1
$\begingroup$

Let $\mathbb{N}$ be the standard model of the natural numbers. For any statement in the language of arithmetic, we can translate into a statement in the language of set theory by asking if it is true of $\mathbb{N}$.

Let's say that a statement in arithmetic is "extraneous" if it is independent of PA. For example, ZFC proves Con(PA), which is extraneous.

My question is, is there a set of statements $S$ (in the language of set theory), such that $S$ proves no extraneous statements, and $S+PA=ZFC$ (or perhaps $S+PA \vdash ZFC$).

Edit: We can also consider the same question, but with PA replaced with the set of arithmetical statements provable in ZFC.

$\endgroup$
  • 1
    $\begingroup$ I think this question is a bit muddled in its language. What does 'independent of PA' mean in a non-arithmetic context? For instance, the statement 'For any two sets $x,y$ there exists the set $\{x,y\}$ is 'independent of $PA$'. Is that not innately an extraneous statement? $\endgroup$ – Steven Stadnicki Sep 5 '17 at 23:24
  • 1
    $\begingroup$ What do you mean by $S+PA=ZFC$? $\mathsf{PA} $ and $\mathsf{ZFC}$ are stated in different languages. Or do you mean the translation of $\mathsf{PA}$ to the language of set theory rather than $\mathsf{PA}$ itself? What precisely is that translation? A straightforward version replaces each $\phi $ with the claim that $\mathbb N $ satisfies $\phi $, but then the translation of $\mathsf{PA}$ witnesses the consistency of $\mathsf{PA}$. $\endgroup$ – Andrés E. Caicedo Sep 6 '17 at 0:39
  • 1
    $\begingroup$ @JoelDavidHamkins I didn't mean for it to have a negative connotation. I just wanted to pick some statements of arithmetic, and PA was the first thing I thought of. The set of tautologies or set of arithmetical theorems of ZFC would also be interesting. $\endgroup$ – PyRulez Sep 6 '17 at 3:25
  • 1
    $\begingroup$ @Andres: We can translate each claim about natural numbers into a claim about finite ordinals, and thus turn sentences in the language of arithmetic into sentences in the language of set theory, to turn PA into a theory in the language of set theory which is strictly weaker than ZFC and in particular which does not itself prove Con(PA) in any standard sense. $\endgroup$ – Sridhar Ramesh Sep 6 '17 at 4:59
  • 1
    $\begingroup$ We could carry out this translation even in the context of a set theory with no axiom of infinity, which would fail to prove Con(PA). $\endgroup$ – Sridhar Ramesh Sep 6 '17 at 5:24
12
$\begingroup$

If $S$ proves no "extraneous" statements, then $S$ cannot prove $X \rightarrow Con(PA)$ for any arithmetic statement $X$ which $PA$ proves. It follows that $S + PA$ cannot prove $Con(PA)$, and therefore $S + PA$ cannot entail all of $ZFC$.

$\endgroup$
  • 2
    $\begingroup$ I would suggest that perhaps more in the spirit of what I would've thought this question would be would be to search for some S and some T such that S proves no extraneous statements, T consists only of arithmetic statements, and S + T = ZFC. Without loss of generality, we may even take T to consist of all arithmetic consequences of ZFC. I suspect no suitable S can be found, but do not know a proof at the moment. $\endgroup$ – Sridhar Ramesh Sep 6 '17 at 0:26
  • $\begingroup$ actually yeah, I like that question better. $\endgroup$ – PyRulez Sep 6 '17 at 1:28
1
$\begingroup$

I think you are asking for ZFC without the Axiom of Infinity, leaving the heriditarily finite sets. This proves the same arithmetic theorems as PA, if I understand it right.

$\endgroup$

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.