Timeline for In NBG (ZFC + classes) if $P$ nonempty predicate of classes must $P$ have definable solution?
Current License: CC BY-SA 3.0
12 events
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| Jun 2, 2017 at 15:29 | comment | added | Joel David Hamkins | You may find the discussion of truth predicates in KM on my blog to be useful: see jdh.hamkins.org/km-implies-conzfc. | |
| Jun 2, 2017 at 15:28 | comment | added | Joel David Hamkins | KM proves that there is a class $T$ that is a truth predicate for first-order truth. The property $\varphi(T)$ of $T$ being such a predicate involves only first-order quantification. So KM proves $\exists T\, \varphi(T)$, but there can be no first-order definable class $X$ with $\varphi(X)$, since that would violate Tarski's theorem. So since you indicate in your revised question that you want to work now in KM rather than NGB, the answer to your question is no, there is always a first-order expressible property that holds of a class, but does not hold of any definable class. | |
| Jun 2, 2017 at 5:17 | comment | added | Peter Gerdes | But if I was wrong about that notation I want to make sure I got what matters: the proof right. I understand to build $P$ with only set quantification in KM so it admits no classes definable via set quantification we can't just use the truth predicate for a model $M$ of KM since that truth predicate would need to use class quantification. Rather you mean to give a truth predicate for $V$ (the sets in $M$) as that gives a definition for $P$ with only set quantification and then any definable via set quantification $C$ with $P(C)$ would give a truth predicate for $V$. | |
| Jun 2, 2017 at 5:10 | comment | added | Peter Gerdes | Because you called the $\Sigma^1_n$ level classes and I assumed you did that because you took that to be a synonym for set. But I was wrong. You did that because you regard the integers quantified over in $\Sigma^0_n$ to be sets rather than simply brute integers (e.g. as in arithmetic with a set sort added) as I'm useed to regarding them. | |
| Jun 2, 2017 at 4:19 | comment | added | Joel David Hamkins | I usually use NBG as a two-sorted theory, and I'm not sure why you thought I use it as one sorted. But it doesn't really matter, since the usual notation for $\Sigma^0_n$ and $\Sigma^1_n$ doesn't depend on that. The superscript $0$ always means first-order, quantification over sets, and the superscript $1$ always means second-order, quantification over classes, whether or not one organizes it into one sort or two. This parallels the usage in arithmetic, where first-order means quantification over numbers and second-order means quantification over sets of numbers, or equivalently, over reals. | |
| Jun 2, 2017 at 4:13 | comment | added | Peter Gerdes | I agree with you if you are working in a theory where set and class are part of the same sort but one at least CAN present NBG where they are two sorts...kinda ugly but that's what I was presuming and then you are moving one type up..hence the motivation.. But yes, I presumed it was wrong...but I still have no idea if there is any notation that distingushes between possibly proper class quantifiers and set (for you non-proper class) quantifiers). | |
| Jun 2, 2017 at 4:12 | comment | added | Peter Gerdes | Also I see where much of the confusion in notation/question came from. I was taking NBG to be a two-sorted theory with separate sorts for class and set while you were talking in terms of (as I now convinced is more elegant) a 1 sorted theory. You are right the second question was essentially what I was after (your answer covers the one minor difference). | |
| Jun 2, 2017 at 4:06 | comment | added | Joel David Hamkins | Well, deliberate or not, I had meant that your notation $\Sigma^1_n$ and $\Sigma^2_n$ is definitely wrong with respect to established usage for your stated meaning. In the usual notation for second- and higher-order set theory, $\Sigma^2_n$ would be understood to refer to quantification over collections of classes, that is, third order set theory. In various set-theoretic contexts, one sees $\Sigma^m_n$, and this notation is fairly universally agreed upon. For this reason, you may want to edit your question to use the standard notation, since otherwise it is confusing. | |
| Jun 2, 2017 at 3:38 | vote | accept | Peter Gerdes | ||
| Jun 2, 2017 at 3:23 | comment | added | Peter Gerdes | Yes, they are DELIBERATELY off by 1 from the usual notion because you go from 0 to 1 when you switch from number to set quantifier and here I switched again to proper class quantifier. | |
| Jun 2, 2017 at 0:37 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 446 characters in body
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| Jun 2, 2017 at 0:31 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |