# example of Aronszajn tree

Construction 1: If $\kappa$ is a singular cardinal then there is a construction of a $\kappa$-Aronszajn tree (http://planetmath.org/KappaAronszjanTree). Let $\langle k_{\beta}\rangle_{\beta<\iota}$ with $\iota<\kappa$ be a sequence cofinal in $\kappa$. Then consider the tree where $T=\{(\alpha,k_{\beta})\mid\alpha with $(\alpha_{1},k_{\beta_{1}})<_{T}(\alpha_{2},k_{\beta_{2}})$ iff $\alpha_{1}<\alpha_{2}$ and $k_{\beta_{1}}=k_{\beta_{2}}$.

Note that this is similar to (indeed, a subtree of) the construction given for a tree with no cofinal branches. It consists of $\iota$ disjoint branches, with the $\beta$-th branch of height $k_{\beta}$. Since $\iota<\kappa$, every level has fewer than $\kappa$ elements, and since the sequence is cofinal in $\kappa$, $T$ must have height and cardinality $\kappa$.

Construction 2: We can construct an Aronszajn tree out of the compact subsets of $\mathbb{Q}^{+}$. $<_{T}$ will be defined by $x<_{T}y$ iff $y$ is an end-extension of $x$. That is, $x\subseteq y$ and if $r\in y\setminus x$ and $s\in x$ then $s.

Let $T_{0}=\{[0]\}$. Given a level $T_{\alpha}$, let $T_{\alpha+1}=\{x\cup\{q\}\mid x\in T_{\alpha}\wedge q>\operatorname{max}x\}$. That is, for every element $x$ in $T_{\alpha}$ and every rational number $q$ larger than any element of $x$, $x\cup\{q\}$ is an element of $T_{\alpha+1}$. If $\alpha<\omega_{1}$ is a limit ordinal then each element of $T_{\alpha}$ is the union of some branch in $T(\alpha)$.

We can show by induction that $|T_{\alpha}|<\omega_{1}$ for each $\alpha<\omega_{1}$. For the case, $T_{0}$ has only one element. If $|T_{\alpha}|<\omega_{1}$ then $|T_{\alpha+1}|=|T_{\alpha}|\cdot|\mathbb{Q}|=|T_{\alpha}|\cdot\omega=\omega<% \omega_{1}$. If $\alpha<\omega_{1}$ is a limit ordinal then $T(\alpha)$ is a countable union of countable sets, and therefore itself countable. Therefore there are a countable number of branches, so $T_{\alpha}$ is also countable. So $T$ has countable levels.

Suppose $T$ has an uncountable branch, $B=\langle b_{0},b_{1},\ldots\rangle$. Then for any $i, $b_{i}\subset b_{j}$. Then for each $i$, there is some $x_{i}\in b_{i+1}\setminus b_{i}$ such that $x_{i}$ is greater than any element of $b_{i}$. Then $\langle x_{0},x_{1},\ldots\rangle$ is an uncountable increasing sequence of rational numbers. Since the rational numbers are countable, there is no such sequence, so $T$ has no uncountable branch, and is therefore Aronszajn.

Title example of Aronszajn tree ExampleOfAronszajnTree 2013-03-22 12:52:39 2013-03-22 12:52:39 Henry (455) Henry (455) 5 Henry (455) Example msc 03E05 msc 05C05