For a tree of height ω1 with every level countable and no uncountable branch (Aronszajn tree), how can I prove it is order-isomorphic (in the sense there existes a bijection which respects the orders) to another such tree in which every level α is a subset of [ωα , ωα+ω) ?
We can suppose every ordinal β has a height α if β ∈ [ωα , ωα+ω) (so ht(β) is well defined and unique)
What is the isomorphism ?
SKETCH: Let $T$ be the original Aronszajn tree. Each level of $T$ is countable, so if $T_\alpha$ is level $\alpha$ of $T$, there is an injection $f_\alpha:T_\alpha\to[\omega\alpha,\omega\alpha+\omega)$. Let
$$T'=\bigcup_{\alpha<\omega_1}f_\alpha[T_\alpha]\,,$$
and use $T$ and the maps $f_\alpha$ to define the tree order on $T'$ so that $\bigcup_{\alpha<\omega_1}f_\alpha$ is an order-isomorphism. In other words, in essence you start with (what is going to be) the isomorphism and use it to construct an isomorphic tree $T'$ whose levels have the desired property.