Does $\mathsf{ZFC} \vdash \operatorname{Con}(\mathsf{ZFC})\rightarrow \operatorname{Con}(\mathsf{ZFC}+V\neq L)$?

I know that by forcing over a transitive model we have $\mathsf{ZFC} \vdash \operatorname{Con} ( \mathsf{ZFC} + \mathsf{SM} ) \rightarrow \operatorname{Con} ( \mathsf{ZFC} + V \neq L )$, where $\mathsf{SM}$ denotes the statement that there is a standard model of $\mathsf{ZFC}$. But does $\mathsf{ZFC}$ also prove the relative consistency statement above?


ZFC proves the completeness theorem - that if $T$ is a consistent theory, it has a model. ZFC also proves the Lowenheim-Skolem theorem - that any structure in a countable language has a countable elementary substructure.

Finally, one last point: ZFC proves that forcing "works" (in the sense that generic filters and extensions exist, and that the forcing theorems hold) over countable models of ZFC, even if those models are ill-founded.

This requires doing a bit more work in the basic development of forcing, but fundamentally not that much: given a countable $M$ and a $P\in M$ such that $M$ thinks "$P$ is a partial order," we have that $P$ (or rather, its translate $P\cap M$) really is a partial order; so we can talk about $P$-generic filters over $M$. Since $M$ is countable, such a filter (call it "$G$") exists. Now, the class of names can be defined in $M$ as usual, and the forcing extension $M[G]$ has as objects equivalence classes of names in $M$, with the equivalence relation coming from the definable forcing relation inside $M$ - even though names may be ill-founded, everything winds up checking out correctly.

Now we reason in ZFC. If ZFC were consistent, it would have (by Completeness and Lowenheim-Skolem) a countable model $M$. This, in turn, means we get a countable model $M[G]$ of ZFC+V$\not=$L as usual.

Incidentally, ZFC is a absolutely terrible upper bound: much less than ZFC, even less than PA, suffices to prove $Con(ZFC)\implies Con(ZFC+V\not=L)$. This requires us to do some serious work, though.

  • $\begingroup$ Interesting,then you can make forcing from any countable model. $\endgroup$
    – luis
    Feb 1 '17 at 6:57
  • $\begingroup$ You identify two names x,y in M iff there is a condition $p\in G$ such that p force $x=y$ and M[G] is the quotient space,it isn't? $\endgroup$
    – luis
    Feb 1 '17 at 7:32
  • $\begingroup$ @Noah I'm just learning somewhat more about forcing and I didn't know about this "approach". Like you said, $\textbf{ZFC}$ is overkill, but still of interest since most mathematicians would assume ambient $\textbf{ZFC}$ either way. Could you expand on how this is achieved? I'm not quite following your argument, mostly because I don't know what absoluteness results we have for non-transitive models. Also, my definition of $M[G]$ does not involve the forcing relation; and for ctms of $\textbf{ZF-P}$ being a $P$-name is absolute, but how do we go about this in the forcing approach you described? $\endgroup$
    – Jori
    Jan 28 '20 at 15:37
  • $\begingroup$ @Noah Maybe I should ask a new question about this... let me know. $\endgroup$
    – Jori
    Jan 28 '20 at 15:38
  • $\begingroup$ @Jori I think you should ask a new question about this. Roughly speaking, though, the issue is this: with a ctm - the key point being "transitive" - we can define $M[G]$ by recursion. However, if $M$ is countable but ill-founded, that recursion doesn't make sense. Instead, we use the formulas which in the ctm case define forcing as our starting point here. When we do this carefully, everything carries over appropriately (for example, $M$ and $M[G]$ will satisfy the same $\Pi^1_2$ sentences). $\endgroup$ Jan 28 '20 at 15:40

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