One way to construct an Aronszajn tree is to build a sequence of functions $\langle e\_\alpha : \alpha < \omega\_1 \rangle$ such that $e\_\alpha$ is an injection from $\alpha$ to $\omega$, and for any $(\alpha, \beta)$, $e\_\alpha$ agrees with $e\_\beta$ except on a finite set.


Is the following generalization possible? There is a family of functions {$e\_x : x \in [\omega\_2]^\omega$} such that each $e\_x$ is an injection from $x$ to $\omega$, and for any $x,y \in [\omega\_2]^\omega$, {$\alpha : e\_x(\alpha) \not= e\_y(\alpha)$} is finite.



larger coherent family of functions
If the continuum hypothesis holds, then this is impossible. To see this, consider a family of $\omega\_2$ many pairwise disjoint subsets of $\omega\_2$, and for each $x$ in this family, consider the range of $e\_x$, which is a subset of $\omega$. By the continuum hypothesis, there are only $\omega\_1$ many possible ranges, and so there must be two disjoint countable subsets $x,y\subset\omega\_2$ with $\text{ran}(e\_x)=\text{ran}(e\_y)$. Let $z=x\cup y$ and consider $e\_z$. It cannot be that $e\_z$ has only finite difference from $e\_x$ and also from $e\_y$, if $e\_z$ is injective.


I'm not yet sure what happens when CH fails... 


Meanwhile, here are a few slightly weaker generalizations that are possible. 


1. There is a family of injective functions $e\_x:x\to \omega$ for all countable $x\subset\omega\_1$, that is, for $x\in [\omega\_1]^\omega$, such that $e\_x$ and $e\_y$ have at most finitely many disagreements on $x\cap y$. To see this, first fix a family of functions $e\_\alpha$ for $\alpha\lt\omega\_1$ as mentioned at the outset of the question, so that $e\_\alpha$ injects $\alpha$ to $\omega$, and any two of them disagree only finitely. The point is that we may extend this family to $e\_x:x\to \omega$ for $x\in[\omega\_1]^\omega$ by defining $e\_x=e\_\alpha\upharpoonright x$, where $\alpha=\sup(x)$. The point is that if $\sup(y)=\beta$, then $e\_\beta$ and $e\_\alpha$ have only finitely many disagreements on their common domain, and so also $e\_x$ and $e\_y$ will therefore also have only finitely many disagreements.
2. There is a family of injective functions $f\_x:x\to \omega$ for all $x\in(\omega\_2)^\omega$, meaning $x\subset\omega\_2$ and $x$ has order type $\omega$, such that $f\_x$ and $f\_y$ have at most finitely many disagreements on $x\cap y$. To see this, fix the $e\_x$ as in the previous paragraph, and also for each $\beta\lt\omega\_2$, fix an injection $\pi\_\beta:\beta\to \omega\_1$, and now for $x\subset\omega\_2$ of order type $\omega$, let $\beta=\sup(x)$, and consider $\pi\_\beta[x]\subset \omega\_1$. We may let $f\_x=(e\_{\pi\_\beta[x]}\circ\pi\_\beta)\upharpoonright x$. That is, we use $\sup(x)$ to tell us how to regard $x$ as really a subset of $\omega\_1$, and then we apply the corresponding function from before. The point is that if $x$ and $y$ are two increasing $\omega$ sequences in $\omega\_2$, then if they have different suprema, then $x\cap y$ is finite anyway and we needn't worry about it, and if they have the same suprema, then $f\_x$ and $f\_y$ agree completely.


