Timeline for truth vs. provability for ordered fields
Current License: CC BY-SA 3.0
25 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Aug 19, 2011 at 4:14 | comment | added | James Propp | @Ricky: You're absolutely right. (I've educated myself a bit about set theory in the two weeks since you posted this comment.) | |
| Aug 1, 2011 at 21:49 | comment | added | user5810 | Davies uses $\omega < \operatorname{cof}(\omega_1)$ to prove that "$\gamma$ must be an ordinal corresponding to a countable set". ZF does not prove that a countable union of countable sets must be countable. | |
| Jul 29, 2011 at 16:10 | comment | added | James Propp | @Ricky: I don't see where Davies' argument (or my recasting of it) uses $\omega < \operatorname{cof} (\omega_1)$; can you be more specific? Also, I have a hard time imagining how this inequality could be independent of ZF. After all, $\omega_1$ is just the set of all countable ordinals. If $\omega_1$ had a countable cofinal subset, we'd have a countable list of countable ordinals whose limit was $\omega_1$. But the union of those countable ordinals is countable; contradiction. Is this argument not formalizable in ZF? Or am I missing something more basic here? | |
| Jul 29, 2011 at 15:58 | comment | added | James Propp | @Ricky: Right you are (about convergence of sequences of Laurent series)! I'll be sure to fix this. | |
| Jul 27, 2011 at 17:40 | comment | added | user5810 | $\langle \epsilon^{-1},\epsilon^{-2},\epsilon^{-3},...\rangle$ does not converge, but does satisfy the definition you gave in the article. I think Davies is assuming $\omega < \operatorname{cof}(\omega_1)$ without saying so (it follows from Countable Choice but is not a theorem of ZF). | |
| Jul 27, 2011 at 15:24 | vote | accept | James Propp | ||
| Jul 27, 2011 at 15:17 | comment | added | James Propp | @Bennet: Thanks for catching the typo; I'll add the $(-1)^n$ to the next draft. | |
| Jul 27, 2011 at 15:12 | comment | added | James Propp | @Ricky: My remarks on page 12 about NIP for surreals were based on Riesenberg and Davies' solution to Monthly problem 5112, which predates Conway's work (see jamespropp.org/surreal-NIP.pdf ); I was trying to adapt their solution by putting it into the surreal number context, where I suspect it belongs. Do you think Riesenberg and Davies' argument is wrong, or is it my attempt at "surrealifying" their argument that's faulty? | |
| Jul 27, 2011 at 15:07 | comment | added | James Propp | I'm delighted by the generous outpouring of comments! First some replies to Ricky's: #1: "Much of a muchness" isn't a typo, but it is obscure; it means "difficult to distinguish". #2: Isn't your definition of convergence of a sequence of Laurent series equivalent to the definition that I give in my article? It seems to me that the only difference is that you lump together the terms with negative exponents and I don't. If you think that they're not equivalent, please explain. #3: Thanks for catching the mistake on page 9. | |
| Jul 27, 2011 at 14:17 | comment | added | Gerald Edgar | From the answers here we see: for such an article as yours, it is best not to mention "provable" at all. Just say: "Every ordered ring R satisfying property P satisfies property P′" and leave it at that. | |
| Jul 27, 2011 at 12:20 | answer | added | Carl Mummert | timeline score: 3 | |
| Jul 26, 2011 at 20:47 | comment | added | Mark Bennet | Your series in the alternating series test (p2) doesn't alternate (as written) | |
| Jul 26, 2011 at 20:43 | comment | added | Kaveh | This is different from Platonist/realist notion of truth. If we think of Platonic/real universe of sets ($V$), then every first order sentence including CH is either true or false. ZFC is not capturing the Platonic/real universe of sets so the notion of Platonic truth does not coincide with the notion of provability in ZFC. | |
| Jul 26, 2011 at 20:42 | comment | added | Kaveh | A small note about CH: AFAIK, a provability only makes sense with respect to a logical system/theory (e.g. w.r.t. ZFC). When we say CH is not provable we mean that CH is not provable in first-order logic from ZFC. We also know that CH is not true in ZFC, i.e. there are models of ZFC in which CH is not true and therefore it is also not provable by the completeness theorem of first order logic (I am assuming that ZFC is consistent). | |
| Jul 26, 2011 at 20:39 | answer | added | François G. Dorais | timeline score: 11 | |
| Jul 26, 2011 at 20:28 | comment | added | user5810 | On page 12, I strongly suspect that NIP for surreals "created on or before day $\omega_1$" is equivalent to $\omega < \operatorname{cof}(\omega_1)$, which is not a theorem of ZF. | |
| Jul 26, 2011 at 20:20 | comment | added | user5810 | I don't believe you explicitly say that completeness will mean Dedekind completeness. I don't know if the first sentence in the second paragraph on page 10 holds for en.wikipedia.org/wiki/Uniform_space#Completeness. | |
| Jul 26, 2011 at 20:19 | answer | added | Kaveh | timeline score: 1 | |
| Jul 26, 2011 at 20:10 | comment | added | user5810 | On page 9, $f$ isn't even defined at $0_R$, so it can't be continuous on $R$. | |
| Jul 26, 2011 at 20:06 | comment | added | user5810 | On page 7, a sequence of Laurent polynomials converges if and only if the principal (en.wikipedia.org/wiki/Principal_part#Laurent_series_definition) part stabilizes and for every non-negative integer $n$, the sequence of coefficients of $\epsilon^n$ stabilizes. | |
| Jul 26, 2011 at 19:55 | comment | added | user5810 | Apparent typo on page 2: "much of a muchness"? | |
| Jul 26, 2011 at 19:53 | comment | added | Gerhard Paseman | No they are not. Consider rephrasing: every ring satisfies P implies P'. Then P implies P' is a true statement, but the chosen axiom system may lead to a logically incomplete theory, and the above statement may lie outside that theory. I think the general theory of rings may be undecidable, while certain extensions such as alg. Closed fields of a given characteristic are complete, and hence decidable. Gerhard "Ask Me About System Design" Paseman, 2011.07.26 | |
| Jul 26, 2011 at 19:46 | comment | added | Qiaochu Yuan | Well, on the one hand, this is true if $P, P'$ are both first-order by the completeness theorem. On the other hand, you seem to be dealing with properties that aren't first-order... | |
| Jul 26, 2011 at 19:17 | history | asked | James Propp | CC BY-SA 3.0 |