On the existence of scattered Suslin trees Assume that there exists a Suslin tree $S$ (that is, an uncountable tree containing neither uncountable chains or antichains). For each $s\in S$, denote $V_s=\{p\in S: p\geq s\}$. Consider $S$ to have the topology given by basic neighborhoods of the form:
$$
V_s\setminus\left(\bigcup_{p\in F} V_p\right),
$$
where $F$ is a finite set of successors of $s$. This is usually called the wedge topology. I would like to know about the possibility of $S$ being scattered.
I started by using Proposition 4.2 of the paper Aurichi, L. F., & Dias, R. R. (2019). A minicourse on topological games. Topology and its Applications, 258, 305-335. which states that $S$ must be Lindelöf. Then, I proved that, since $S$ admits elements with height arbitrarily close to $\omega_1$, the Cantor-Bendixson derivative $S^{(\alpha)}$ is non-empty for any $\alpha<\omega_1$. Then, I noticed that $(S^{(\alpha)})_{\alpha<\omega_1}$ is a decreasing net of closed subsets of $S$ and used the fact that $\omega_1$ has uncountable cofinality to conclude that $(S^{(\alpha)})_{\alpha<\omega_1}$ has the countable intersection property, so, since $S$ is Lindelöf,
$$
S^{(\omega_1)} = \bigcap_{\alpha<\omega_1} S^{(\alpha)} \neq \emptyset.
$$
Now, I would like to conclude that if $S$ were scattered, then it would happen that any point in $S^{(\omega_1)}$ must have height greater than or equal to $\omega_1$, which would generate a contradiction. I'm still stuck in this part.
I would appreciate if someone had any suggestion on how to go through, or even if someone knows another proof or a counterexample for the result that I'm seeking.
 A: I found the answer to this question on Lemma 9.3 on Jech's book "Set Theory". He proves that, if there exists a Suslin tree, then there exists a normal Suslin tree. It turns out that normal Suslin trees are dense-in-itself, and here is why.
The definition of normal trees has to do with a list of five or six properties, from which we highlight:

*

*Each non-maximal $x\in S$ has infinitely many successors;


*For each $x\in S$ and any ordinal $\beta$ satisfying ${\rm height}(x)<\beta<{\rm height}(T)$, there exists $y> x$ with height $\beta$.
Considering a normal Suslin tree, since ${\rm height}(x)<{\rm height}(S)$ for any $x\in S$, it follows by 2) that $S$ has no maximal elements. Therefore, by 1), any $x\in S$ has infinitely many successors. But notice that the isolated points in the wedge topology are precisely the ones that have a finite amount of successors. Therefore, $S$ has no isolated points, so it is dense-in-itself.
So far, this only proves that there exists non-scattered Suslin trees (the normal ones). But actually, by the construction in the proof of Jech's Lemma, we conclude that any Suslin tree contains a Suslin tree satisfying 1) and 2) above. So it actually follows that no Suslin tree endowed with the wedge topology is scattered.
