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 with be a sequence cofinal in $\kappa$. Then consider the tree where with iff 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 , 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 ${\mathbb{Q}}^{+}$. will be defined by 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 .

Let $T_0=\{[0]\}$. Given a level $T_{\alpha }$, let $T_{\alpha +1}=\{x\cup \{q\}\mid x\in T_{\alpha }\wedge q>\max x\}$. 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 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 for each . For the case, $T_0$ has only one element. If then . If 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, $B=⟨b_0,b_1,\mathrm{\dots }⟩$. Then for any , $b_i\subset b_j$. 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 $⟨x_0,x_1,\mathrm{\dots }⟩$ 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.