Skip to main content
11 events
when toggle format what by license comment
Apr 13, 2023 at 15:48 comment converted from answer Andreas Weiermann I think being able to prove that a lower bound for the resulting order type is given by $\varphi\omega0$. It is not clear whether one can reach $\Gamma_0$ at the end.
Apr 6, 2023 at 15:22 comment converted from answer Andreas Weiermann I think that the suggested problem is very interesting. My guess would be that the resulting order type (after Harry's modification) should be about $\Gamma_0$.
Mar 15, 2023 at 14:37 answer added Harry Altman timeline score: 1
Dec 31, 2011 at 21:52 vote accept Vladimir Reshetnikov
Dec 7, 2011 at 21:18 comment added Harry Altman Yikes -- not certain my proof of the above comment is correct, or that the comment is correct. Same applies where I stated the same in my answer below, of course. Of course, since I've since solved the problem (as stated), it's kind of irrelevant...
Dec 4, 2011 at 0:51 comment added Harry Altman Hm... the obvious way to contradict well-ordering would be to construct the predecessor function, but this is impossible; for any $f\in S$, $\mathbb{N}$ can be split into two sets A and B such that f is bounded on A and satisfies $f(n)\ge n$ on B. (Omitting proof because once you think to make the statement, it's straightforward.)
Dec 3, 2011 at 23:49 comment added Joel David Hamkins Harry, I would think that for a lower bound of $\alpha$ Vladimir would simply want us to exhibit a collection of functions having order type $\alpha$ under eventual domination. And the bigger the $\alpha$, the more nontrivial the bound.
Dec 3, 2011 at 22:18 answer added Harry Altman timeline score: 7
Dec 3, 2011 at 21:26 comment added Harry Altman I think it's obvious what would count as a nontrivial upper bound, but what counts as a nontrivial lower bound?
Dec 3, 2011 at 20:35 history edited Vladimir Reshetnikov CC BY-SA 3.0
added 85 characters in body
Dec 1, 2011 at 19:06 history asked Vladimir Reshetnikov CC BY-SA 3.0