# 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 ${T}_{\alpha +1}=\{x\cup \{q\}\mid x\in {T}_{\alpha}\wedge q>\mathrm{max}x\}$. 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, $B=\u27e8{b}_{0},{b}_{1},\mathrm{\dots}\u27e9$. 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 $\u27e8{x}_{0},{x}_{1},\mathrm{\dots}\u27e9$ 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 |