Infinite trees with finite levels, generalization of König's Lemma

Question

EDIT: this question is due to a confusion. Noah answered it appropriately. However, a solution of the more general theorem has been added as an answer. In particular, it is a proof of the theorem that I thought was proven in Drake's reference.

I was reading Set Theory, An introduction to Large Cardinals, by Frank Drake (1974), and I found the following proof of König's Lemma for arbitrarily large trees (in particular, trees on any cardinal with finite levels):

For clarification: "finitary tree" is defined there as a tree with finite elements in the first level and such that any element has finitely many successors and $T_\alpha$ denotes the subset of the tree whose elements are below the $\alpha$-th level. In addition, the concept of "infinite tree" is not defined explicitly, but it is interpreted by the context (for example, the last sentence of the proof), that they refer to trees on an arbitrary infinite cardinal.

My first problem is that I don't see how this proof (from the second paragraph onwards) fails for trees with non-finite levels. For example, if $|T|=\omega_1$ and the levels are countable, I think we could still say "pick $t_0\in T_1$ such that $t_0$ has $\omega_1$-many successors in $T$, there must be one since $|T|=\omega_1$ and $|T_1|=\omega$". And similarly for the $n$-th step. Therefore, this would be against Aronszajn's theorem (this is, there exist an Aronszajn tree).

Secondly, I thought that the critical step for this prove was the limit step, but only the successor step is taken into account during the proof...

Regarding the two previous issues, how does this proof work? Thanks in advance!!

Answer 1

The point is that the proof is only trying to build an infinite path, not a path of length $\omega_1$. If you want a path of the latter type you need to figure out how to "continue past $\omega$," but if you want a path of the former type you just need to figure out how to keep going for each natural number.

In particular (and re: your proposed counterexample), the following is true: if $T$ is an uncountable tree with each level countable, then $T$ has an infinite path. However, this doesn't contradict the existence of Aronszajn trees, since the requirement for the latter is that they have no uncountable (not just infinite) path. An Aronszajn tree $T$ does have lots of infinite paths, it's just that all the infinite paths through such a $T$ are "much shorter than" $T$ itself.

Answer 2

The original objective of this post was to ask for clarity on the following theorem:

Theorem : if $\langle T,<_T\rangle$ is a tree of height $\kappa$, (with $\kappa$ any infinite cardinal) whose levels are finite, then, it has an infinite branch (with length $\kappa$).

It turned out that, as Noah rightly pointed out, this was not the theorem being proved in Drake's reference. However, I think it is convenient to present a proof of the above theorem. It is worth mentioning that there are some more general forms of this theorem, in which "finite" can be substituted for a cardinal sufficiently smaller than $\kappa$.

Proof: pick $t_0$ at the first level of the tree such that $t_0$ has $\kappa$-many successors in $T$. This element must exist, because the tree has height $\kappa$. Now, for any $\alpha<\kappa$, if $t_\alpha\in T$ has $\kappa$-many successors, we can pick some $t_{\alpha+1}$ at level $\alpha+1$ such that $t_{\alpha+1}$ has $\kappa$-many successors. This is possible because $t_\alpha$ has $\kappa$-many successors and the level $\alpha+1$, by hypothesis, is finite (the argument is the same as for $t_0$). Now, let's consider that $\alpha<\kappa$ is a limit ordinal and we have a chain $C=\{t_\beta\in T:\beta<\alpha\}$ where $t_\beta$ is at level $\beta$, $t_{\beta_1}<t_{\beta_2}$ if $\beta_1<\beta_2$ and each element has $\kappa$-many successors. Since the level $\alpha$ is, by hypothesis, finite, we can enumerate its elements: $\{s_0,\dots,s_{n-1}\}=A$. And again, it is clear that there is some $s_i$ with $\kappa$-many successors. But we need to prove that one of those is bigger than $t_\gamma$ for any $t_\gamma\in C$. We can assume, towards a contradiction, that for any $s_i'\in\{s\in A:s \text{ has $\kappa$-many successors}\}=\{s_0',\dots,s_{m-1}'\}=B$, $s_i'$ is not successor of all the elements in $C$. Let's write $P(s_i')=\{t\in T:t<_Ts_i', \ s_i'\in B\}$. By assumption, there must be some element $t_{\beta_i}$ such that $t_\beta<t_{\beta_i}$ for any $t_\beta\in P(s_i')\cap C$ (if not, then $s_i'$ is bigger than any element in $C$, which is not our assumption). For any $s_i'\in B$ we can pick an element like that and consider $\max_T\{t_{\beta_0},\dots,t_{\beta_{m-1}}\}=t'\in C$. Therefore, for any $t_\gamma\in C$ such that $t'<t_\gamma $, there are two options: $t_\gamma$ cannot be a predecessor of any $s_i'\in B$, because $t_\beta<_Tt'$ for any $t_\beta\in P(s_i')\cap C$ and any $s_i'\in B$; $t_\gamma$ cannot be a predecessor of any $s_i\in A\backslash B$, because $t_\gamma$ has $\kappa$-many successors and the elements in $A\backslash B$ don't satisfy that. Therefore, there are elements in $C$ which are not predecessors of any element in $A$, but this is absurd, because any element in $A$ must have $\kappa$-many successors. So, there is some element in $B$ which is successor of every element in $C$.

Finally, this proves that we can build an infinite branch with length $\kappa$.