorphic normal Aronszajn trees of height $\aleph_{\alpha+1}$ # 1. Introduction. >[!note] Notation >Let $T=\langle A, \leqq\rangle$ be a partially ordered system. > For every $x \in A$ denote $P_x=\{y \mid y<x\}$. > $T$ is a [[Trees|tree]] if every $P_{x}$ is well-ordered. > The *rank of $x\in A$ $\rho(x)$,* is the order type of $P_x$ (this is an ordinal number). > The *rank of the tree* $\rho(T)$ is the least upper bound of $\rho(x), x \in A$. > For every ordinal $\alpha$ > - $R_\alpha(T)$ is the set of all elements of rank $\alpha$ > - $S_{\alpha}(T)$ is the set of all elements of rank lt;\alpha$ > - $T(\alpha)=\left< S_{\alpha}(T),\leqq \right>$ > A subset $A^{\prime}$ of $A$ is *full* if for every $x \in A^{\prime}$ we have $P_x \subseteq A^{\prime}$ > A *node* of $T$ is a subset $N$ of $A$ of the form $\{x \mid x \in A$ and $\left.P_x=P_y\right\}$ for some $y \in A$. > A *path* of $T$ is a subset of $A$ which is full and linearly (totally) ordered by $\leqq$. >[!info] Definition > A *normal $\aleph_\alpha$ tree* is a tree $\langle A, \leqq\rangle$ of rank $\omega_{\alpha+1}$ having the following properties: > (N1) $R_{\xi}$ is of power $\aleph_\alpha$ for all $\xi, 0<\xi<\omega_{\alpha+1}$. > (N2) Nodes of rank $\xi+1$ are of power $\aleph_\alpha$ and nodes whose rank is a limit ordinal (including 0 ) are of power 1. > (N3) If $x \in A$ and $\rho(x)<\eta<\omega_{\alpha+1}$ then there is a $y \in R_{\eta}$ for which $x<y$. > (N4) Every path is of power lt;\aleph_{\alpha+1}$. (Or, equivalently, every path has order type lt;\omega_{\alpha+1}$.) > (N5) If $A^{\prime}$ is a path of power lt;\aleph_\alpha$ then there is an $x \in A$ such that $y \leqq x$ for all $y \in A^{\prime}$. >[!warning] >This is different from the contemporary conventions [[Trees#Types of trees|here]]. >In modern terms, their normal $\aleph_{\alpha}$ tree would be a normal, Hausdorff $\aleph_{\alpha+1}$ tree which is also $\aleph_{\alpha}$-splitting, $\aleph_{\alpha+1}$-Aronszajn and satisfies (N5). **Lemma.** Given any two normal $\boldsymbol{\aleph}_\alpha$ trees, $T_i=\left\langle A_i, \leqq i\right\rangle, i=1,2$, and any $\xi<\omega_{\alpha+1}$, the trees $T_i(\xi), i=1,2$, are isomorphic. Moreover if $\xi \leqq \eta<\omega_{\alpha+1}$ then any isomorphism between $T_1(\xi)$ and $T_2(\xi)$ can be extended to an isomorphism between $T_1(\eta)$ and $T_2(\eta)$. **Theorem.** If $\aleph_{\alpha}^{\aleph_{\xi}}=\aleph_\alpha$, for all $\xi<\alpha$, then there are exactly $2^{\aleph_{\alpha+1}}$ different isomorphism types of normal $\boldsymbol{\aleph}_\alpha$ trees. >[!note] Remark >In the paper they prove the weaker statement replacing "exactly $2^{\aleph_{\alpha+1}}quot; with "at least $\aleph_{\alpha+2}quot;, and claim without proof that it implies the general statement even without GCH. **Main Lemma.** If $\aleph_\alpha^{\aleph_{\xi}}=\aleph_\alpha$, for all $\xi<\alpha$, then one can associate with every subset $X$ of $\omega_{\alpha+1}$ of power $\boldsymbol{\aleph}_{\alpha+1}$ a normal $\boldsymbol{\aleph}_\alpha$ tree, $T(X)$, so that $T(X)$ and $T\left(X^{\prime}\right)$ are not isomorphic if $\left|X \cap X^{\prime}\right|<\aleph_{\alpha+1}$. **Fact ([[Sierpiński - Cardinal and ordinal numbers]]).** There is a class $C$ of subsets of $\omega_{\alpha+1}$ such that $|C|=\aleph_{\alpha+2},|X|=\aleph_{\alpha+1}$ for all $X \in C$ and $\left|X \cap X^{\prime}\right|<\aleph_{\alpha+1}$ for all $X, X^{\prime} \in C$ which are different. The main lemma (proved at subsequent sections) plus the fact imply the weaker version of the theorem. # 2. Sequential trees >[!info] Definition >A *sequential tree* is a tree $\langle S, \leqq\rangle$ which satisfies the following conditions: > (ST1) $S$ is a set of sequences. > (ST2) If $s \in S$ then $s \mid \beta \in S$ for all $\beta \leqq l(s)$. > (ST3) $s \leqq s^{\prime}$ iff for some $\beta$ we have $s=s^{\prime} \mid \beta$. ^sequential >[!info] Definition >Let $S_\alpha$ be the set of all sequences $s$ satisfying the following conditions: > (S1) $l(s)<\omega_{\alpha+1}$. > (S2) All the members of $s$ are ordinals lt;\omega_\alpha$. > (S3) $\left\{\alpha \mid s_\alpha \neq 0\right\}$ is of power lt;\boldsymbol{\aleph}_\alpha$. > Defining " $\leqq$ " according to (ST3) it is immediate that $T_\alpha=\left\langle S_\alpha \leqq\right\rangle$ is a sequential tree. **Lemma 1.** If, for all $\xi<\alpha, \aleph_\alpha^{\aleph_\xi}=\aleph_\alpha$ then the tree $T_\alpha$ satisfies the conditions (N1), (N2), (N3), and (N5). **Lemma 2.** If, for all $\xi<\alpha, \aleph_\alpha^{\aleph_\xi}=\aleph_\alpha$ then every full subset of $S_\alpha$ of power $\aleph_{\alpha+1}$ contains a path of power $\boldsymbol{\aleph}_{\alpha+1}$. # 3. Composition of trees. >[!info] Definition >Let $s$ be a sequence and $X$ a set of ordinals. Then > - $s \mid X$ is subsequence of $s$ obtained by letting the index range over $X$ only > - $s \mid \bar{X}$ is defined as $s \mid(l(s)-X)$ . > - If $s \mid X=s^1$ and $s \mid \bar{X}=s^2$ then we write $s=s^1 *_X s^2$. > - > **Claim.** For every $s^1, s^2$ and $X$ there is at most one $s$ such that $s=s^1 *_X s^2$. > > Let $T^{i}=\left< S^{i},\leqq \right>$ ($i=1,2$) sequential trees, and $X$ a set of ordinals. > $T^{1} \ast_{X} T^{2}$ is the sequential tree whose elements are of the form $s^{1} \ast_{X} s^{2}$ for $s^{i} \in S^{i}$ ($i=1,2$). **Lemma 3.** Let $T^1=\left\langle S^1, \leqq\right\rangle$ and $T^2=\left\langle S^2, \leqq\right\rangle$ be sequential trees of rank $\omega_{\alpha+1}$ and let $X$ be a subset of $\omega_{\alpha+1}$, then: (I) If both $T^1$ and $T^2$ satisfy any of the five conditions ( $\mathrm{N} j$ ), $j=1, \cdots, 5$, then $U=T^1 *_X T^2$ satisfies the same condition. (II) If $|X|=\aleph_{\alpha+1}$ and $T^1$ satisfies (N4) so does $U$. # 4. A class of normal trees. >[!info] Definition >Let $\alpha$ be such that for all $\xi<\alpha, \aleph_\alpha^{\aleph_\xi}=\aleph_\alpha$, $T_{\alpha}$ the tree defined above, and $T$ a fixed normal $\aleph_{\alpha}$ tree. For every $Y \subseteq \omega_{\alpha+1}$ let $T(Y)=T \ast_{Y} T_{\alpha}$. **Lemma 4.** If $X \subseteq \omega_{\alpha+1}$ is of power $\aleph_{\alpha+1}$ then $T(X)$ is a normal $\aleph_{\alpha}$ tree. Lemmas 5-6 lead to: **Corollary.** If $Y, Y^{\prime} \subseteq \omega_{\alpha+1}$, $|Y|=\left|Y^{\prime}\right|=\aleph_{\alpha+1}$ and $\left|Y \cap Y^{\prime}\right|<\boldsymbol{\aleph}_{\alpha+1}$ then $T(Y)$ and $T\left(Y^{\prime}\right)$ are not isomorphic. This implies the Main Lemma.