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My question is very imprecise, as I know very little about descriptive set theory.

In a problem I am thinking about I have a family of well-founded trees (finite sequences on $\cup_n X^n$ closed under taking predecessors, such that every path has a maximal element) on a Polish space (separable, completely metrizable) space. These trees have several additional proprieties/constraints. I am trying to construct for each $\alpha<\omega_1$ a tree that has order at least $\alpha$. The order of the tree is defined inductively by removing maximal elements until nothing is left. The least ordinal for which this happens is the order of the tree.

My questions, as I said very imprecise, is what properties of a tree imply that the order is larger than a prescribed ordinal? Are there any standard techniques of constructing such trees? I know this is a long shot as I have several other constraints on these trees, but if there is some relevant literature where I can start looking, that would be appreciated. I know of the standard reference of Kechris book, but I couldn't find something that I can use (or perhaps I did not know where to look in that book).

Thank you and if you think this question is too vague, please just delete it.

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    $\begingroup$ Trees of any order $\alpha < \omega_1$ can easily be constructed by induction: For the successor step assume there is a tree $T$ of order $\alpha$ and define $T'$ by simply putting a new root below $T$. Then $T'$ is well-founded and has order $\alpha+1$. Similar for the limit case $\lambda$: Since $\lambda < \omega_1$ there is a sequence $(\alpha_n)_{n \in \omega}$ such that $\lambda= \sup_{n \in \omega} \alpha_n$ and there are trees $T_n$ of order $\alpha_n$. Define $T'$ simply by putting the $T_n$'s side by side and a new root below them. Again, $T'$ is well-founded and has order $\lambda$. $\endgroup$
    – Johannes Schürz
    Commented Jul 10, 2020 at 21:14
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    $\begingroup$ @JohannesSchürz I apologize, could you please explain in a bit more detail what you mean by "a new root below $T$". These trees are all single rooted at $\emptyset$. Perhaps I understand something different by a root? $\endgroup$
    – MariaM
    Commented Jul 11, 2020 at 2:19
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    $\begingroup$ The order of a tree, as you call it, is more commonly known as its rank, at least among descriptive set theorists. You can find the definition in section I.2.E (Well-founded trees and ranks) in Kechris's book. Perhaps it will help you to know that the rank of a tree $T$ is completely determined by the partial order $\prec_T$ defined on $T$ by $s\prec_T t$ iff $s\supsetneq t$. Thus the rank of $T$ is fairly robust to changes on $T$ itself. $\endgroup$
    – Paul McKenney
    Commented Jul 19, 2020 at 17:21
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    $\begingroup$ As for how you put "a new root below $T$": pick an arbitrary point $x_0$ in $X$, and define $T' = \{\emptyset\} \cup \{\langle x_0 \rangle^\frown s \; | \; s\in T\}$. Similarly, to put a new root below several trees at once, you'll need to pick an arbitrary point for each of them. $\endgroup$
    – Paul McKenney
    Commented Jul 19, 2020 at 17:26
  • $\begingroup$ @MariaM The above "$T$ can be constructed by appending roots via transfinite induction up to $\alpha$" implies that the rank of $T$ is less than an ordinal $\alpha$, not greater. In order to give a better picture of which kind of conditions you're looking for (e.g. Ramsey-theoretic, graph-theoretic, etc), do go have examples of some of the several additional properties/constraints? $\endgroup$
    – C7X
    Commented Oct 6, 2022 at 10:14

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