Wikipedia gives the following construction of a special Aronszajn tree. Supposedly, this tree has $\aleph_1$ nodes, as each level and each branch is countable. However, it seems to me that this construction actually produces $\frak{c}$-many nodes after stage $\omega$, where $\omega$ is the first infinite ordinal:
The elements of the tree are certain well-ordered sets of rational numbers with supremum that is rational or $−∞$. If $x$ and $y$ are two of these sets then we define $x ≤ y$ (in the tree order) to mean that $x$ is an initial segment of the ordered set $y$. For each countable ordinal $α$ we write $U_α$ for the elements of the tree of level $α$, so that the elements of $U_α$ are certain sets of rationals with order type $α$. The special Aronszajn tree $T$ is the union of the sets $U_α$ for all countable $α$. We construct the countable levels $U_α$ by transfinite induction on $α$ as follows starting with the empty set as $U_0$:
- If $α + 1$ is a successor, then $U_{α+1}$ consists of all extensions of a sequence $x$ in $U_α$ by a rational greater than $\sup x$. $U_{α+1}$ is countable as it consists of countably many extensions of each of the countably many elements in $U_α$.
- If $α$ is a limit, then let $T_α$ be the tree of all points of level less than $α$. For each $x \in T_α$ and for each rational number $q > \sup x$, choose a level $α$ branch of $T_α$ containing $x$ with supremum $q$. Then $U_α$ consists of these branches. $U_α$ is countable as it consists of countably many branches for each of the countably many elements in $T_α$.
I bolded out the parts that I have trouble understanding.
I buy that $U_n$ is countable at each finite stage $n \geq 0$, since it's the union of all subsets of $\Bbb{Q}$ of size $n$.
I also buy that $T_\omega$ will be countable, since $T_\omega = \bigcup_{n \geq 0} U_n$ is a countable union of countable sets (in fact, $T_\omega$ is the set of all finite subsets of $\Bbb{Q}$).
But a level $\omega$ branch of $T_\omega$ with finite supremum must be a countably infinite increasing subset of $\Bbb{Q}$, and aren't there $\frak{c}$-many of those?
For instance, $T_\omega$ should have a distinct branch for every infinite subset of $\{1 - 1/n\}_{n \geq 1} = \{0, 1/2, 2/3, 3/4, ...\}$, and there's $2^{\aleph_0} = \frak{c}$ such subsets. So it seems like $U_{\omega}$ should have $2^{\aleph_0}$ nodes already.
What am I missing? Did they just flub the transcription when describing this construction?
Never mind, I figured it out. I’m not adding every branch of $T_\omega$ as a node in $U_\omega$, I’m picking a specific branch for every $x \in T_\omega$ and every rational $q > \sup x$. This is a countable number of branches, so $U_\omega$ has countably many nodes.