0
$\begingroup$

In his article "THE INDEPENDENCE OF THE CONTINUUM HYPOTHESIS", Paul Cohen writes in the beginning:

We shall work with the usual axioms for Zermelo-Fraenkel set theory, and by Z-F we shall denote these axioms without the Axiom of Choice, (but with the Axiom of Regularity).

Thus the meta-theory he is working in is the Zermelo-Fraenkel set theory without AC, but with the axiom of regularity. In this meta-theory, Cohen proves his famous result that CH is indepedent of ZFC. Now I wonder:

Did he use the regularity axiom of his meta-theory in his proof?

$\endgroup$
  • 1
    $\begingroup$ You can actually do forcing in an arithmetic meta theory. Even something as weak as PRA. Are you asking about Cohen's proof specifically, or about forcing in general? (Also, as Cohen worked with countable models, you can probably get away with just separation and power set if you replace "transitive" with "well founded", then replacement is only necessary for the Mostowski collapse.) $\endgroup$ – Asaf Karagila Aug 21 '16 at 14:10
  • $\begingroup$ I see that my question can be interpreted in many ways. Is it okay if I don't specify exactly, but just say "My question can be viewed as asking about Cohen's proof specifically or about forcing in general, and I am interested in both corresponding answers."? $\endgroup$ – 6787 Aug 21 '16 at 14:20
  • $\begingroup$ It shouldn't be too hard to come up with a forcing result that fails without replacement. Is that the 'general' case you are asking about? $\endgroup$ – Stefan Mesken Aug 21 '16 at 14:21
  • $\begingroup$ @Stefan: By the way, my question is only concerned with the question of needing the replacement axiom in the meta-theory, and not the question whether the (in this meta-theory) investigated models satisfy replacement. Is your comment consistent with this point? $\endgroup$ – 6787 Aug 21 '16 at 14:25
  • $\begingroup$ @notSignedUp My comment was actually addressed to the question in your comment and I was thinking about replacement in our countable model. Now I'm not sure what you actually meant in your first comment. The proof of the relative independence of CH from ZFC takes place in some weak arithmetical theory - say PRA. I don't see how replacement comes into play here. $\endgroup$ – Stefan Mesken Aug 21 '16 at 14:33
3
$\begingroup$

If he did use Regularity in some of the argument, then he still didn't have to, because he could just have carried out everything relative to WF, the class of hereditarily well-founded set. WF satisfies ZF with Regularity even if the ambient universe does not satisfy Regularity.

Becuase the consistency of this theory or that is just a matter of arithmetic (thanks to Gödel) and $\omega$ is the same within WF as outside it, this would lift any relative consistency result from WF to a weaker meatheory.

$\endgroup$
  • 1
    $\begingroup$ Also, you can code a countable structure into a subset of $\omega$, which lies in WF, and therefore you can collapse it to a transitive set there. (Assuming of course that we start with a well founded model...) $\endgroup$ – Asaf Karagila Aug 21 '16 at 16:04
0
$\begingroup$

When $M$ is a ground model and $P\in M$ is a poset and $A$ is a $P$-name over $M, $ then $A$ is a set of ordered pairs $(x,y')$ where $x\in P$ and $y'$ is a $P$-name over $M.$ If $G$ is a $P$-generic filter over $M$ then $A_G=\{y'_G: \exists x\in G\;((x,y')\in A\}.$ Now we can't assume $A_G$ exists without Replacement: By Comprehension we have the existence of $B=\{(x,y')\in A: x\in G\}.$ And $\forall (x,y')\in B\;\exists ! z \;(z=y'_G\}.$ But we must apply Replacement to this formula to conclude that $A_G$ exists.

Remark: Cohen's paper considered only forcing over $P$ \ min$(P)$ when $P$ is a complete Boolean algebra. But all forcing can be done this way; it is not always convenient to do so.

$\endgroup$
  • $\begingroup$ I haven't looked at Cohen's paper for a while now. But if my memory serves me right the idea of using complete Boolean algebras is due to Scott and Solovay, and came out later. $\endgroup$ – Asaf Karagila Aug 21 '16 at 15:52
  • $\begingroup$ "Now we can't assume $A_G$ exists without Replacement" This seems to contradict (does it really contradict?) Asaf's assertion that "working with or without regularity is entirely irrelevant". $\endgroup$ – 6787 Aug 21 '16 at 15:52
  • $\begingroup$ No. It means there are different axiom schemas and different forcing def'ns within them. In ZF we need Replacement to get $A_G$. $\endgroup$ – DanielWainfleet Aug 21 '16 at 15:59
  • $\begingroup$ @AsafKaragila. You may be right about the history. I have not really studied the history of it. Cohen also has a long paper (a memoir) about the process of his discovery. but I read only a little of it. $\endgroup$ – DanielWainfleet Aug 21 '16 at 16:02

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.