3
$\begingroup$

Are there any easy relative consistency proofs in set theory that can be done using proper class models, rather than set models?

The only easy one I can think of is proving the consistency of $\mathsf{ZF}$ relative to $\mathsf{ZF} - \text{Foundation}$ using the class $\text{WF} = \bigcup_{\alpha \in \text{Ord}} V_\alpha$ of well-founded (i.e. hereditary) sets.

The next ones I can think of are the consistency of $\mathsf{ZFC}$ relative to $\mathsf{ZF}$, using $\text{L}$ or $\text{HOD}$, and the consistency of $\mathsf{ZF} + \mathsf{CH}$ relative to $\mathsf{ZF}$, using $\text{L}$. But these are a bit harder. Are there any more that are as easy as the one using $\text{WF}$?


EDIT: By "easy" here I mean that it does not involve encoding formulas as natural numbers, or formalizing the satisfaction relation. I consider showing one theory to be consistent relative to another theory using proper class models and relativization to be easier than proving formalized consistency statements using set models and the formalized satisfaction relation. For this reason, I am teaching it first in my class. However, this advantage of using proper class models is nullified if I have to use the formalized satisfaction relation to construct $\text{L}$ or $\text{HOD}$.

$\endgroup$
  • 2
    $\begingroup$ You cannot do this for sentences $\sigma$ false in $L$, as if you could create a proper class model $M$ in $\mathsf{ZF}$ of $\sigma$, we could still create it in $\mathsf{ZF+V=L}$, but as $Ord\subseteq M$, $M=L$. $\endgroup$ – Camilo Arosemena-Serrato Nov 25 '13 at 4:43
  • $\begingroup$ @CamiloArosemena Good point. But perhaps one can still do something interesting where the desired sentence $\sigma$ is simply an axiom of $\mathsf{ZF}$ but we are only assuming some fragment of $\mathsf{ZF}$, as in the example with $\text{WF}$. $\endgroup$ – Trevor Wilson Nov 25 '13 at 4:49
  • $\begingroup$ But there isn't much you can do, the other interesting axioms would be replacement and power set, but then again this cannot be done as $\mathsf{ZF-P}$ and $\mathsf{Z}$ are stricly weaker than $\mathsf{ZF}. $\endgroup$ – Camilo Arosemena-Serrato Nov 25 '13 at 5:20
  • 1
    $\begingroup$ Something which is not very difficult, but can be "less easy" is to show the consistency of anti-foundation axioms. If you start with models with atoms you can construct a permutation model in which foundation fails, and you have an exact control on how it fails. So you can show the consistency of anti-foundation axioms like that. $\endgroup$ – Asaf Karagila Nov 25 '13 at 6:07
  • 2
    $\begingroup$ The simplest, then, would be failure of choice or the existence of atoms. $\endgroup$ – Asaf Karagila Nov 25 '13 at 6:09
2
$\begingroup$

The relativized contructible universes $L[A]$ can be used to establish consistency results of the form: Con$(2^{\aleph_0} = \aleph_{n})$ implies Con $(2^{\aleph_0} = \aleph_m)$ for any $0 < m < n < \omega$ - This maybe due to Hajnal, I don't have a reference.

Of course in the post-forcing set theory, these results seem nothing more than curiosities.

$\endgroup$
  • $\begingroup$ By "easy" I meant easier than $\operatorname{Con}(\mathsf{ZFC}) \implies \operatorname{Con}(\mathsf{ZFC} + \mathsf{CH})$, although the comments convinced me that this is probably impossible. $\endgroup$ – Trevor Wilson Dec 13 '16 at 19:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.