Aronszajn, Suslin and special trees

Use ${\omega }_1$ from The first uncountable ordinal ${\omega }_1:=\mathrm{\aleph }(\omega )$ and “countable” to include finite sets as in Finite, countably infinite, countable, uncountable. An Aronszajn tree is a tree of height ${\omega }_1$ whose every level is countable and which has no cofinal branch. A Suslin tree is an Aronszajn tree with no uncountable antichain. Normality and splitting are not implicit. This direct formulation does not presuppose at the point of definition that ${\omega }_1$ has already been proved to be an infinite cardinal.

A tree $T$ is special if there is $f:T\to \omega$ with $f(s)\mathrm{\ne }f(t)$ whenever $s{<}_{T}t$. Equivalently, $T$ is a countable union of antichains. Indeed, a witnessing $f$ gives antichains $A_n=f^{-1}(\{n\})$ and $T={\bigcup }_{n<\omega }{A}_n$. Conversely, given antichains $A_n$ covering $T$, set $f(t)=\min \{n:t\in A_n\}$. This minimum exists for each $t$; comparable distinct nodes cannot have the same minimum because they would belong to the same antichain. No choice is used in this equivalence, and overlaps among the $A_n$ cause no difficulty.

A strictly increasing rational labeling $q:T\to \mathbb{Q}$ suffices for specialness. Fix an injection $j:\mathbb{Q}\to \omega$, available from $\mathbb{Q}$ is countably infinite, and put $f=j\circ q$. If $s{<}_{T}t$, then $q(s)<q(t)$, so $f(s)\mathrm{\ne }f(t)$. No converse about increasing rational labelings is asserted. The empty and singleton trees are special (use the empty map and the constant-zero map respectively), but are not Aronszajn or Suslin trees since they lack height ${\omega }_1$.