Timeline for Number of paths through infinite trees with given "growth rates"
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 26, 2021 at 15:31 | vote | accept | usul | ||
| Nov 7, 2014 at 11:45 | answer | added | Emil Jeřábek | timeline score: 6 | |
| Nov 7, 2014 at 3:50 | comment | added | usul | Thanks to all the commenters for the points/clarifications! I think at this point that Joel David Hamkins' answer covers most of the interesting possibilities of the question, other than talking about different structures of paths which is probably out of scope. | |
| Nov 6, 2014 at 23:48 | comment | added | David Handelman | The path space is a separable compact totally disconnected space; hence it is a Cantor set (no atoms), or is atomic (the second case above), or is the product of a Cantor set with an atomic one (the latter possibly finite). Only in the atomic case is the path space countable, and it should be possible to decide precisely when this occurs. | |
| Nov 6, 2014 at 23:08 | comment | added | Emil Jeřábek | That’s right. I didn’t intend my comments to be positive or negative, but rather to make the OP think more about the question and clarify what he or she really wants. | |
| Nov 6, 2014 at 22:55 | comment | added | Joel David Hamkins | The comments are correct, but there are some positive things to say, aren't there? For example, very high growth rate, near $2^n$, seems to force the tree to have $2^\omega$ many branches. We can interpret the question as: what can one say about sufficient rates of growth to ensure continuum many branches? For example, probably even superpolynomial growth does not suffice? | |
| Nov 6, 2014 at 22:51 | answer | added | Joel David Hamkins | timeline score: 13 | |
| Nov 6, 2014 at 22:31 | comment | added | S. Carnahan♦ | Another way to phrase Emil Jeřábek's first comment is that for any growth function $f$, there is a contracting homotopy from a "stretched" tree, given by inserting vertices of degree 2 so that only one branch appears at each level, that induces a bijection on the set of infinite paths. | |
| Nov 6, 2014 at 22:28 | comment | added | Emil Jeřábek | Also, I’m not sure how you intended the reference to the continuum hypothesis, but anyway the number of paths in any such tree is either countable or $2^\omega$ by the Cantor–Bendixson theorem. | |
| Nov 6, 2014 at 22:16 | comment | added | Emil Jeřábek | The growth rate by itself doesn’t determine the number of paths. You can easily construct a tree with growth rate $f(n)=n+1$ and $2^\omega$ paths by choosing more liberally who gets the new child at each level. | |
| Nov 6, 2014 at 22:05 | history | asked | usul | CC BY-SA 3.0 |