How can an $\omega_1$-tree possibly be normal and yet not have any $\omega_1$-branch?

Question

An $\omega_1$-tree is a tree of height $\omega_1$.

An $\omega_1$-tree $T$ is normal if:

1. $T$ has a unique least point (the root);
2. every level of $T$ is at most countable;
3. if $x \in T$ then there are infinitely many nodes at the successor level;
4. if $x \in T$ then there is $y>x$ at each higher level less than $\omega_1$;
5. the order $<$ is extensional within each level $U_\gamma$ such that $\gamma < \omega_1$ is a limit ordinal, that is: for all $x,y \in U_\gamma$, if $\{z \in T \mid z<x\} = \{z \in T \mid z<y\}$ then $x=y$.

A Suslin tree is an $\omega_1$-tree whose antichains are at most countable and which has no $\omega_1$-branch.

Fact. If there exists a Suslin tree, then there exists a normal Suslin tree.

Question: I can't understand how this can possibly be true. How can an $\omega_1$-tree satisfy condition (4) and still have no $\omega_1$-branch? Doesn't condition (4) trivially allow us to build an $\omega_1$-branch by recursion? Where does this very easy proof fail?

Attempt of proof. Define $f \colon \omega_1 \to T$ strictly increasing by recursion as follows: $f(0)$ is the root, $f(\alpha+1)$ is one arbitrarily chosen immediate successor of $f(\alpha)$. For the limit case (suppose $\gamma$ is limit) I claim that there is a node $x$ at level $\gamma$ such that

$$f(\alpha) < x \text{ for all } \alpha < \gamma.$$

Assume towards a contradiction that $\beta < \gamma$ is such that there exists no $x$ at level $\gamma$ such that $f(\alpha) < x$ for all $\alpha \leq \beta$. But by condition (4) we have that there exists $y>f(\beta)$ at level $\gamma$. But then $y$ is greater than any $f(\alpha)$ for $\alpha \leq \beta$, which is a contradiction.

Hence define $f(\alpha) = x$. Then $\langle f(\alpha) \mid \alpha < \omega_1 \rangle$ is an $\omega_1$-branch in $T$.

Definitions are taken from chapter 9 of Set theory by T. Jech.

Answer 1

Straightforward use of condition (4) would allow you to build an $\omega$-chain in $T$: $t_0 < t_1 < \cdots < t_n < \cdots$. The problem, then, is extending this to an $(\omega+1)$-chain.

While we can take $\alpha \geq \sup \{ \operatorname{ht}(t_n) : n \in \omega \}$ and find an extension of each $t_n$ on the $\alpha$th level of $T$, there is no way to guarantee that a single node on the $\alpha$th level extends each $t_n$.

Consider the tree $$T = \{ t \in \omega^{<\omega_1} : t\text{ has finitely many }0\text{s} \}.$$ This is fairly easily seen to be a normal $\omega_1$-tree, using the end-extension order. (This tree also obviously has $\omega_1$-chains, but I think it will be instructive nonetheless.) If $t_n = \overbrace{0\ldots 0}^{n\text{ zeroes}}$ for each $n$, then clearly we have constructed an $\omega$-chain in $T$. Also, each $t_n$ has an extension on the $\omega$th level of $T$. However no node on the $\omega$th (or any higher) level of $T$ extends each $t_n$, since such a node would have to contain infinitely many zeroes.

Answer 2

Every such attempt to construct a branch necessarily dies out. The best approach is probably to look at some actual constructions. This PDF on Aronszajn trees gives a couple of constructions in very full detail. The construction in this PDF should also be helpful.

Answer 3

Every node has a continuation. But when you get to a limit level not every branch which "tickles" the limit level gets realized.

The whole idea in the usual constructions of normal Suslin trees is that we use some combinatorial principle to guess in advance which branches can be realized along the way without threading a cofinal branch.