Timeline for Can we prove the epsilon theorems without the axiom of choice?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Mar 10, 2018 at 1:23 | comment | added | Noah Schweber | @AndreasBlass Either way, the mosquito is dead. | |
| Mar 10, 2018 at 1:02 | comment | added | Andreas Blass | @NoahSchweber Oops; I intended to write "not" before "very". I agree it's overkill, but I wanted to point out that it could get even worse. | |
| Mar 9, 2018 at 16:05 | comment | added | Noah Schweber | @AndreasBlass Oh yes, I never intended to imply it wasn't. (But SA is even worse!) | |
| Mar 9, 2018 at 16:01 | comment | added | Andreas Blass | @NoahSchweber Arithmetical absoluteness is very massive overkill. I've seen people invoke Shoenfield absoluteness in similar situations. | |
| Mar 9, 2018 at 8:36 | comment | added | Asaf Karagila♦ | @Joel: Your comment here matches your answer on the question I mention in my first comment. | |
| Mar 9, 2018 at 8:35 | comment | added | Emil Jeřábek | It is reasonably clear from the linked SEP article that the original proofs were finitistic (that was kind of the whole point of the enterprise), so they wouldn’t use the axiom of choice in the first place. | |
| Mar 9, 2018 at 2:36 | comment | added | Noah Schweber | @PyRulez Well, you can't have that unless your language is uncountable. But actually, since "$\vdash$" is compact, even that doesn't matter: a statement of the form "$\Gamma\vdash\varphi$" is absolute, regardless of how big $\Gamma$ is. | |
| Mar 9, 2018 at 2:33 | comment | added | Christopher King | @NoahSchweber what if it's even greater than the cardinality of the reals? | |
| Mar 9, 2018 at 2:32 | comment | added | Noah Schweber | @PyRulez That corresponds to a real parameter, and doesn't affect anything. | |
| Mar 9, 2018 at 1:09 | comment | added | Christopher King | @NoahSchweber actually, wait mind. What about an uncountable set of formulae? | |
| Mar 9, 2018 at 1:05 | comment | added | Joel David Hamkins | Since you are asking about set-theoretic uses of $\varepsilon$, it seems natural to interpret $\varepsilon$ in the context of second-order set theory as providing a global choice function. In this case, if one has the comprehension axiom in the language with $\varepsilon$, then it would not be provable in GB+AC, since it would be equivalent to global choice. For example, see jdh.hamkins.org/…. | |
| Mar 9, 2018 at 0:47 | comment | added | Christopher King | @NoahSchweber oh, I forgot about absoluteness. Sorry. | |
| Mar 9, 2018 at 0:33 | comment | added | Noah Schweber | I'm a bit confused. The epsilon theorems are about derivability in a fixed formal system; these are purely combinatorial (or broadly number-theoretic) facts, and have nothing to do with the axiom of choice. (Massive overkill: the epsilon theorems are arithmetic statements, and we have absoluteness between $V$ and $L$ for such statements in ZF ...) So is this really what you're asking? | |
| Mar 9, 2018 at 0:20 | comment | added | Christopher King | @AsafKaragila fixed | |
| Mar 9, 2018 at 0:20 | history | edited | Christopher King | CC BY-SA 3.0 |
edited body
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| Mar 9, 2018 at 0:14 | comment | added | Asaf Karagila♦ | Also, in the second inference rule you have $p(x)$ and $q(x)$, and then $p(x)$ and $p(y)$. Something is off... | |
| Mar 8, 2018 at 22:51 | comment | added | Asaf Karagila♦ | Also mathoverflow.net/questions/281301/… | |
| Mar 8, 2018 at 22:40 | history | asked | Christopher King | CC BY-SA 3.0 |