Timeline for A weak fragment of analysis?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Jun 9, 2018 at 8:06 | vote | accept | Alex Gavrilov | ||
| Jun 6, 2018 at 12:38 | comment | added | user44143 | Any theory of analysis should be able to state the Navier-Stokes equations, and the Clay question about their solutions. So even without sets and numbers, I don’t think any theory of analysis will be decidable. | |
| Jun 6, 2018 at 8:20 | history | edited | Alex Gavrilov | CC BY-SA 4.0 |
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| Jun 6, 2018 at 8:14 | history | edited | Alex Gavrilov | CC BY-SA 4.0 |
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| Jun 5, 2018 at 16:11 | answer | added | Emil Jeřábek | timeline score: 5 | |
| Jun 5, 2018 at 13:34 | comment | added | Alex Gavrilov | This sounds interesting. Can you write it as an answer (to one of the questions)? | |
| Jun 5, 2018 at 13:31 | comment | added | Emil Jeřábek | (1) It is undecidable even for formulas without set quantifiers. (By which I assume you mean that there may be free set variables, implicitly universally quantified outside. Otherwise there is no way to refer to sets whatsoever.) (2) The only assumptions on the theory I need are that it includes the theory of ordered rings, and that it does not disprove the existence of standard sets of the form $\{0,1,2,\dots,42\}$. Even this can be weakened (e.g., I don't care if the set contains other garbage outside the interval $[0,42]$.) | |
| Jun 5, 2018 at 13:20 | history | edited | Alex Gavrilov | CC BY-SA 4.0 |
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| Jun 5, 2018 at 13:03 | comment | added | Noah Schweber | @MikhailKatz Building off of Emil's comment, note that even Robinson arithmetic is undecidable, indeed essentially undecidable. | |
| Jun 5, 2018 at 12:56 | history | edited | Alex Gavrilov | CC BY-SA 4.0 |
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| Jun 5, 2018 at 11:44 | history | edited | Alex Gavrilov | CC BY-SA 4.0 |
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| Jun 5, 2018 at 10:35 | comment | added | Emil Jeřábek | @MikhailKatz All theories used in reverse mathematics include fairly strong fragments of arithmetic, and are undecidable as well. | |
| Jun 5, 2018 at 10:31 | comment | added | Emil Jeřábek | The thing is, the formula $N(x)$ given as $\forall X\,(0\in X\land\forall y\,(y\in X\to y+1\in X)\to x\in X)$ defines natural numbers. It is not hard to show this way that any consistent theory in your language that includes the theory of ordered rings and comprehension for quantifier-free formulas is undecidable. I have no idea how to avoid this. | |
| Jun 5, 2018 at 10:21 | comment | added | Mikhail Katz | Following Friedman, Simpson, and others working in reverse mathematics, there are now many ways of developing weak systems where fragments of analysis and other fields can be handled. It may be helpful to reorient your question in this context, and also update the relevant tags. | |
| Jun 5, 2018 at 9:03 | history | edited | Alex Gavrilov | CC BY-SA 4.0 |
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| Jun 5, 2018 at 8:44 | comment | added | Alex Gavrilov | Not a problem, just declare the whole ${\mathbb R}^n$ a set. (For each $n$ . By the way, I did not mean to list all the axioms, so it is not a precise description of a theory anyway.) | |
| Jun 5, 2018 at 8:20 | comment | added | Emil Jeřábek | With these axioms alone, you cannot prove the existence of any set at all. (And if you add the axiom that a set exists, you cannot prove the existence of any nonempty set.) So, clearly you need more than that. | |
| Jun 5, 2018 at 7:57 | history | asked | Alex Gavrilov | CC BY-SA 4.0 |