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

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

• Ah, but does it imply the existence of a non-special $\omega_2$-Aronszajn tree? :-) – 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.