Timeline for Trees of prescribed ordinal
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| Oct 6, 2022 at 10:14 | comment | added | C7X | @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? | |
| Jul 19, 2020 at 17:26 | comment | added | Paul McKenney | 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. | |
| Jul 19, 2020 at 17:21 | comment | added | Paul McKenney | 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. | |
| Jul 11, 2020 at 2:19 | comment | added | MariaM | @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? | |
| Jul 10, 2020 at 21:14 | comment | added | Johannes Schürz | 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 $\lambda$. | |
| Jul 10, 2020 at 17:05 | history | edited | MariaM |
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| Jul 10, 2020 at 16:26 | history | asked | MariaM | CC BY-SA 4.0 |