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Implicit in the technique of forcing is the following relative consistency result:

If $\mathfrak M$$\vDash$$T$, and therefore $T$ is consistent (where $\mathfrak M$ is the ground model) then if $\mathfrak M$$[$$G$$]$$\vDash$$T^{'}$, $T^{'}$ is consistent (since I am assuming $T$ and $T^{'}$ are first-order theories, this seems an easy consequence of the Goedel completeness theorem).

Also, it is known that class forcings do not always preserve the axioms of $\mathfrak M$$\vDash$$T$ in $\mathfrak M$$[$ $G$ $]$ $\vDash$ $T^{'}$.

The theory $T$ I am specifically interested in is $ZF$ $-$ Infinity, that is, $ZF$ with the Axiom of Infinity dropped.

Question: Is there a class forcing extension of $\mathfrak M$$\vDash$ $ZF$ $-$ Infinity such that $\mathfrak M$$[$ $G$ $]$$\vDash$$ZF$ $-$ Infinity $+$ Infinity (or in the alternative, where $\mathfrak M$$\vDash$$ZF$$-$ Infinity $+$ $\lnot$Infinity , is there a class forcing producing a forcing extension $\mathfrak M$$[$ $G$ $]$ in which $\lnot$Infinity fails)? If there is no such class forcing, show why there cannot be such.

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  • $\begingroup$ Why the downvote? If whoever downvoted my question can give a good reason for the downvote, I will delete the question. $\endgroup$
    – Thomas Benjamin
    Commented Jan 9, 2017 at 9:28
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    $\begingroup$ I didn't downvote,but the question seems unclear to me. My first question concerns "$\mathfrak M\models T$ is consistent." This could mean that it's consistent that $\mathfrak M\models T$, or it could men that $\mathfrak M$ satisfies the formula "$T$ is consistent." Which of those do you mean? Or do you perhaps mean that $\mathfrak M\models T$, and therefore $T$ is consistent? Or something else? $\endgroup$
    – Andreas Blass
    Commented Jan 9, 2017 at 15:43
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    $\begingroup$ If you're looking for a proof of the consistency of ZF from the consistency of ZF$-$Infinity, then there is no such proof (at least none that can be formalized in ZFC), because ZF proves the consistency of ZF$-$Infinity. As for proving the consistency, relative to ZF$-$Infinity, of ZF$-$Infinity$+\neg$Infinity, that should be possible by interpreting the latter theory in the former, but I don't see how to bring forcing to bear on that matter. $\endgroup$
    – Andreas Blass
    Commented Jan 9, 2017 at 15:48
  • $\begingroup$ @AndreasBlass: I mean, "if $\mathfrak M$$\vDash$$T$, and therefore $T$ is consistent, then $\mathfrak M$$[$ $G$ $]$$\vDash$$T^{'}$, and therefore $T^{'}$ is consistent" (apologies all around for the vagueness of my question--I definitely concede that, given the helpful comments, the question could have been better written). It was my hope that one could use class forcing to show that Infinity was independent of $ZF$ $-$ Infinity ($ZF$ $-$ Infinity should be able to be proven consistent in $PRA$ $+$ $TI({\epsilon_0})$), but Noah's answer suggests that such a project is doomed to fail. $\endgroup$
    – Thomas Benjamin
    Commented Jan 10, 2017 at 7:17

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Leaving aside the issues around consistency (which I don't really follow - see Andreas' comments), it seems to me that the mathematical question you're asking is:

If $M\models ZF-Inf$, can there be a class forcing extension $M[G]$ of $M$ which satisfies full $ZF$?

(Note that $ZF$ and $ZF-Inf+Inf$ are the same thing.)

Let's first think about this in the context of the usual model $V_\omega$ of hereditarily finite sets. We can set up the machinery of class forcing as usual: however, note that every name is finite! (Since names are just sets of a certain form in the ground model, and every element of $V_\omega$ is finite.) This means in particular that, no matter what $\mathbb{P}$ and $G$ we pick, in $V_\omega[G]$ every set will be finite. So in particular, $V_\omega[G]$ will not satisfy Inf.

I believe that this generalizes to all models of $ZF-Inf+\neg Inf$+"transitive closures exist" (this is classically proved via Replacement + Infinity), via a similar argument: given a putative name for $\omega$ in the extension, we may recover $\omega$ from its transitive closure in the ground model, contradicting $\neg Inf$ there. I am uncertain as to whether TCE can be dropped here; I don't have much experience with finite set theories. However, I suspect that it can be.

Note that set forcing in finite set theories is trivial: $ZF-Inf+\neg Inf$ proves "in every (set) partial order, the set of minimal elements is dense."

Going back to consistency issues, it sounds like you are trying to prove $Con(ZF-Inf)\implies Con(ZF)$ in a reasonably weak base theory (say, a subtheory of ZF). This cannot succeed unless $ZF$ is inconsistent: since $ZF$ does prove $Con(ZF-Inf)$, this would imply $ZF\vdash Con(ZF)$.

If that's not what you're trying to do, what are you trying to do?

Incidentally, if you're trying to prove $Con(ZF-Inf)\implies Con(ZF)$ in a possibly stronger base theory, note that this is easily doable: ZF+"there is an inaccessible cardinal" proves $Con(ZF)$, so proves $Con(ZF-Inf)\implies Con(ZF)$. And by the reasoning above, this is the only way to have this occur: if $T$ is a theory containing $ZF$, then $T\vdash Con(ZF-Inf)\implies Con(ZF)$ iff $T\vdash Con(ZF)$.

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  • $\begingroup$ Considering the nature of the comments given, I am willing to delete this question. Your answer was helpful to me, but if you think your answer will help others, I will not delete. What do you wish to do? $\endgroup$
    – Thomas Benjamin
    Commented Jan 10, 2017 at 7:25
  • $\begingroup$ Since $ZF$ $+$ "there is an inaccessible cardinal" proves $CON(ZF)$, does the "inacessible cardinal" in question need to be uncountable? I am given to understand that $\omega$ is countable yet inaccessible. $\endgroup$
    – Thomas Benjamin
    Commented Jan 12, 2017 at 12:19
  • $\begingroup$ @ThomasBenjamin If you consider $\omega$ to be inaccessible, then it is not true that ZF+"there is an inaccessible" proves Con(ZF). You do indeed need the inaccessible to be uncountable. (Think about the proof - given $\kappa$ inaccessible, we argue that $V_\kappa\models ZF$. But in order for $V_\kappa$ to satisfy $ZF$, it has to satisfy in particular Infinity - so we need $\kappa>\omega$.) My understanding, in fact, is that most authors do not consider $\omega$ to be inaccessible, but I could be wrong. $\endgroup$
    – Noah Schweber
    Commented Jan 12, 2017 at 15:41
  • $\begingroup$ So in $ZF$ $+$ $AD$ (since $\omega_1$ is inaccessible), one could prove the consistency of $ZF$? $\endgroup$
    – Thomas Benjamin
    Commented Jan 12, 2017 at 15:49
  • $\begingroup$ @ThomasBenjamin ZF+AD does indeed prove Con(ZF) (and much, much more), but not quite the way you think. In my comment above, by "inaccessible" I meant "strongly inaccessible" - in order to conclude that $V_\kappa\models ZF$ in the most direct possible way, we need to show that $V_\kappa$ is closed under powersets. Now, large cardinal notions behave weirdly without choice, but there is no good sense in which $\omega_1$ is strongly inaccessible assuming AD: we will always have a surjection from $\mathbb{R}$ to $\omega_1$. (cont'd) $\endgroup$
    – Noah Schweber
    Commented Jan 12, 2017 at 16:56

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