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 $T=\{(\alpha,k_{\beta})\mid\alpha<k_{\beta}\wedge\beta<\iota\}$ 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 $s<r$ .

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 $i<j<\omega_{1}$ , $b_{i}\subset b_{j}$ . 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
Canonical name	ExampleOfAronszajnTree
Date of creation	2013-03-22 12:52:39
Last modified on	2013-03-22 12:52:39
Owner	Henry (455)
Last modified by	Henry (455)
Numerical id	5
Author	Henry (455)
Entry type	Example
Classification	msc 03E05
Classification	msc 05C05