So, I've been trying to learn as much as I can about forcing. I know that a model provides its own (trivial) forcing extension. What I'm curious about is whether there is a way to think of the iterative hierarchy in terms of (trivial?) forcing extensions?

The iterative hierarchy for $\mathsf{ZF}$ can be given in the traditional manner:

The first level, $V_0$, is defined to be the empty set, $\emptyset$ (if the set theory is impure, this is also where the urelements, i.e., the non-set individuals, reside). Subsequent levels are formed by taking the powerset of the previous stage such that the level immediately following $V_n$ (for finite ordinals $n$), the successor level $V_{n+1}$, is defined as $\mathcal{P}(V_n)$. Once you have formed all of the finite stages, you form the first limit level $V_\omega$ by taking the infinite union of the preceding levels, $\bigcup_{n<\omega} V_n$. The process then repeats for the successor levels of $V_\omega$, $V_{\omega + 1}, V_{\omega + 1},\dots$, with the successor levels of $V_\omega$ unionized to form the next limit level $V_{\omega \cdot 2}$. The universe of sets, $V$, is the union of all the levels: $V = \bigcup_{\alpha \in O} V_\alpha$, where $O$ is the class of all ordinals.

Is there a way to understand the various levels as forcing extensions of preceding levels? I'm asking because it's common, in talk of the iterative conception, to allow that sets formed at previous levels can be used as parameters in defining new sets. These parameters sort of reminded me of the names that get introduced when you add constants for your forcing language.

Is there any interesting connection between the levels of $\mathsf{ZF}$ and the forcing extensions of a theory? Can we understand the levels as something akin to forcing extensions of preceding levels?


Not really - they are fundamentally different types of extension. One of the crucial properties of forcing is that a forcing extension $V\subset V[G]$ is an end extension - no set "gets new elements" - but not a top extension (forcing adds no new ordinals). A top extension is one in which every new set has higher rank than every old set - that is, exactly as $V_{\alpha+1}$ is to $V_\alpha$. So in a sense, forcing is really "orthogonal" to the cumulative hierarchy - iterating powersets builds "up," forcing builds "sideways.

Building off a comment below: note that nontrivial forcing extensions are never elementary extensions. Now, iterating powerset doesn't yield elementary extensions level-by-level either - that is, $V_\alpha$ is never an elementary substructure of $V_{\alpha+1}$ (since $V_{\alpha+1}$ satisfies "there is a set $X$ of maximal rank," so must $V_\alpha$, so $\alpha=\beta+1$; but then $V_\alpha$ and $V_\beta$ disagree on what the largest ordinal is!) - but we can have $V_\alpha\prec V_\beta$ for certain $\alpha<\beta$.

To find such $\alpha$ and $\beta$: Fix $\alpha_0$. Now by iteratively closing under Skolem functions, we can find an $\alpha>\alpha_0$ such that $V_\alpha\prec V$. Fix $\beta_0>\alpha$, and similarly find $\beta>\beta_0$ such that $V_\beta\prec V$. Then we use the general fact from model theory that $$A\preccurlyeq C, B\preccurlyeq C, A\subseteq B\implies A\preccurlyeq B.$$

(This argument actually takes more than ZFC - in particular, as written it implies the consistency of ZFC! - but not too much more.)

  • $\begingroup$ Yes they sure are different. But maybe there is an analogy: the Boolean valued model of the forcing gives a cumulative hierarchy $V^B_{\alpha}$, the "B-valued" versions of $V_{\alpha}$, so in that sense yeah. $\endgroup$ – BrianO Dec 18 '15 at 12:32
  • $\begingroup$ I still don't think that counts - the OP is asking for an analogy between "take another powerset" and "take a forcing extension." The fact that the forcing extension has a cumulative hierarchy doesn't (to me at least) seem to fit the bill. $\endgroup$ – Noah Schweber Dec 18 '15 at 18:30
  • $\begingroup$ Agreed. I just mentioned the Booelan-valued $V_{\alpha}$ because that's a valid analogy, within one model. It's not some class in the hyperverse (and one with a very different ordering), which is a bad analogy that OP asked about. $\endgroup$ – BrianO Dec 18 '15 at 21:10
  • $\begingroup$ I just came across this answer by Joel David Hamkins which says the following: $\endgroup$ – Dennis Dec 24 '15 at 7:47
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    $\begingroup$ @Dennis Sorry for the late reply, I just noticed this. Yes, every elementary end-extension is a top extension. But (nontrivial :P) forcing extensions are never elementary, for this exact reason! It is never the case that forcing adds ordinals. (Note that forcing extensions may preserve the theory - that is, we may have $M\equiv M[G]$ - but this does not mean $M\prec M[G]$!) $\endgroup$ – Noah Schweber Dec 26 '15 at 18:07

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