In a set theory book, I read a proof of

There exists an Aronszajn tree

It makes use of the following lemma.

Let $[\omega_1]^2$ be the collection of all increasing pairs of countable ordinals. Suppose e is a function from $[\omega_1]^2$ into some set $X$. Then, for each $\alpha < \omega_1$ we write $e_\alpha : \alpha \rightarrow X$ for the function with domain $\alpha$, given by $e_\alpha(\beta ) = e(\langle \beta,\alpha\rangle)$.

Then there is $e : [\omega_1]^2\rightarrow\omega$ satisfying the following conditions

  1. For every $\alpha < \omega_1$, $e_\alpha$ is one-to-one.
  2. For all $\alpha < \beta < \omega_1$, $e_\alpha$ and $e _\beta\upharpoonright\alpha$ disagree in only nitely-many places.

I understand the proof, but later it says, that a similar argument can be used to show that

CH implies the existence of an $\omega_2$-Aronszajn tree.

I was giving a try. I think I should try toprove first a similar version of the lemma by replacing $\omega_1$ by $\omega_2$, $\omega$ by $\omega_1$, and "finitely-many" by "countably many". But I can't reach the conclusion.

Any help is welcome, thanks

  • $\begingroup$ Ah, but does it imply the existence of a non-special $\omega_2$-Aronszajn tree? :-) $\endgroup$ – Asaf Karagila May 29 at 17:27

Your line of thought is correct. In my opinion, the notation you use is unneccesarrily complicated for the purpose of this question, so I will phrase it a little different. Show that there is a sequence of functions $\langle f_\alpha\mid\alpha<\omega_2\rangle$ so that :

  1. $f_\alpha:\alpha\rightarrow\omega_1$ is one-to-one
  2. For $\alpha<\beta<\omega_2$, $f_\alpha$ and $f_\beta\vert\alpha$ differ only on countably many places.

I would recommend to build this sequence by induction. To make the construction a little simpler, make sure that you always have enough room left, i.e. $\omega_1\setminus\operatorname{ran}(f_\alpha)$ should always be uncountable. Note that other than for $e$, you now have to deal with two different limit cases: One where $\operatorname{cof}(\alpha)$ is $\omega$ and one where it is $\omega_1$. Try to adapt the construction of $e$ in the second case (I do not know for sure that that works with the construction known to you, as you have not provided a source). The first case is not hard to deal with.

Now under the assumption of $CH$, show that $\{f_\alpha\vert\beta\mid\beta\leq\alpha<\omega_2\}$ ordered by inclusion is an $\omega_2$-Aronszajn-tree. If you get stuck, try to use this lemma.

One final remark: The assumption of CH is necessary here. It is consistent, relative to the existence of a weakly compact cardinal, that there are no $\omega_2$-Aronszajn trees.


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