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
- For every $\alpha < \omega_1$, $e_\alpha$ is one-to-one.
- 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
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 :
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.