Timeline for Proper classes and their consequences
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| May 22 at 16:49 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
http -> https (the question was bumped anyway)
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| Jan 11, 2012 at 6:47 | comment | added | The Mathemagician | Addendum: Before the set theorists and logicians jump all over me,I should make a clarifying qualification: I'm referring of course to the finite axiomization of NGB using Godel class function constructions. The infinite axomatization is quite a bit more problematic as a first order logical theory. It might be workable with properly chosen axioms,but the finite axiom version seems to be the more conservative extension of ZFC and I'm more comfortable with that,as many mathematicians should be. | |
| Jan 11, 2012 at 6:36 | comment | added | The Mathemagician | To be honest,I've never thought this approach is a huge improvement over simply including proper classes and being done with it. NBG set theory shows how this can be done in a straightforward-and careful- manner with the result of a model that is equivalent to ZFC but without it's limitations.I can't see how Grothendieck universes represent an improvement over this-to me,it looks more like terminological slight-of-hand that doesn't really improve the ontology for foundational discussions. And isn't that really why mathematicians construct axiomatic set-theories in the first place? | |
| Dec 6, 2010 at 7:56 | comment | added | Jason | As a point of clarification, $\aleph_0$ is technically not an inaccessible cardinal since it is not uncountable. The reason this uncountability condition is in place is because inaccessibility is considered a large cardinal notion, and it should be relatively consistent with ZFC that any given large cardinal does not exist. Since the existence of $\aleph_0$ is provable from ZFC, it should not be characterized as a large cardinal. However, as noted, it could be considered a large cardinal in the absence of the axiom of infinity. | |
| Dec 6, 2010 at 1:43 | comment | added | Harry Gindi | @Adam: I've heart such universes called "uncountable universes". | |
| Dec 6, 2010 at 0:08 | comment | added | Adam | By the way, as defined on the Wikipedia page a "universe" is NOT a model of ZFC -- it fails the Axiom of Infinity. I believe the standard definition is the one given by Bergman in definition 6.4.1 of his book (avaliable online here: math.berkeley.edu/~gbergman/245/Ch.6.ps ). In that definition, "$\omega\in U$" is part of the definition, ensuring that every universe is a transitive model of ZFC. | |
| Dec 6, 2010 at 0:02 | comment | added | Adam | (some people use ZFCU for ZFC with a weak extensionality axiom which allows for "urelemente"; others use ZFCA (ZFC with Atoms) for this) | |
| Dec 6, 2010 at 0:00 | comment | added | Adam | @Harry: if by ZFCU you mean ZFC+"there exists a set which is a universe" then they are strictly linearly ordered by consistency strength: $Con(ZFCU)\Rightarrow Con(ZFC)\Rightarrow Con(ZFC^{\text{fin}})$. The requirement "must contain $\omega$" is usually part of the definition "is a universe", although @Greg didn't list it above; this accounts for the first implication. The second is because $V_\omega$ is a model of $ZFC^{\text{fin}}$ and $V_\omega$ is a set of ZFC. | |
| Dec 5, 2010 at 18:40 | comment | added | Harry Gindi | @Adam: I was wondering: How do ZFC and ZFCU compare in consistency strength relative to $ZFC^{fin}$ (I actually asked this question earlier on MO, but decided to delete it)? | |
| Dec 5, 2010 at 5:49 | vote | accept | Avi Steiner | ||
| Dec 5, 2010 at 5:49 | |||||
| Dec 5, 2010 at 5:48 | vote | accept | Avi Steiner | ||
| Dec 5, 2010 at 5:48 | |||||
| Dec 4, 2010 at 22:09 | comment | added | Adam | The problem is that only one strongly inaccessible cardinal can be constructed in ZFC; the countable cardinal $\aleph_0$ -- but note that it's only in ZFC because we explicitly assume so via the Axiom of Infinity. Any and every strongly inaccessible cardinal ($\aleph_0$ included) in our set theory winds up there because we assume so (relative to Peano Arithmetic/$\text{ZFC}^{\text{fin}}$). | |
| Dec 4, 2010 at 21:49 | history | answered | Greg Muller | CC BY-SA 2.5 |