2

Assuming consistency of ZF, ZF$\setminus ${Power Set axiom} is consistent. But can we prove consistency of ZF$\setminus ${Power Set axiom} without assuming consistency of ZF.

Andrés E. Caicedo
  • 81,687
Sushil
  • 2,881
  • 1
    You'll have to assume the consistency of something, be it $\mathrm{ZF}$ or some other system. – Clive Newstead Apr 21 '16 at 16:53
  • @CliveNewstead Sorry but your comment is completely confusing me. Why we have to assume consistency of something? And you said some other system, what it mean to say system? I mean what kind of thing this system is? – Sushil Apr 21 '16 at 17:15

2 Answers2

8

The one thing to stress in all these discussions is that proofs don't live in vacuum. There is very little you can actually prove without any assumptions. You have to start somewhere, with some axioms (and usually, the assumptions that your axioms are not inconsistent) and work from there.

The theory of $\sf ZF-Pwr$, or $\sf ZF^-$, is quite a strong theory which still falls prey to Goedel's theorem. In particular, it cannot prove its own consistency. You can prove its consistency from $\sf ZFC$, but you don't have to assume that $\sf ZFC$ is consistent for this proof to go through.

You can also prove this from some finite fragment of $\sf PA$, augmented by the number theoretic statement whose interpretation is $\operatorname{Con}\sf (ZF^-)$. This is a very weak theory, but it will prove the consistency of $\sf ZF^-$ nonetheless. For something less silly, you can use $\sf ZF^-+\mathcal P^6(\omega)\textrm{ exists}$ to construct the set of all hereditarily countable sets, which will then prove the consistency of $\sf ZF^-$ (I'm being excessive with $6$, you could probably get away with $2$ or so).

So in conclusion, you cannot discuss about formal proofs without some formal background. But yes, weaker theories than $\sf ZF$ can prove the consistency of $\sf ZF^-$.

Asaf Karagila
  • 405,794
  • Let me justa sk another question. Assuming consistency of $ZF-Pwr$ can we say anything about consistency of $ZF$ – Sushil Apr 22 '16 at 07:19
  • 1
    No, of course not. That would amount to saying something about the consistency of $\sf ZF^-$, being a subtheory. – Asaf Karagila Apr 22 '16 at 08:00
  • 'fall prey to Goedel Theorem', after long time understood it properly after understanding Goedel Theorem, how it can be prey to Goedel theorem when Goedel theorem is ZF result and I want to discrad ZF and work in $ZF^-$ only – Sushil Jan 14 '18 at 10:31
3

Elaborating on Asaf's answer: in fact, in a meaningful sense the consistency of ZF$^-$ is strictly weaker than the consistency of ZF. Specifically, assuming that $(i)$ ZF$^-$ is consistent and $(ii)$ PA is sound, the theory PA+Con(ZF$^-$) does not prove Con(ZF).

(The relevant phrase to look up is consistency strength.)

In many cases a result of this form is quite difficult. Here, however, it's easy:

First, it's easy to see that ZF contains (an appropriately interpreted form of) PA.

Second, and more interestingly, ZF proves the consistency of ZF$^-$. Specifically, ZF proves that there is a model of ZF$^-$ - it's a good exercise to show that ZF proves that the set $H(\omega_1)$ of hereditarily countable sets exists and is a model of ZF$^-$ - and is strong enough to prove that having a model implies consistency.

So if PA+Con(ZF$^-$) were to prove Con(ZF), we could translate that into a ZF-proof of Con(ZF) - showing that ZF is inconsistent. Since Con(ZF$^-$) is true and PA is sound, this means PA doesn't prove the false statement Con(ZF$^-$) implies Con(ZF), hence PA+Con(ZF$^-$) doesn't prove Con(ZF).

The use of soundness above was critical: it's conceivable after all that PA proves $\neg$Con(ZF$^-$) even if ZF$^-$ is consistent (remember that Godel implies that PA+$\neg$Con(PA) is even consistent!). In this case PA+Con(ZF$^-$) would trivially prove Con(ZF).

Noah Schweber
  • 260,658
  • Sorry didn't get when you said that PA proves negation of Con(ZF^-), How? – Sushil Jan 13 '18 at 08:45
  • @Sushil I didn't say that it does prove $\neg$Con(ZF$^-$), I said "it's conceivable after all that PA proves $\neg$Con(ZF$^-$)" - in the same way that by Godel it's conceivable that PA proves $\neg$Con(PA). Of course we (almost) all believe it doesn't since we're confident in the soundness of PA. – Noah Schweber Jan 13 '18 at 08:48
  • I was confused because you wrote 'Godel implies' – Sushil Jan 13 '18 at 08:54
  • @Sushil I wrote "Godel implies that PA+$\neg$Con(PA) is even consistent," not true/sound. – Noah Schweber Jan 13 '18 at 08:57
  • sorry is 'PA +\neg Con(PA)$ Consistent? – Sushil Jan 13 '18 at 09:03
  • @Sushil Yes, it is. This is a consequence of Godel's theorem: if PA+$\neg$Con(PA) were inconsistent, PA would prove Con(PA), and it can't do that (again, assuming PA is in fact consistent). (Note that there's nothing special about PA here - any theory to which Godel applies will do the same thing.) – Noah Schweber Jan 13 '18 at 09:05
  • Ohh but here Con(PA) is a particular formula which Godel Used. What I mean to say about consistency is about metatheorem about theory. – Sushil Jan 13 '18 at 09:09
  • @Sushil I don't understand what you're asking then. If PA is consistent then so is PA+$\neg$Con(PA). – Noah Schweber Jan 13 '18 at 09:11
  • Yes I agree. And I got your argument also . But there is one more question which I have from so much time: https://math.stackexchange.com/questions/2603355/clarification-about-godel-second-incompleteness-theorem – Sushil Jan 13 '18 at 09:24
  • Let us continue this discussion in chat. – Sushil Jan 13 '18 at 09:24