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



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> If $\mathfrak M$$\vDash$$T$, and therefore $T$ is consistent (where $\mathfrak M$ is the ground model) then if $\mathfrak M$$[$$G$$]$$\vDash$$T^{'}$, $T^{'}$ is consistent (since I am assuming $T$ and $T^{'}$ are first-order theories, this seems an easy consequence of the Goedel completeness theorem).
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Also, it is known that class forcings do not always preserve the axioms of $\mathfrak M$$\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. 



Forcing the consistency of $ZF$ from a fragment of $ZF$
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:



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




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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 $T\vdash Con(ZF)$.*


